# LitMind Dictionary: An Open-Source Online Dictionary

Cunliang Kong†, Xuezhi Fang†, Liner Yang†, Yun Chen‡, Erhong Yang†
†School of Information Science, Beijing Language and Culture University
‡School of Information Management & Engineering,
Shanghai University of Finance and Economics
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
<EMAIL_ADDRESS><EMAIL_ADDRESS>Indicates corresponding author

###### Abstract

Dictionaries can help language learners to learn vocabulary by providing
definitions of words. Since traditional dictionaries present word senses as
discrete items in predefined inventories, they fall short of flexibility,
which is required in providing specific meanings of words in particular
contexts. In this paper, we introduce the LitMind Dictionary
(https://dictionary.litmind.ink), an open-source online generative dictionary
that takes a word and context containing the word as input and automatically
generates a definition as output. Incorporating state-of-the-art definition
generation models, it supports not only Chinese and English, but also Chinese-
English cross-lingual queries. Moreover, it has a user-friendly front-end
design that can help users understand the query words quickly and easily. All
the code and data are available at https://github.com/blcuicall/litmind-
dictionary.

##  1 Introduction

Helping language learners understand words in doubt is an important topic in
the field of Intelligent Computer-Assisted Language Learning (ICALL) (Segler
et al., 2002; Enayati and Gilakjani, 2020; Lolita et al., 2020). Most
dictionaries present word senses as discrete items in predefined inventories
to help language learners understand new words. Nevertheless, this form is
suffering from several limitations and may bring users inconveniences in many
cases. First, many commonly used words are polysemous, and it’s difficult for
language learners to distinguish different word senses because of the
cognitively inaccurate nature of discrete sense boundaries (Rosch and Mervis,
1975; Kilgarriff, 1997; Tyler and Evans, 2001). As Kilgarriff (2007) argued,
there are no decisive ways of identifying where one sense of a word ends and
the next begins. In addition, the predefined inventories need to be updated
manually by lexicographers, which is time-consuming and causes dictionaries to
lag behind the ever-changing language usage. For example, many new words have
emerged along with the change in people’s lifestyles, such as tweet (make a
post on Twitter) and geekery (enthusiasm for a subject). However, these words
and senses didn’t appear in traditional dictionaries until they were used for
a long time.

Figure 1: The main functionality and architecture of LitMind Dictionary.

We overcome these limitations by developing LitMind Dictionary,111LitMind is
derived from The Literary Mind and the Carving of Dragons, which is the first
systematic literary theory work in China, written around 502 BC. an open-
source online generative dictionary. It takes a word and the context
containing the word as input and provides automatically generated definitions
for the word. In this way, the word’s definition in current context can be
given directly, saving users from selecting from a variety of senses. This
approach breaks away from the limitation of predefined inventories, with
potential to generate correct definitions of new words according to the
contexts.

In LitMind Dictionary, the context-aware definitions are generated using a
range of NLP and machine learning techniques mainly based on our previous work
of definition modeling (Yang et al., 2020; Fan et al., 2020; Kong et al.,
2020). The proposed dictionary uses improved versions of these models, and
incorporates some engineering tricks to handle extreme cases. Moreover, the
dictionary supports Chinese and English queries as well as Chinese-English
cross-lingual queries, all of which are realized in a generative fashion for
the first time. Finally, the user-friendly interface design can help users
understand the query words as quickly and easily as possible.

##  2 Related Work

Our work is related to the task of definition modeling, which aims at
automatic definition generation. This task is first introduced by Noraset et
al. (2017). They used word embeddings as encoded information and a two-layer
LSTM as the decoder.

However, their method failed to account for polysemous words, and subsequent
work proposed different solutions for this problem. Gadetsky et al. (2018)
released a dataset containing example sentences and computed the AdaGram
vector (Bartunov et al., 2016) of input words, which is a non-parametric
Bayesian extension of skip-gram capable to learn numbers of representations at
desired semantic resolution. Chang et al. (2018) proposed to project the given
words to high-dimensional sparse vectors, and picked different dimensions for
different meanings. Mickus et al. (2019) implemented a self-attention based
model, and proposed several masking strategies for the input words and example
sentences. Li et al. (2020) explicitly decomposed the meaning of words into
semantic components, and modeled them with discrete latent variables. Yang et
al. (2020) explored definition modeling for Chinese and incorporated sememes
(Dong and Dong, 2006), minimal semantic units, as part of the representation
of given words. Zheng et al. (2021) proposed to enhance the definition
generation with word formation features in parataxis languages like Chinese.
Besides, Ishiwatari et al. (2019) extended this task to describe unknown
phrases. They replaced the target word in context with a placeholder, and used
a character-level CNN together with static embedding for word representation.

Recent years have witnessed the application of pretrained language models in
definition modeling (Chang and Chen, 2019). Reid et al. (2020) initialized
encoders with BERT (Devlin et al., 2019) and employed variational inference
for estimation and leverage contextualized word embeddings for improved
performance. Bevilacqua et al. (2020) employed a novel span-based encoding
scheme to fine-tune a pre-trained English Encoder-Decoder system to generate
definitions. Huang et al. (2021) leveraged the T5 model Raffel et al. (2019)
for this task and introduced a re-ranking mechanism to model specificity in
definitions.

Our approach follows previous work in using pretrained language models for two
reasons: (1) due to the data scarcity in the definition modeling task, it’s
difficult for other models to obtain better performance; (2) using contextual
word embeddings can solve the ambiguity of polysemous words. Moreover, we
employ cross-lingual pre-trained language models (Lample and Conneau, 2019;
Conneau et al., 2019) as an extension of the task to support Chinese-English
queries.

Figure 2: The overall workflow of LitMind Dictionary.

##  3 System Description

Figure 1 shows an overview of the main functionality and architecture of
LitMind Dictionary. In this section, we first describe its overall workflow,
then we detail the definition generation model, and finally we introduce its
user interface design.

###  3.1 Overall Workflow

The workflow of LitMind Dictionary is illustrated in Figure 2. When using the
dictionary, a user needs to provide the query word and context. We require the
word to appear in the context. If the query word refers to a named entity, the
dictionary will directly retrieve the corresponding definition from a
predefined list, such as Name, State or Province, Organization, etc.
Otherwise, the dictionary will feed the query word and context into the
monolingual or cross-lingual definition generation model, depending on the
user’s selection.

Figure 3: Overall structure of the definition generation model.

###  3.2 Definition Generation Model

The definition generation model (DGM) shown in Figure 3 is the core component
of LitMind Dictionary. It is used to automatically generate the definition of
a given word. The whole model is a transformer (Vaswani et al., 2017) based
encoder-decoder model, and we use pretrained language models to initialize the
parameters in experiments.

For a given word $w$ and its corresponding context $\bm{c}$, we segment them
into subwords and use each subword as a token. To facilitate the word and
context encoding, we concatenate then into a whole sequence $(w,$
[SEP]$,\bm{c})$. In order to represent the position of different tokens, we
add a positional embedding and a segment embedding to each token as Vaswani et
al. (2017)’s original setting. For the $i$-th token in the sequence, we add
its token embedding $tok_{i}$, position embedding $pos_{i}$, and segment
embedding $seg_{i}$ together as $\mathbf{x}_{i}=tok_{i}+pos_{i}+seg_{i}$. We
input the obtained embeddings into the definition generation model, and train
the parameters by minimizing the cross-entropy loss between the generated
definitions and the true distributions underlying the dataset:

$\bm{\theta^{*}}=\operatorname*{arg\,max}_{\bm{\theta}}\sum_{\bm{d}\in
D}\log({P(\bm{d}|\mathbf{x},\bm{\theta})}),$ (1)

where $\bm{\theta}$ refers to all trainable parameters, and $\bm{d}$ is the
predicted definition.

###  3.3 Monolingual Mode

LitMind Dictionary supports English and Chinese monolingual modes (Yang et
al., 2020; Fan et al., 2020). Monolingual refers to that the language of the
definition is the same as that of the word. For English mode, users input
English word and context, and our system will return English definitions
($<w^{en},\bm{c}^{en}>\rightarrow\bm{d}^{en}$).

###  3.4 Cross-lingual Mode

Now we introduce LitMind Dictionary in the cross-lingual scenario. The only
difference from the monolingual mode is that the language of the input word
and context is different from that of the output definition (Kong et al.,
2020). In Chinese-English mode, the input word and context are in Chinese, and
the output definition is in English
($<w^{zh},\bm{c}^{zh}>\rightarrow\bm{d}^{en}$).

As high quality Chinese-English dictionary resources are difficult to obtain,
we don’t train the model on Chinese-English parallel dataset. Instead, we only
train the model on English-English dataset, and then directly transfer the
model to Chinese-English scenario in a zero-shot manner. Since multilingual
PLMs is capable of encoding sequences in various languages, this zero-shot
method shows effective results in our manual evaluation. In addition, this
approach can be extended to other low-resource languages to help more language
learners.

Figure 4: User interface of LitMind Dictionary in the English monolingual
mode.

###  3.5 User Interface

As shown in figure 4, the interface of LitMind Dictionary is friendly designed
and very easy to use. To make a query, users need to input a word and a
sentence into the textboxes, and then click the search button. Our system will
automatically generate and display the corresponding definition. Under the
definition, some example sentences are listed to help users better understand
the query word.

We also designed a feedback channel to collect real-world data. Users can
click the feedback button to write the definition they think is appropriate.
Besides, if users have overall suggestions for LitMind Dictionary, they can
click the Make Suggestions button to give their advice.

##  4 Evaluation

In this section, we evaluate the performance of LitMind Dictionary. We conduct
both monolingual (Chinese and English) and cross-lingual(Chinese-English)
evaluations.

###  4.1 Datasets

For English monolingual experiments, we use the Oxford dataset published by
Gadetsky et al. (2018) as training, validation and test dataset. The dataset
has 97,855 entries in the training set, 12,232 entries in the validation set
and test set respectively.

As for Chinese, we build a dataset from the Contemporary Chinese Learner’s
Dictionary (CCLD),222CCLD: https://www.cp.com.cn/book/9554d669-7.html which is
specially designed for Chinese learners. We first convert each page of the
book into text using the optical character recognition (OCR) technology. We
then recruit a group of annotators to proofread the text and correct the
errors generated in the conversion process. Finally, we structure the text
into the json format and extract words, example sentences and definitions and
obtain the entire dataset of 6,284 words, 89,065 entries. The dataset is then
split into three subsets as training set, validation set, and test set by the
ratio of 8:1:1.

For the Chinese-English cross-lingual definition generation, we train the
model using the above mentioned Oxford dataset. For evaluation, we randomly
sampled 200 entries from the CCLD test set, and perform zero-shot generation
on it. Since there are no golden English definitions in the test set, we
organize manual evaluation to score the results generated by models.

Tabel 1 lists more detailed statistics of the above datasets.

Dataset |  Words |  Entries |  Exp. |  Def.
---|---|---|---|---
Oxford |  |  |  |
Train |  33,128 |  97,855 |  17.74 |  11.02
Valid |  8,867 |  12,232 |  17.80 |  10.99
Test |  8,850 |  12,232 |  17.56 |  10.95
CCLD-Mono
Train |  5,028 |  71,328 |  7.06 |  13.41
Valid |  628 |  8,700 |  6.89 |  13.43
Test |  628 |  9,037 |  7.37 |  13.47
CCLD-Cross
Test |  163 |  200 |  7.19 |  14.38
|  |  |  |
Table 1: Statistics of the Oxford dataset, CCLD (Monolingual) dataset, and
CCLD (Cross-lingual) dataset. The columns are the number of words and entries,
the average length of example sentences and definitions.

###  4.2 Models

In this work, we mainly compare models of three different categories: non-
pretrained model, pretrained masked LM, and pretrained encoder-decoder. We
then present the detailed settings of these models.

#### Non-Pretrained Model

We choose the LOG-CaD Ishiwatari et al. (2019) model as a baseline for
monolingual experiments. This model is an encoder-decoder model proposed to
describe unknown phrases, which can also be used to generate definitions for
given words and contexts. The model has three different encoders, which are
(1) a global context encoder to lookup pretrained embeddings of the given
word; (2) a local context encoder (Bi-LSTM) to encode the context; (3) a CNN
layer to encode character-level features of the given word. The decoder is a
two-layer LSTM, which receives the above three encoded information and
dynamically weighs them at each time step. We set the hyper-parameters exactly
like the original paper for a fair comparison.

| English | Chinese
---|---|---
BLEU |  NIST |  BLEU |  NIST
LOG-CaD |  18.76 |  43.97 |  20.71 |  44.57
BERT-Base |  22.84 |  65.24 |  23.64 |  41.91
BERT-Large |  27.35 |  87.32 |  – |  –
(m)BART |  29.51 |  98.92 |  22.36 |  49.53
|  |  |  |
Table 2: BLEU and NIST scores of monolingual experiments. Note that BART is
used for English and mBART is used for Chinese.

#### Pretrained Masked LM

The masked LMs are pretrained by the MLM Devlin et al. (2019) task, which aims
to predict masked text pieces based on surrounded context. We use the masked
LMs to initialize parameters in the transformer encoder. For monolingual
experiments, we compare the BERT-Base and BERT-Large models. We only evaluate
the effectiveness of English BERT-Large model since no Chinese model
available. For cross-lingual experiments, we compare the mBERT333mBERT:
https://github.com/google-research/bert/blob/master/multilingual.md and XLM-R-
Large Conneau et al. (2019) models. In practice, we randomly initialize a
transformer decoder and feed the output of masked LMs into the cross-attention
mechanism. The decoder architecture is set the same as Vaswani et al. (2017).
We train the entire model in two phases. The first phase fix the parameters in
the encoder and train the decoder from scratch by the learning rate of 1e-4.
And the second phase use a smaller learning rate of 1e-5 to fine-tune the
entire model. We report test results after the fine-tuning phase in Section
4.4.

#### Pretrained Encoder-Decoder

The encoder-decoder model leverages a left-to-right LM to generate a sentence
conditioned on a separate encoder for a given text. We fine-tune the BART
Lewis et al. (2020) model on the Oxford dataset for English experiments. Since
no Chinese BART model available, we fine-tune the mBART Liu et al. (2020)
model on the CCLD-Mono dataset and set both input and output language prompts
as zh_CN. For cross-lingual experiments, we fine-tune the mBART model on the
Oxford dataset and set both input and output language prompts as en_XX. And
then alter the input language prompt as zh_CN for cross-lingual inference. The
fine-tuning learning rate is set to 1e-4.

|  | #1 | #2 | #3 | Avg.
---|---|---|---|---|---
Acc. | XLM-R-L | 1.10 | 1.04 | 1.12 | 1.09
mBART | 2.39 | 1.77 | 2.28 | 2.15
mBERT | 2.59 | 2.20 | 2.90 | 2.56
Flu. | XLM-R-L | 3.39 | 4.27 | 4.83 | 4.16
mBART | 4.01 | 4.53 | 4.95 | 4.50
mBERT | 3.93 | 4.50 | 4.80 | 4.41
|  |  |  |  |
Table 3: Manually evaluated accuracy and fluency scores of the cross-lingual
dictionary and baseline models.

###  4.3 Evaluation

For monolingual methods, we use the BLEU Papineni et al. (2002) and NIST
Doddington (2002) as automatic evaluation metrics. NIST focuses on content
words by giving more weightage to them. This makes NIST more informative than
solely assigning an equal weight to each n-grams as BLEU Huang et al. (2021).

For cross-lingual methods, since there are no golden standard English
definitions, we let 3 scorers to manually evaluate the generated results. We
randomly shuffle a total of 600 definitions generated by 3 models and let
scorers rate them independently. Specifically, each scorer evaluates a
definition on two criteria of accuracy and fluency. Both criteria range from 1
to 5, with 1 being the lowest and 5 being the highest.

English Monolingual Mode
---
Word |  prominence
Context |  By the close of the 1870s, Homer had achieved national  prominence.
Reference |  the state of being important, famous, or noticeable
LOG-CaD |  the state of being protuberant or [unk]
BERT-Base |  the state or fact of being prominent
BERT-Large |  the quality or state of being recognized or prominent
BART |  the state or fact of surpassing all others; superiority in rank or status
Chinese Monolingual Mode
Word |  惦记 (thinking about)
Context |  我在这边过得很好，您别总惦记着。  (I have a good time here, don’t worry about it all the time.)
|
|
Reference |  因想念、担心、期盼等而心里老想着 (Always thinking about sth. because of missing, worrying, looking forward to, etc.)
|
LOG-CaD |  用在名词或作品上 (Used in nouns or works.)
|
BERT-Base |  无法回忆起曾经发生过的事情；过去知道的事不记得了 (Unable to recall what happened; don’t remember old things in the past.)
|
mBART |  心里记挂着某人或某事 (Keep thinking about sb. or sth.)
|
Chinese-English Cross-lingual Mode
Word |  基础 (foundation)
Context |  楼越高，基础越要坚实。  (The higher the building, the more solid the foundation.)
|
|
Reference |  地面以下用来支撑建筑物的部分 (The part below the ground that supports a building.)
|
XLM-R-L |  (with reference to the temperature) free from normal temperatures
mBART |  the basic level of difficulty
mBERT |  the base of something
|

Table 4: Definition generation cases in three different modes.

###  4.4 Results

Table 2 illustrates results of monolingual experiments. We observe that BART
performs best among all the models. On the English test set, BART yields
significantly better results than the other three methods. On the Chinese test
set, although BART performs slightly worse on BLEU score, it still outperforms
other model on NIST score significantly. Therefore, the LitMind Dictionary
uses BART for both English and Chinese monolingual definition generation.

Table 3 shows the manual evaluation results of the cross-lingual experiments.
We observe that mBERT get the highest accuracy score and mBART get the highest
fluency score. We use the mBERT in LitMind Dictionary because: (1) these two
models have almost the same level of fluency, and (2) we believe that the
accuracy of dictionary definitions are more important than fluency.

###  4.5 Case Study

Table 4.3 shows the generated definitions in English, Chinese and Chinese-
English modes. The models we chose to serve in LitMind Dictionary successfully
generate accurate and fluent definitions.

For the English mode, LOG-CaD erroneously uses protuberant to explain
prominence, and generates a special [unk] token. Both BERT-Base and BERT-Large
use the given word in the definitions, and fail to explain the meaning. BART
defines the given word as surpassing and superiority, which is more close to
the reference semantically.

For the Chinese mode, LOG-CaD generates a completely irrelevant sentence and
fails to explain the given word. Bert-Base generates the definition of 忘记
(forget) rather than 惦记 (thinking about), which basically have the opposite
meanings. The mBART generates the most relevant definition compared to other
models.

For the Cross-lingual mode, temperature generated by XLM-R-L has nothing to do
with the given word foundation. The mBART generate a keyword of basic, but
difficulty also fails to explain the meaning of given word. In contrast, the
definition generated by mBERT is the most relevant.

##  5 Conclusion and Future Work

In this paper, we present LitMind Dictionary, an open-source online generative
dictionary, which can generate context-aware definitions of a given word. Our
system supports Chinese and English monolingual queries as well as Chinese-
English cross-lingual queries, all of which are realized in a generative
fashion for the first time. In the future, we will try to control the
difficulty of the generated definitions to make it more suitable for language
learners. We will also work on how to match more appropriate example sentences
for the query words.

## References

  * Bartunov et al. (2016) Sergey Bartunov, Dmitry Kondrashkin, Anton Osokin, and Dmitry Vetrov. 2016.  Breaking sticks and ambiguities with adaptive skip-gram.  In _Proceedings of of the 2016 Conference on Artificial Intelligence and Statistics_ , pages 130–138.
  * Bevilacqua et al. (2020) Michele Bevilacqua, Marco Maru, and Roberto Navigli. 2020.  Generationary or “how we went beyond word sense inventories and learned to gloss”.  In _Proceedings of the 2020 Conference on Empirical Methods in Natural Language Processing_ , pages 7207–7221.
  * Chang and Chen (2019) Ting-Yun Chang and Yun-Nung Chen. 2019.  What Does This Word Mean? Explaining Contextualized Embeddings with Natural Language Definition.  In _Proceedings of the 2019 Conference on Empirical Methods in Natural Language Processing and the 9th International Joint Conference on Natural Language Processing (EMNLP-IJCNLP)_ , pages 6064–6070.
  * Chang et al. (2018) Ting-Yun Chang, Ta-Chung Chi, Shang-Chi Tsai, and Yun-Nung Chen. 2018.  xsense: Learning sense-separated sparse representations and textual definitions for explainable word sense networks.  _arXiv preprint arXiv:1809.03348_.
  * Conneau et al. (2019) Alexis Conneau, Kartikay Khandelwal, Naman Goyal, Vishrav Chaudhary, Guillaume Wenzek, Francisco Guzmán, Edouard Grave, Myle Ott, Luke Zettlemoyer, and Veselin Stoyanov. 2019.  Unsupervised cross-lingual representation learning at scale.  _arXiv preprint arXiv:1911.02116_.
  * Devlin et al. (2019) Jacob Devlin, Ming-Wei Chang, Kenton Lee, and Kristina Toutanova. 2019.  BERT: Pre-training of deep bidirectional transformers for language understanding.  In _Proceedings of the 2019 Conference of the North American Chapter of the Association for Computational Linguistics: Human Language Technologies_ , pages 4171–4186.
  * Doddington (2002) George Doddington. 2002.  Automatic evaluation of machine translation quality using n-gram co-occurrence statistics.  In _Proceedings of the Second International Conference on Human Language Technology Research_ , page 138–145. Morgan Kaufmann Publishers Inc.
  * Dong and Dong (2006) Zhendong Dong and Qiang Dong. 2006.  Hownet and the computation of meaning.
  * Enayati and Gilakjani (2020) Fatemeh Enayati and Abbas Pourhosein Gilakjani. 2020.  The impact of computer assisted language learning (CALL) on improving intermediate EFL learners’ vocabulary learning.  _International Journal of Language Education_ , 4(2):96–112.
  * Fan et al. (2020) Qinan Fan, Cunliang Kong, Liner Yang, and Erhong Yang. 2020.  Chinese definition modeling based on BERT and beam seach.  In _Proceedings of the 19th Chinese National Conference on Computational Linguistics_ , pages 336–348.
  * Gadetsky et al. (2018) Artyom Gadetsky, Ilya Yakubovskiy, and Dmitry P. Vetrov. 2018.  Conditional generators of words definitions.  In _Proceedings of the 56th Annual Meeting of the Association for Computational Linguistics (Volume 2: Short Papers)_ , volume abs/1806.10090, pages 266–271.
  * Huang et al. (2021) Han Huang, Tomoyuki Kajiwara, and Yuki Arase. 2021.  Definition modelling for appropriate specificity.  In _Proceedings of the 2021 Conference on Empirical Methods in Natural Language Processing_ , pages 2499–2509.
  * Ishiwatari et al. (2019) Shonosuke Ishiwatari, Hiroaki Hayashi, Naoki Yoshinaga, Graham Neubig, Shoetsu Sato, Masashi Toyoda, and Masaru Kitsuregawa. 2019.  Learning to describe unknown phrases with local and global contexts.  In _Proceedings of the 2019 Conference of the North American Chapter of the Association for Computational Linguistics: Human Language Technologies_ , pages 3467–3476.
  * Kilgarriff (1997) Adam Kilgarriff. 1997.  I don’t believe in word senses.  _Computers and the Humanities_ , 31(2):91–113.
  * Kilgarriff (2007) Adam Kilgarriff. 2007.  Word senses.  In _Word Sense Disambiguation_ , pages 29–46. Springer.
  * Kong et al. (2020) Cunliang Kong, Liner Yang, Tianzuo Zhang, Qinan Fan, Zhenghao Liu, Yun Chen, and Erhong Yang. 2020.  Toward cross-lingual definition generation for language learners.  _arXiv preprint arXiv:2010.05533_.
  * Lample and Conneau (2019) Guillaume Lample and Alexis Conneau. 2019.  Cross-lingual language model pretraining.  _arXiv preprint arXiv:1901.07291_.
  * Lewis et al. (2020) Mike Lewis, Yinhan Liu, Naman Goyal, Marjan Ghazvininejad, Abdelrahman Mohamed, Omer Levy, Veselin Stoyanov, and Luke Zettlemoyer. 2020.  BART: Denoising sequence-to-sequence pre-training for natural language generation, translation, and comprehension.  In _Proceedings of the 58th Annual Meeting of the Association for Computational Linguistics_.
  * Li et al. (2020) Jiahuan Li, Yu Bao, Shujian Huang, Xinyu Dai, and Jiajun Chen. 2020.  Explicit semantic decomposition for definition generation.  In _Proceedings of the 58th Annual Meeting of the Association for Computational Linguistics_ , pages 708–717.
  * Liu et al. (2020) Yinhan Liu, Jiatao Gu, Naman Goyal, Xian Li, Sergey Edunov, Marjan Ghazvininejad, Mike Lewis, and Luke Zettlemoyer. 2020.  Multilingual denoising pre-training for neural machine translation.  _Transactions of the Association for Computational Linguistics_ , 8:726–742.
  * Lolita et al. (2020) Yuri Lolita, Endry Boeriswati, and Ninuk Lustyantie. 2020.  The impact of computer assisted language learning (CALL) use of english vocabulary enhancement.  _Linguistic, English Education and Art (LEEA) Journal_ , 4(1):206–221.
  * Mickus et al. (2019) Timothee Mickus, Denis Paperno, and Matthieu Constant. 2019.  Mark my word: A sequence-to-sequence approach to definition modeling.  In _Proceedings of the First NLPL Workshop on Deep Learning for Natural Language Processing_ , pages 1–11.
  * Noraset et al. (2017) Thanapon Noraset, Chen Liang, Lawrence Birnbaum, and Doug Downey. 2017.  Definition modeling: Learning to define word embeddings in natural language.  In _Proceedings of the 2017 AAAI Conference on Artificial Intelligence_ , volume 31.
  * Papineni et al. (2002) Kishore Papineni, Salim Roukos, Todd Ward, and Wei-Jing Zhu. 2002.  Bleu: A method for automatic evaluation of machine translation.  In _Proceedings of the 40th Annual Meeting on Association for Computational Linguistics_ , page 311–318. Association for Computational Linguistics.
  * Raffel et al. (2019) Colin Raffel, Noam Shazeer, Adam Roberts, Katherine Lee, Sharan Narang, Michael Matena, Yanqi Zhou, Wei Li, and Peter J. Liu. 2019.  Exploring the limits of transfer learning with a unified text-to-text transformer.  _CoRR_ , abs/1910.10683.
  * Reid et al. (2020) Machel Reid, Edison Marrese-Taylor, and Yutaka Matsuo. 2020.  VCDM: Leveraging Variational bi-encoding and Deep contextualized Word Representations for Improved Definition Modeling.  In _Proceedings of the 2020 Conference on Empirical Methods in Natural Language Processing_ , pages 6331–6344.
  * Rosch and Mervis (1975) Eleanor Rosch and Carolyn B Mervis. 1975.  Family resemblances: Studies in the internal structure of categories.  _Cognitive Psychology_ , 7(4):573–605.
  * Segler et al. (2002) Thomas M Segler, Helen Pain, and Antonella Sorace. 2002.  Second language vocabulary acquisition and learning strategies in ICALL environments.  _Computer Assisted Language Learning_ , 15(4):409–422.
  * Tyler and Evans (2001) Andrea Tyler and Vyvyan Evans. 2001.  Reconsidering prepositional polysemy networks: The case of over.  _Language_ , pages 724–765.
  * Vaswani et al. (2017) Ashish Vaswani, Noam Shazeer, Niki Parmar, Jakob Uszkoreit, Llion Jones, Aidan N. Gomez, Lukasz Kaiser, and Illia Polosukhin. 2017.  Attention is all you need.  In _Proceedings of the 2017 Conference on Neural Information Processing Systems_ , pages 6000–6010.
  * Yang et al. (2020) Liner Yang, Cunliang Kong, Yun Chen, Yang Liu, Qinan Fan, and Erhong Yang. 2020\.  Incorporating sememes into chinese definition modeling.  _IEEE/ACM Transactions on Audio, Speech, and Language Processing_ , 28:1669–1677.
  * Zheng et al. (2021) Hua Zheng, Damai Dai, Lei Li, Tianyu Liu, Zhifang Sui, Baobao Chang, and Yang Liu. 2021.  Decompose, fuse and generate: A formation-informed method for Chinese definition generation.  In _Proceedings of the 2021 Conference of the North American Chapter of the Association for Computational Linguistics: Human Language Technologies_ , pages 5524–5531.


# SOME GENERALIZED JORDAN MAPS ON TRIANGULAR RINGS FORCE ADDITIVITY

Sk Aziz  Department of Mathematics, Indian Institute of Technology Patna,
Patna-801106<EMAIL_ADDRESS>,  Arindam Ghosh  Department of
Mathematics, Government Polytechnic Kishanganj, Kishanganj-855116  E-mail:
<EMAIL_ADDRESS>and  Om Prakash⋆ Department of Mathematics, Indian
Institute of Technology Patna, Patna-801106<EMAIL_ADDRESS>
###### Abstract.

In this paper, we show that a map $\delta$ over a triangular ring
$\mathcal{T}$ satisfying
$\delta(ab+ba)=\delta(a)b+a\tau(b)+\delta(b)a+b\tau(a)$, for all
$a,b\in\mathcal{T}$ and for some maps $\tau$ over $\mathcal{T}$ satisfying
$\tau(ab+ba)=\tau(a)b+a\tau(b)+\tau(b)a+b\tau(a)$, is additive. Also, it is
shown that a map $T$ on $\mathcal{T}$ satisfying $T(ab)=T(a)b=aT(b)$, for all
$a,b\in\mathcal{T}$, is additive. Further, we establish that if a map $D$ over
$\mathcal{T}$ satisfies $(m+n)D(ab)=2mD(a)b+2naD(b)$, for all
$a,b\in\mathcal{T}$ and integers $m,n\geq 1$, then $D$ is additive.

###### Key words and phrases:

Additivity, Jordan derivation, Two-sided centralizer, $(m,n)$-derivation,
Triangular ring

###### 2020 Mathematics Subject Classification:

16W10, 16S50

* Corresponding author

##  1\. Introduction

Let $R$ be a ring. A map $f:R\rightarrow R$ is said to be additive if

$\displaystyle f(a+b)=f(a)+f(b)$

for all $a,b\in R$. In 1957, Herstein [5] introduced Jordan derivation over
rings as an additive map $\tau:R\rightarrow R$ satisfies

$\displaystyle\tau(a^{2})=\tau(a)a+a\tau(a)$

for all $a\in R$. He also proved that any Jordan derivation over a prime ring
is a derivation with some torsion restrictions. In 2003, Jing and Lu [6]
introduced generalized Jordan derivation and proved that every generalized
Jordan derivation is a generalized derivation over a $2$-torsion-free prime
ring. Recall that an additive map $\delta:R\rightarrow R$ is said to be a
generalized Jordan derivation if

$\displaystyle\delta(a^{2})=\delta(a)a+a\tau(a)$

for all $a\in R$ and for some Jordan derivation $\tau:R\rightarrow R$. In
1952, Wendel [13] introduced the concept of a centralizer. An additive map
$T:R\rightarrow R$ is said to be a left centralizer if

$\displaystyle T(ab)=T(a)b$

for all $a,b\in R$. Similarly, we define the right centralizer over a ring. An
additive map $T:R\rightarrow R$ is said to be a two-sided centralizer if $T$
is both left and right centralizer. In 2014, Ali and Fošner [1] introduced the
concept of $(m,n)$-derivation. Let $m,n\geq 0$ be integers. Then an additive
map $D:R\rightarrow R$ is said to be an $(m,n)$-derivation if

$\displaystyle(m+n)D(ab)=2mD(a)b+2naD(b),$

for all $a,b\in R$. In fact, for $m=1$ and $n=1$, $D$ is a normal derivation
on a $2$-torsion free ring $R$.

Let $n\geq 2$ be any integer. Then a ring $R$ is said to be $n$-torsion free
if $na=0$ for some $a\in R$ implies $a=0$. It is an interesting question that
when a map over a ring (satisfying some functional equations) is additive. The
question was first raised by Rickart [9] in 1948. He showed that any bijective
and multiplicative mapping of a Boolean ring B onto any arbitrary ring S is
additive. Later, Johnson made an extensive study of these results in [7]. In
1969, Martindale III [8] proved that any multiplicative isomorphism of a ring
$R$ onto an arbitrary ring $S$ is additive under the existence of a family of
idempotent elements in $R$ satisfying certain conditions. In 1991, by assuming
Martindale’s conditions, Daif [2] proved that any multiplicative derivation of
a ring $R$ is additive. Later, in 2009, Wang [11] proved that any
$n$-multiplicative derivation $d$ of $R$ is additive by using Martindale’s
conditions where $n>1$ is an integer. He proved that any $n$-multiplicative
isomorphism or $n$-multiplicative derivation of a standard operator algebra
$A$ over a Banach space $X$ with $dim(X)\geq 2$, is additive.

Again, in 2011, Wang [12] proved that any $n$-multiplicative derivation $d$ of
a triangular ring $\mathcal{T}$ is additive, with some assumptions on
$\mathcal{T}$. In 2014, Ferreira [3] proved that every $m$-multiplicative
isomorphism from a $n$-triangular matrix ring $\mathcal{T}$ onto any ring
$\mathcal{S}$ is additive with some assumptions on $\mathcal{T}$ where $n,m>1$
are integers. He also showed that every $m$-multiplicative derivation of
$\mathcal{T}$ is additive. In 2015, Ferreira [4] again revisit and proved that
a map $\tau$ on a triangular ring $\mathcal{T}$ satisfying

$\displaystyle\tau(ab+ba)=\tau(a)b+a\tau(b)+\tau(b)a+b\tau(a)$

for all $a,b\in\mathcal{T}$, is additive with some assumptions on
$\mathcal{T}$. Moreover, if $\mathcal{T}$ is $2$-torsion free, then $\tau$ is
a Jordan derivation. Ferreira’s result [4] motivated us to work to find out
the conditions under which a multiplicative generalized Jordan derivation is
additive on a triangular matrix ring. This paper provides an affirmative
answer to the above question.

In 2014, El-Sayed et al. [10] proved that every map $T$ on a prime ring with a
non-trivial idempotent satisfying

(1.1) $T(ab)=T(a)b$

for all $a,b\in R$, is additive. Note that a ring $R$ is said to be a prime
ring if $aRb=0$ for some $a,b\in R$ implies either $a=0$ or $b=0$ and is said
to be a semi-prime ring if $aRa=0$ for some $a\in R$ implies that $a=0$. Thus,
every prime ring is semi-prime, but the converse is not true. Motivated by the
work of El-Sayed et al. [10], we prove that a map $T$ on $\mathcal{T}$
satisfying

$\displaystyle T(ab)=T(a)b=aT(b),$

for all $a,b\in\mathcal{T}$, is additive.

Further, we prove that any map $D$ on $\mathcal{T}$ which satisfies

$\displaystyle(m+n)D(ab)=2mD(a)b+2naD(b),$

for all $a,b\in\mathcal{T}$ and for some integers $m,n\geq 1$, is additive.

Throughout the work, we use the definition of triangular algebra given by Wang
[12].

######  Definition 1.1.

Let $A$ and $B$ be two rings, and $M$ be an $(A,B)$-bimodule such that

(a) $M$ is faithful as left $A$-module and right $B$-module;

(b) If $m\in M$ is such that $AmB=0$ implies $m=0$.

Then the ring

$\displaystyle\mathcal{T}=\text{Tri}(A,M,B)=\Biggl{\\{}\left(\begin{array}[]{ccc}a&m\\\
&b\end{array}\right)|\enspace a\in A,\,b\in B,\,m\in M\Biggl{\\}},$

under usual matrix addition and multiplication is said to be a triangular
ring.

Let $\mathcal{T}_{11}=\Biggl{\\{}\left(\begin{array}[]{ccc}a&0\\\
&0\end{array}\right)|\enspace a\in A\Biggl{\\}}$,
$\mathcal{T}_{12}=\Biggl{\\{}\left(\begin{array}[]{ccc}0&m\\\
&0\end{array}\right)|\enspace m\in M\Biggl{\\}}$ and

$\mathcal{T}_{22}=\Biggl{\\{}\left(\begin{array}[]{ccc}0&0\\\
&b\end{array}\right)|\enspace b\in B\Biggl{\\}}$.

Then
$\mathcal{T}=\mathcal{T}_{11}\oplus\mathcal{T}_{12}\oplus\mathcal{T}_{22}$.
Henceforth, $t_{ij}\in\mathcal{T}_{ij}$. Also, $t_{ij}t_{kl}=0$, if $j\neq k$
where $j,~{}k\in\\{1,2\\}$.

##  2\. GENERALIZED JORDAN DERIVATIONS ON TRIANGULAR RINGS

######  Theorem 2.1.

Let $\mathcal{T}$ be a triangular ring with conditions:

(i) $aA=0$ for some $a\in A$ implies $a=0$;

(ii) $bB=0$ for some $b\in B$ implies $b=0$;

(iii) $A$ and $B$ are rings with identity.

If a function $\delta:\mathcal{T}\rightarrow\mathcal{T}$ satisfies

$\displaystyle\delta(ab+ba)=\delta(a)b+a\tau(b)+\delta(b)a+b\tau(a),$

for all $a,b\in\mathcal{T}$ where $\tau:\mathcal{T}\rightarrow\mathcal{T}$
satisfies

$\displaystyle\tau(ab+ba)=\tau(a)b+a\tau(b)+\tau(b)a+b\tau(a),$

for all $a,b\in\mathcal{T}$, then $\delta$ is additive. Moreover, if
$\mathcal{T}$ is $2$-torsion free, then $\delta$ is a generalized Jordan
derivation.

Since $\tau:\mathcal{T}\rightarrow\mathcal{T}$ satisfies all the conditions in
Theorem $3.1$ of [4], $\tau$ is additive. In this section, we frequently use
all three conditions for the triangular ring $\mathcal{T}$, the conditions on
the maps $\delta$ and $\tau$ described in Theorem 2.1, and the additivity of
$\tau$ without mentioning them. Note that $\delta(0)=0$. Before proving
Theorem 2.1, we first discuss several needful lemmas.

######  Lemma 2.2.

Let $a\in\mathcal{T}_{11}$, $b\in\mathcal{T}_{22}$ and $m\in\mathcal{T}_{12}$.
Then

(2.1) $\displaystyle(i)~{}~{}\delta(a+m)=\delta(a)+\delta(m),$
$\displaystyle(ii)~{}~{}\delta(b+m)=\delta(b)+\delta(m).$

###### Proof.

Let $t_{2}\in\mathcal{T}_{22}$. Then

(2.2) $\displaystyle\delta[(a+m)t_{2}+t_{2}(a+m)]$
$\displaystyle=\delta(a+m)t_{2}+(a+m)\tau(t_{2})+\delta(t_{2})(a+m)+t_{2}\tau(a+m).$

On the other hand,

(2.3) $\displaystyle\delta[(a+m)t_{2}+t_{2}(a+m)]$
$\displaystyle=\delta(mt_{2})=\delta(0)+\delta(mt_{2}+0)$
$\displaystyle=\delta(at_{2}+t_{2}a)+\delta(mt_{2}+t_{2}m)$
$\displaystyle=\delta(a)t_{2}+a\tau(t_{2})+\delta(t_{2})a+t_{2}\tau(a)$
$\displaystyle+\delta(m)t_{2}+m\tau(t_{2})+\delta(t_{2})m+t_{2}\tau(m).$

Since $\tau$ is additive, comparing (2.2) and (2.3), we have

(2.4) $\displaystyle[\delta(a+m)-\delta(a)-\delta(m)]t_{2}=0$
$\displaystyle\implies[\delta(a+m)-\delta(a)-\delta(m)]_{12}t_{2}=0$
$\displaystyle\text{and}~{}~{}[\delta(a+m)-\delta(a)-\delta(m)]_{22}t_{2}=0.$

Using conditions (b) and (ii) of Definition 1.1 and Theorem 2.1, respectively,
we have

(2.5) $\displaystyle[\delta(a+m)-\delta(a)-\delta(m)]_{12}=0$
$\displaystyle\text{and}~{}~{}[\delta(a+m)-\delta(a)-\delta(m)]_{22}=0.$

Now, in order to prove $[\delta(a+m)-\delta(a)-\delta(m)]_{11}=0$, let
$n\in\mathcal{T}_{12}$. Then

(2.6) $\displaystyle\delta[(a+m)n+n(a+m)]=\delta(an)$
$\displaystyle=\delta(an+na)+\delta(mn+nm)$
$\displaystyle=\delta(a)n+a\tau(n)+\delta(n)a+n\tau(a)+\delta(m)n+m\tau(n)+\delta(n)m+n\tau(m).$

Also, we have

(2.7) $\displaystyle\delta[(a+m)n+n(a+m)]$
$\displaystyle=\delta(a+m)n+(a+m)\tau(n)+\delta(n)(a+m)+n\tau(a+m)$
$\displaystyle=\delta(a+m)n+(a+m)\tau(n)+\delta(n)(a+m)+n(\tau(a)+\tau(m))~{}(\text{By
additivity of}~{}\tau).$

Comparing the equalities (2.6) and (2.7),

(2.8) $\displaystyle[\delta(a+m)-\delta(a)-\delta(m)]_{11}n=0$
$\displaystyle\implies[\delta(a+m)-\delta(a)-\delta(m)]_{11}=0~{}(\text{by
condition (a) of Definition \ref{def1}}).$

Hence, $\delta(a+m)=\delta(a)+\delta(m)$. Similarly, we can prove that
$\delta(b+m)=\delta(b)+\delta(m)$. ∎

######  Lemma 2.3.

Let $a\in\mathcal{T}_{11}$, $b\in\mathcal{T}_{22}$ and
$m,n\in\mathcal{T}_{12}$. Then

(2.9) $\delta(am+nb)=\delta(am)+\delta(nb).$

###### Proof.

Since $am+nb=(a+n)(m+b)+(m+b)(a+n)$, by Lemma 2.2 and the additivity of
$\tau$, we have

(2.10) $\displaystyle\delta(am+nb)$
$\displaystyle=\delta[(a+n)(m+b)+(m+b)(a+n)]$
$\displaystyle=\delta(a+n)(m+b)+(a+n)\tau(m+b)$
$\displaystyle+\delta(m+b)(a+n)+(m+b)\tau(a+n)$
$\displaystyle=(\delta(a)+\delta(n))(m+b)+(a+n)(\tau(m)+\tau(b))$
$\displaystyle+(\delta(m)+\delta(b))(a+n)+(m+b)(\tau(a)+\tau(n))$
$\displaystyle=\delta(am+ma)+\delta(nb+bn)+\delta(mn+nm)+\delta(ab+ba)$
$\displaystyle=\delta(am)+\delta(nb).$

∎

######  Lemma 2.4.

Let $m,n\in\mathcal{T}_{12}$. Then

(2.11) $\delta(m+n)=\delta(m)+\delta(n).$

###### Proof.

Since we assume both $A$ and $B$ are rings with identity in Theorem 2.1,
putting $a=1$ and $b=1$ in Lemma 2.3, we have the desired result. ∎

######  Lemma 2.5.

Let $a_{1},a_{2},\in\mathcal{T}_{11}$ and $b_{1},b_{2},\in\mathcal{T}_{22}$.
Then

(2.12) $\displaystyle(i)~{}\delta(a_{1}+a_{2})=\delta(a_{1})+\delta(a_{2}),$
$\displaystyle(ii)~{}\delta(b_{1}+b_{2})=\delta(b_{1})+\delta(b_{2}).$

###### Proof.

Let $t_{2}\in\mathcal{T}_{22}$. Then

(2.13)
$\displaystyle\delta(a_{1}+a_{2})t_{2}+(a_{1}+a_{2})\tau(t_{2})+\delta(t_{2})(a_{1}+a_{2})+t_{2}\tau(a_{1}+a_{2})$
$\displaystyle=\delta[(a_{1}+a_{2})t_{2}+t_{2}(a_{1}+a_{2})]=\delta(0)=0$
$\displaystyle=\delta(a_{1}t_{2}+t_{2}a_{1})+\delta(a_{2}t_{2}+t_{2}a_{2})$
$\displaystyle=\delta(a_{1})t_{2}+a_{1}\tau(t_{2})+\delta(t_{2})a_{1}+t_{2}\tau(a_{1})$
$\displaystyle+\delta(a_{2})t_{2}+a_{2}\tau(t_{2})+\delta(t_{2})a_{2}+t_{2}\tau(a_{2}).$

Since $\tau$ is additive, from (2.13), we have

(2.14) $\displaystyle[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]t_{2}=0$
$\displaystyle\implies[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]_{12}t_{2}=0$
$\displaystyle\text{and}~{}~{}[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]_{22}t_{2}=0.$

Using conditions $(b)$ and $(ii)$ of Definition 1.1 and Theorem 2.1,
respectively, we have

(2.15) $\displaystyle[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]_{12}=0$
$\displaystyle\text{and}~{}~{}[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]_{22}=0.$

Now, to prove $[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]_{11}=0$, let
$m\in\mathcal{T}_{12}$. Then

(2.16)
$\displaystyle\delta(a_{1}+a_{2})m+(a_{1}+a_{2})\tau(m)+\delta(m)(a_{1}+a_{2})+m\tau(a_{1}+a_{2})$
$\displaystyle=\delta[(a_{1}+a_{2})m+m(a_{1}+a_{2})]$
$\displaystyle=\delta[(a_{1}m+ma_{1})+(a_{2}m+ma_{2})]$
$\displaystyle=\delta(a_{1}m+ma_{1})+\delta(a_{2}m+ma_{2})~{}(\text{By Lemma
\ref{lem3}})$
$\displaystyle=\delta(a_{1})m+a_{1}\tau(m)+\delta(m)a_{1}+m\tau(a_{1})+\delta(a_{2})m+a_{2}\tau(m)+\delta(m)a_{2}+m\tau(a_{2}).$

By (2.16) and the additivity of $\tau$, we see that

(2.17) $\displaystyle[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]m=0$
$\displaystyle\implies[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]_{11}m=0$
$\displaystyle\implies[\delta(a_{1}+a_{2})-\delta(a_{1})-\delta(a_{2})]_{11}=0~{}(\text{by
condition (a) of Definition \ref{def1}}).$

Hence, $\delta(a_{1}+a_{2})=\delta(a_{1})+\delta(a_{2})$. In a similar way, we
can prove that $\delta(b_{1}+b_{2})=\delta(b_{1})+\delta(b_{2}).$ ∎

######  Lemma 2.6.

Let $a\in\mathcal{T}_{11}$, $b\in\mathcal{T}_{22}$ and $m\in\mathcal{T}_{12}$.
Then

(2.18) $\delta(a+m+b)=\delta(a)+\delta(m)+\delta(b).$

###### Proof.

Let $t_{2}\in\mathcal{T}_{22}$. Then

(2.19)
$\displaystyle\delta(a+m+b)t_{2}+(a+m+b)\tau(t_{2})+\delta(t_{2})(a+m+b)+t_{2}\tau(a+m+b)$
$\displaystyle=\delta[(a+m+b)t_{2}+t_{2}(a+m+b)]$
$\displaystyle=\delta(mt_{2}+bt_{2}+t_{2}b)$
$\displaystyle=\delta(bt_{2}+t_{2}b)+\delta(mt_{2})~{}(\text{By Lemma
\ref{lem1}})$
$\displaystyle=\delta(bt_{2}+t_{2}b)+\delta(t_{2}m+mt_{2})+\delta(at_{2}+t_{2}a)$
$\displaystyle=\delta(b)t_{2}+b\tau(t_{2})+\delta(t_{2})b+t_{2}\tau(b)+\delta(t_{2})m+t_{2}\tau(m)$
$\displaystyle+\delta(m)t_{2}+m\tau(t_{2})+\delta(a)t_{2}+a\tau(t_{2})+\delta(t_{2})a+t_{2}\tau(a)$
$\displaystyle\implies[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]t_{2}=0~{}(\text{Using
additivity of}~{}\tau)$
$\displaystyle\implies[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]_{12}t_{2}=0$
$\displaystyle\text{and}~{}[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]_{22}t_{2}=0.$

Using conditions $(b)$ and $(ii)$ of Definition 1.1 and Theorem 2.1, we have

(2.20) $\displaystyle[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]_{12}=0$
$\displaystyle\text{and}~{}[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]_{22}=0.$

Let $t_{1}\in\mathcal{T}_{11}$. Then

(2.21)
$\displaystyle\delta(a+m+b)t_{1}+(a+m+b)\tau(t_{1})+\delta(t_{1})(a+m+b)+t_{1}\tau(a+m+b)$
$\displaystyle=\delta[(a+m+b)t_{1}+t_{1}(a+m+b)]$
$\displaystyle=\delta(at_{1}+t_{1}a+t_{1}m)$
$\displaystyle=\delta(at_{1}+t_{1}a)+\delta(t_{1}m)~{}(\text{By Lemma
\ref{lem1}})$
$\displaystyle=\delta(at_{1}+t_{1}a)+\delta(t_{1}m+mt_{1})+\delta(bt_{1}+t_{1}b)$
$\displaystyle=\delta(a)t_{1}+a\tau(t_{1})+\delta(t_{1})a+t_{1}\tau(a)+\delta(t_{1})m+t_{1}\tau(m)$
$\displaystyle+\delta(m)t_{1}+m\tau(t_{1})+\delta(b)t_{1}+b\tau(t_{1})+\delta(t_{1})b+t_{1}\tau(b)$
$\displaystyle\implies[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]t_{1}=0$
$\displaystyle\implies[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]_{11}t_{1}=0$
$\displaystyle\implies[\delta(a+m+b)-\delta(a)-\delta(m)-\delta(b)]_{11}=0$
$\displaystyle(\text{By condition (i) of Theorem \ref{der1}}).$

Hence, by (2.20) and (2.21), $\delta(a+m+b)=\delta(a)+\delta(m)+\delta(b)$. ∎

Now, we are ready to prove Theorem 2.1.

###### Proof of Theorem 2.1.

Let $t\in\mathcal{T}$ and $u\in\mathcal{T}$. Then $t=t_{11}+t_{12}+t_{22}$ and
$u=u_{11}+u_{12}+u_{22}$ where $t_{ij},u_{ij}\in\mathcal{T}_{ij}$ and
$i,~{}j\in\\{1,2\\}$. Now,

(2.22) $\displaystyle\delta(t+u)$
$\displaystyle=\delta((t_{11}+t_{12}+t_{22})+(u_{11}+u_{12}+u_{22}))$
$\displaystyle=\delta((t_{11}+u_{11})+(t_{12}++u_{12})+(t_{22}+u_{22}))$
$\displaystyle=\delta(t_{11}+u_{11})+\delta(t_{12}+u_{12})+\delta(t_{22}+u_{22})~{}(\text{By
Lemma \ref{lem5}})$
$\displaystyle=\delta(t_{11})+\delta(u_{11})+\delta(t_{12})+\delta(u_{12})+\delta(t_{22})+\delta(u_{22})$
$\displaystyle~{}(\text{By Lemma \ref{lem3} and \ref{lem4}})$
$\displaystyle=\delta(t_{11})+\delta(t_{12})+\delta(t_{22})+\delta(u_{11})+\delta(u_{12})+\delta(u_{22})$
$\displaystyle=\delta(t_{11}+t_{12}+t_{22})+\delta(u_{11}+u_{12}+u_{22})~{}(\text{By
Lemma \ref{lem5}})$ $\displaystyle=\delta(t)+\delta(u).$

Therefore, $\delta$ is additive.

Let $\mathcal{T}$ be $2$-torsion free. By Theorem $3.1$ of [4], $\tau$ is a
Jordan derivation. For any $t\in\mathcal{T}$,

(2.23) $\displaystyle
2\delta(t^{2})=\delta(2t^{2})=\delta(tt+tt)=2[\delta(t)t+t\tau(t)]$
$\displaystyle\implies\delta(t^{2})=\delta(t)t+t\tau(t).$

Thus, $\delta$ is a generalized Jordan derivation. ∎

The following example shows that dropping condition $(iii)$ of Theorem 1.1
does not make any difference in the conclusion of this theorem.

######  Example 2.7.

Let
$\mathcal{T}=\text{Tri}(2\mathbb{Z},\mathbb{Z},3\mathbb{Z})=\Biggl{\\{}\left(\begin{array}[]{ccc}a&m\\\
&b\end{array}\right)|\enspace a\in 2\mathbb{Z},\,b\in
3\mathbb{Z},\,m\in\mathbb{Z}\Biggl{\\}}$ where $2\mathbb{Z}$ is the ring of
even integers, $3\mathbb{Z}$ is the ring of integers multiplied by $3$ and
$\mathbb{Z}$ is the $(2\mathbb{Z},3\mathbb{Z})$-bimodule of integers. Note
that $2\mathbb{Z}$ and $3\mathbb{Z}$ are rings without identity. Define
$\tau:\mathcal{T}\rightarrow\mathcal{T}$ as

$\displaystyle\tau\left(\begin{array}[]{ccc}a&m\\\
&b\end{array}\right)=\left(\begin{array}[]{ccc}0&\alpha a+\beta m-\alpha b\\\
&0\end{array}\right),$

where $\alpha,\beta\in\mathbb{Z}$. Then it is easy to check that $\tau$
satisfies all the assumptions of Theorem $3.1$ of [4]. Hence, $\tau$ is
additive and also a Jordan derivation. Define
$\delta:\mathcal{T}\rightarrow\mathcal{T}$ as

$\displaystyle\delta\left(\begin{array}[]{ccc}a&m\\\
&b\end{array}\right)=\left(\begin{array}[]{ccc}a\gamma&\alpha
a+(\beta+\gamma)m+\mu b\\\ &wb\end{array}\right),$

where $\gamma,\mu$ and $w\in\mathbb{Z}$. Then it is a routine exercise to
check that $\delta$ satisfies all the conditions except (iii) of Theorem 2.1
with the associated map $\tau$ just mentioned. Also, we can quickly check that
$\delta$ is additive and hence a generalized Jordan derivation.

Example 2.7 motivates us to post Theorem 2.1 without assuming condition
$(iii)$ as a conjecture.

######  Conjecture 2.8.

Let $\mathcal{T}$ be a triangular ring with conditions:

(i) $aA=0$ for some $a\in A$ implies $a=0$;

(ii) $bB=0$ for some $b\in B$ implies $b=0$.

If a map $\delta:\mathcal{T}\rightarrow\mathcal{T}$ satisfies

$\displaystyle\delta(ab+ba)=\delta(a)b+a\tau(b)+\delta(b)a+b\tau(a),$

for all $a,b\in\mathcal{T}$ where $\tau:\mathcal{T}\rightarrow\mathcal{T}$
satisfies

$\displaystyle\tau(ab+ba)=\tau(a)b+a\tau(b)+\tau(b)a+b\tau(a),$

for all $a,b\in\mathcal{T}$, then $\delta$ is additive. Moreover, if
$\mathcal{T}$ is $2$-torsion free, then $\delta$ is a generalized Jordan
derivation.

##  3\. TWO-SIDED CENTRALIZERS ON TRIANGULAR RINGS

######  Theorem 3.1.

Let $\mathcal{T}$ be a triangular ring with conditions:

(i) $Aa=0$ for some $a\in A$ implies $a=0$,

(ii) $bB=0$ for some $b\in B$ implies $b=0$,

(iii) $aA=0$ for some $a\in A$ implies $a=0$.

If a map $T:\mathcal{T}\rightarrow\mathcal{T}$ satisfies

$\displaystyle T(ab)=T(a)b=aT(b),$

for all $a,b\in\mathcal{T}$, then $T$ is additive. Moreover, $T$ is a two-
sided centralizer.

######  Lemma 3.2.

Let $a\in\mathcal{T}_{11}$, $b\in\mathcal{T}_{22}$ and $m\in\mathcal{T}_{12}$.
Then

(3.1) $\displaystyle(i)~{}T(a+m)=T(a)+T(m),$
$\displaystyle(ii)~{}T(b+m)=T(b)+T(m).$

###### Proof.

Let $t_{2}\in\mathcal{T}_{22}$. Now, we have

(3.2) $\displaystyle T[(a+m)t_{2}]=T(a+m)t_{2}.$

Also,

(3.3) $\displaystyle T[(a+m)t_{2}]=T(mt_{2})$ $\displaystyle=0+T(mt_{2})$
$\displaystyle=T(at_{2})+T(mt_{2})$ $\displaystyle=T(a)t_{2}+T(m)t_{2}.$

Comparing identities (3.2) and (3.3),

(3.4) $\displaystyle[T(a+m)-T(a)-T(m)]t_{2}=0$
$\displaystyle\implies[T(a+m)-T(a)-T(m)]_{12}t_{2}=0$
$\displaystyle\text{and}~{}[T(a+m)-T(a)-T(m)]_{22}t_{2}=0$
$\displaystyle\implies[T(a+m)-T(a)-T(m)]_{12}=0~{}(\text{By condition (b) of
Definition \ref{def1}})$
$\displaystyle\text{and}~{}[T(a+m)-T(a)-T(m)]_{22}=0~{}(\text{By condition
(ii) of Theorem \ref{thm3.1}}).$

Let $n\in\mathcal{T}_{12}$. We have

(3.5) $\displaystyle T[(a+m)n]=T(a+m)n.$

Also,

(3.6) $\displaystyle T[(a+m)n]=T(an)+0=T(an)+T(mn)=T(a)n+T(m)n.$

Comparing (3.5) and (3.6), we have

(3.7) $\displaystyle[T(a+m)-T(a)-T(m)]n=0$
$\displaystyle\implies[T(a+m)-T(a)-T(m)]_{11}n=0$
$\displaystyle\implies[T(a+m)-T(a)-T(m)]_{11}=0~{}(\text{By Condition (a) of
Definition \ref{def1}}).$

Hence, by using (3.4) and (3.7), we get $(i)$ of Lemma 3.2. Similarly, we have
$(ii)$ of Lemma 3.2. ∎

######  Lemma 3.3.

Let $a\in\mathcal{T}_{11},b\in\mathcal{T}_{22}$ and $m,n\in\mathcal{T}_{12}$.
Then

(3.8) $T(am+nb)=T(am)+T(nb).$

###### Proof.

Since $am+nb=(a+n)(b+m)$, we have

(3.9) $\displaystyle T(am+nb)$ $\displaystyle=T((a+n)(b+m))$
$\displaystyle=(a+n)T(b+m)$ $\displaystyle=(a+n)(T(b)+T(m))~{}(\text{By (ii)
of Lemma \ref{lem3.2}}).$

Also,

(3.10) $\displaystyle T(am)+T(nb)$ $\displaystyle=T(am)+T(nm)+T(nb)+T(ab)$
$\displaystyle=aT(m)+nT(m)+nT(b)+aT(b)$ $\displaystyle=(a+n)T(m)+(a+n)T(b)$
$\displaystyle=(a+n)(T(m)+T(b)).$

Comparing (3.9) and (3.10), we have the desired result. ∎

######  Lemma 3.4.

Let $b\in\mathcal{T}_{22}$ and $m,n\in\mathcal{T}_{12}$. Then

(3.11) $T(mb+nb)=T(mb)+T(nb).$

###### Proof.

Let $a\in\mathcal{T}_{11}$. Then

(3.12) $\displaystyle T[a((m+n)b)]=aT((m+n)b).$

Also,

(3.13) $\displaystyle T[a((m+n)b)]$ $\displaystyle=T(a(mb)+(an)b)$
$\displaystyle=T(a(mb))+T((an)b)~{}(\text{By Lemma \ref{lem3.3}})$
$\displaystyle=aT(mb)+aT(nb).$

Comparing (3.12) and (3.13), we have

(3.14) $\displaystyle a[T(mb+nb)-T(mb)-T(mb)]=0$ $\displaystyle\implies
a[T(mb+nb)-T(mb)-T(nb)]_{11}=0$
$\displaystyle\&~{}a[T(mb+nb)-T(mb)-T(nb)]_{12}=0$
$\displaystyle\implies[T(mb+nb)-T(mb)-T(nb)]_{11}=0~{}(\text{By condition (i)
of Theorem \ref{thm3.1}})$
$\displaystyle\&~{}[T(mb+nb)-T(mb)-T(nb)]_{12}=0~{}(\text{By condition (b) of
Definition \ref{def1}}).$

Let $p\in\mathcal{T}_{12}$. Then

(3.15) $\displaystyle T[p((m+n)b)]=pT((m+n)b).$

Also,

(3.16) $\displaystyle T[p((m+n)b)]$ $\displaystyle=T[p(mb+nb)]$
$\displaystyle=T(0)=0$ $\displaystyle=T(p(mb))+T(p(nb))$
$\displaystyle=pT(mb)+pT(nb).$

Comparing (3.15) and (3.16),

(3.17) $\displaystyle p[T(mb+nb)-T(mb)-T(nb)]=0$ $\displaystyle\implies
p[T(mb+nb)-T(mb)-T(nb)]_{22}=0$
$\displaystyle\implies[T(mb+nb)-T(mb)-T(nb)]_{22}=0~{}(\text{By condition (a)
of Definition \ref{def1}}).$

Thus the result follows from (3.14) and (3.17). ∎

######  Lemma 3.5.

Let $m,n\in\mathcal{T}_{12}$. Then

(3.18) $T(m+n)=T(m)+T(n).$

###### Proof.

Let $b\in\mathcal{T}_{22}$. Then

(3.19) $\displaystyle T((m+n)b)=T(m+n)b.$

Now,

(3.20) $\displaystyle T((m+n)b)$ $\displaystyle=T(mb+nb)$
$\displaystyle=T(mb)+T(nb)~{}(\text{By Lemma \ref{lem3.4}})$
$\displaystyle=T(m)b+T(n)b.$

By (3.19) and (3.20),

(3.21) $\displaystyle[T(m+n)-T(m)-T(n)]b=0$
$\displaystyle\implies[T(m+n)-T(m)-T(n)]_{12}b=0$
$\displaystyle\&~{}[T(m+n)-T(m)-T(n)]_{22}b=0$
$\displaystyle\implies[T(m+n)-T(m)-T(n)]_{12}=0~{}(\text{By condition (b) of
Definition \ref{def1}})$
$\displaystyle\&~{}[T(m+n)-T(m)-T(n)]_{22}=0~{}(\text{By condition (ii) of
Theorem \ref{thm3.1}}).$

Let $p\in\mathcal{T}_{12}$. Then

(3.22) $\displaystyle T((m+n)p)=T(m+n)p.$

Also,

(3.23) $\displaystyle T((m+n)p)=T(0)=0=T(mp)+T(np)=T(m)p+T(n)p.$

Comparing (3.22) and (3.23),

(3.24) $\displaystyle[T(m+n)-T(m)-T(n)]p=0$
$\displaystyle\implies[T(m+n)-T(m)-T(n)]_{11}p=0$
$\displaystyle\implies[T(m+n)-T(m)-T(n)]_{11}=0~{}(\text{By condition (a) of
Definition \ref{def1}}).$

By (3.21) and (3.24), we get the result. ∎

######  Lemma 3.6.

Let $a_{1},a_{2}\in\mathcal{T}_{11}$ and $b_{1},b_{2}\in\mathcal{T}_{22}$.
Then

(3.25) $\displaystyle(i)~{}T(a_{1}+a_{2})=T(a_{1})+T(a_{2})$
$\displaystyle(ii)~{}T(b_{1}+b_{2})=T(b_{1})+T(b_{2}).$

###### Proof.

Let $t_{2}\in\mathcal{T}_{22}$. Then

(3.26) $\displaystyle T((a_{1}+a_{2})t_{2})=T(a_{1}+a_{2})t_{2}.$

Now,

(3.27) $\displaystyle
T((a_{1}+a_{2})t_{2})=T(0)=T(a_{1}t_{2})+T(a_{2}t_{2})=T(a_{1})t_{2}+T(a_{2})t_{2}.$

Comparing (3.26) and (3.27),

(3.28) $\displaystyle[T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]t_{2}=0$
$\displaystyle\implies[T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]_{12}t_{2}=0$
$\displaystyle\&~{}[T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]_{22}t_{2}=0$
$\displaystyle\implies[T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]_{12}=0~{}(\text{By
condition (b) of Definition \ref{def1}})$
$\displaystyle\&~{}[T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]_{22}=0~{}(\text{By
condition (ii) of Theorem \ref{thm3.1}}).$

Let $m\in\mathcal{T}_{12}$. Then

(3.29) $T((a_{1}+a_{2})m)=T(a_{1}+a_{2})m.$

Also,

(3.30) $\displaystyle T((a_{1}+a_{2})m)$ $\displaystyle=T(a_{1}m+a_{2}m)$
$\displaystyle=T(a_{1}m)+T(a_{2}m)~{}(\text{By Lemma \ref{lem3.5}})$
$\displaystyle=T(a_{1})m+T(a_{2})m.$

Comparing (3.29) and (3.30),

(3.31) $\displaystyle T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]m=0$
$\displaystyle\implies[T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]_{11}m=0$
$\displaystyle\implies[T(a_{1}+a_{2})-T(a_{1})-T(a_{2})]_{11}=0~{}(\text{By
condition (a) of Definition \ref{def1}}).$

By (3.28) and (3.31), we have $(i)$ of Lemma 3.6. Similarly, we can prove
$(ii)$ of Lemma 3.6. ∎

######  Lemma 3.7.

Let $a\in\mathcal{T}_{11},b\in\mathcal{T}_{22}$ and $m\in\mathcal{T}_{12}$.
Then

(3.32) $T(a+m+b)=T(a)+T(m)+T(b).$

###### Proof.

Let $a_{1}\in\mathcal{T}_{11}$. Then

(3.33) $\displaystyle T((a+m+b)a_{1})=T(a+m+b)a_{1}.$

Also,

(3.34) $\displaystyle T((a+m+b)a_{1})$ $\displaystyle=T(aa_{1}+ma_{1}+ba_{1})$
$\displaystyle=T(aa_{1})$ $\displaystyle=T(aa_{1})+T(ma_{1})+T(ba_{1})$
$\displaystyle=T(a)a_{1}+T(m)a_{1}+T(b)a_{1}.$

By (3.33) and (3.34),

(3.35) $\displaystyle[T(a+m+b)-T(a)-T(m)-T(b)]a_{1}=0$
$\displaystyle\implies[T(a+m+b)-T(a)-T(m)-T(b)]_{11}a_{1}=0$
$\displaystyle\implies[T(a+m+b)-T(a)-T(m)-T(b)]_{11}=0$
$\displaystyle~{}(\text{By condition (iii) of Theorem \ref{thm3.1}}).$

Let $t_{2}\in\mathcal{T}_{22}$. Then

(3.36) $\displaystyle T((a+m+b)t_{2})=T(a+m+b)t_{2}$

Also,

(3.37) $\displaystyle T((a+m+b)t_{2})$ $\displaystyle=T(at_{2}+mt_{2}+bt_{2})$
$\displaystyle=T(mt_{2}+bt_{2})$
$\displaystyle=T(mt_{2})+T(bt_{2})\quad[\text{By (ii) of Lemma \ref{lem3.2}
}]$ $\displaystyle=T(at_{2})+T(mt_{2})+T(bt_{2})$
$\displaystyle=T(a)t_{2}+T(m)t_{2}+T(b)t_{2}.$

Comparing (3.36) and (3.37),

(3.38) $\displaystyle[T(a+m+b)-T(a)-T(m)-T(b)]t_{2}=0$
$\displaystyle\implies[T(a+m+b)-T(a)-T(m)-T(b)]_{12}t_{2}=0$
$\displaystyle\&~{}[T(a+m+b)-T(a)-T(m)-T(b)]_{22}t_{2}=0$
$\displaystyle\implies[T(a+m+b)-T(a)-T(m)-T(b)]_{12}=0~{}(\text{By condition
(b) of Definition \ref{def1} })$
$\displaystyle\&~{}[T(a+m+b)-T(a)-T(m)-T(b)]_{22}=0~{}(\text{By condition (ii)
of Theorem \ref{thm3.1}}).$

Thus, by (3.35) and (3.38), we have $T(a+m+b)=T(a)+T(m)+T(b)$. ∎

Now, we are ready to prove Theorem 3.1.

###### Proof of Theorem 3.1.

Let $t\in\mathcal{T}$ and $u\in\mathcal{T}$. Then $t=t_{11}+t_{12}+t_{22}$ and
$u=u_{11}+u_{12}+u_{22}$ where $t_{ij},u_{ij}\in\mathcal{T}_{ij}$ and
$i,~{}j\in\\{1,2\\}$. We have

(3.39) $\displaystyle T(t+u)$
$\displaystyle=T((t_{11}+t_{12}+t_{22})+(u_{11}+u_{12}+u_{22}))$
$\displaystyle=T((t_{11}+u_{11})+(t_{12}+u_{12})+(t_{22}+u_{22}))$
$\displaystyle=T(t_{11}+u_{11})+T(t_{12}+u_{12})+T(t_{22}+u_{22})~{}(\text{By
Lemma \ref{lem3.7}})$
$\displaystyle=T(t_{11})+T(u_{11})+T(t_{12})+T(u_{12})+T(t_{22})+T(u_{22})$
$\displaystyle~{}(\text{By Lemma \ref{lem3.5} and \ref{lem3.6}})$
$\displaystyle=T(t_{11})+T(t_{12})+T(t_{22})+T(u_{11})+T(u_{12})+T(u_{22})$
$\displaystyle=T(t_{11}+t_{12}+t_{22})+T(u_{11}+u_{12}+u_{22})~{}(\text{By
Lemma \ref{lem3.7}})$ $\displaystyle=T(t)+T(u).$

Hence, $T$ is additive and $T$ is a two-sided centralizer. ∎

Now, we provide an example which shows that every triangular ring
$\mathcal{T}$ is not always a prime ring.

######  Example 3.8.

Let $\mathcal{T}=\left(\begin{array}[]{ccc}2\mathbb{Z}&\mathbb{Z}\\\
&3\mathbb{Z}\end{array}\right)$. Then

$\displaystyle\left(\begin{array}[]{ccc}0&1\\\
&0\end{array}\right)\left(\begin{array}[]{ccc}2\mathbb{Z}&\mathbb{Z}\\\
&3\mathbb{Z}\end{array}\right)\left(\begin{array}[]{ccc}0&1\\\
&0\end{array}\right)=\left(\begin{array}[]{ccc}0&0\\\ &0\end{array}\right),$

but $\left(\begin{array}[]{ccc}0&1\\\
&0\end{array}\right)\neq\left(\begin{array}[]{ccc}0&0\\\
&0\end{array}\right)$. Therefore, $\mathcal{T}$ is not a semi-prime ring and
hence not a prime ring.

##  4\. $(m,n)$-DERIVATIONS ON TRIANGULAR RINGS

######  Theorem 4.1.

Let $m>0,n>0$ be integers and $\mathcal{T}$ be a triangular ring with
conditions: (i) $aA=0$ for some $a\in A$ implies $a=0$;

(ii) $bB=0$ for some $b\in B$ implies $b=0$;

(iii) $Aa=0$ for some $a\in A$ implies $a=0$;

(iv) $\mathcal{T}$ is $mn(m+n)$-torsion free.

If a mapping $D:\mathcal{T}\rightarrow\mathcal{T}$ satisfies

$\displaystyle(m+n)D(ab)=2mD(a)b+2naD(b)$

for all $a,b\in\mathcal{T}$, then $D$ is additive. Moreover, $D$ is a
$(m,n)$-derivation.

In this section, we frequently use all four conditions for the triangular ring
$\mathcal{T}$, the condition on the map $D$ described in Theorem 4.1 without
mentioning them. Note that $D(0)=0$. Before proving Theorem 4.1, we have some
lemmas.

######  Lemma 4.2.

Let $a\in\mathcal{T}_{11},p\in\mathcal{T}_{12}$ and $b\in\mathcal{T}_{22}$.
Then

(4.1) $\displaystyle(i)~{}D(a+p)=D(a)+D(p)$
$\displaystyle(ii)~{}D(b+p)=D(b)+D(p).$

###### Proof.

Let $c\in\mathcal{T}_{22}$. Then

(4.2) $\displaystyle(m+n)D((a+p)c)=2mD(a+p)c+2n(a+p)D(c).$

Also,

(4.3) $\displaystyle(m+n)D((a+p)c)$ $\displaystyle=(m+n)D(pc)$
$\displaystyle=(m+n)[D(ac)+D(pc)]$
$\displaystyle=2mD(a)c+2naD(c)+2mD(p)c+2npD(c).$

Comparing (4.2) and (4.3),

(4.4) $\displaystyle 2m[D(a+p)-D(a)-D(p)]c=0$
$\displaystyle\implies[D(a+p)-D(a)-D(p)]c=0~{}(\text{By condition (iv) of
Theorem \ref{der3}})$ $\displaystyle\implies[D(a+p)-D(a)-D(p)]_{12}c=0$
$\displaystyle\&~{}[D(a+p)-D(a)-D(p)]_{22}c=0$
$\displaystyle\implies[D(a+p)-D(a)-D(p)]_{12}=0~{}(\text{By condition (b) of
Definition \ref{def1}})$
$\displaystyle\&~{}[D(a+p)-D(a)-D(p)]_{22}=0~{}(\text{By condition (ii) of
Theorem \ref{der3}}).$

Let $a_{1}\in\mathcal{T}_{11}$. Then

(4.5) $\displaystyle(m+n)D((a+p)a_{1})=2mD(a+p)a_{1}+2n(a+p)D(a_{1}).$

Also,

(4.6) $\displaystyle(m+n)D((a+p)a_{1})$ $\displaystyle=(m+n)D(aa_{1})$
$\displaystyle=(m+n)[D(aa_{1})+D(pa_{1})]$
$\displaystyle=2mD(a)a_{1}+2naD(a_{1})+2mD(p)a_{1}+2npD(a_{1}).$

Comparing (4.5) and (4.6), we have

(4.7) $\displaystyle 2m[D(a+p)-D(a)-D(p)]a_{1}=0$
$\displaystyle\implies[D(a+p)-D(a)-D(p)]a_{1}=0~{}(\text{By condition (iv) of
Theorem \ref{der3}})$ $\displaystyle\implies[D(a+p)-D(a)-D(p)]_{11}a_{1}=0$
$\displaystyle\implies[D(a+p)-D(a)-D(p)]_{11}=0~{}(\text{By condition (i) of
Theorem \ref{der3}}).$

By (4.4) and (4.7), we have $(i)$ of Lemma 4.2. Similarly, we can prove $(ii)$
of Lemma 4.2. ∎

######  Lemma 4.3.

Let $a\in\mathcal{T}_{11}$, $b\in\mathcal{T}_{22}$ and
$p,q\in\mathcal{T}_{12}$. Then

(4.8) $D(ap+qb)=D(ap)+D(qb).$

###### Proof.

Since $(ap+qb)=(a+q)(p+b)+(p+b)(a+q)$,

(4.9) $\displaystyle(m+n)D(ap+qb)$ $\displaystyle=D((a+q)(p+b)+(p+b)(a+q))$
$\displaystyle=2mD(a+q)(p+b)+2n(a+q)D(p+b)$
$\displaystyle=2m(D(a)+D(q))(p+b)+2n(a+q)(D(p)+D(b))$
$\displaystyle~{}(\text{By Lemma \ref{lem4.2}})$
$\displaystyle=2mD(a)p+2naD(p)+2mD(q)b+2nqD(b)$
$\displaystyle+2mD(q)p+2nqD(p)+2mD(a)b+2naD(b)$
$\displaystyle=(m+n)[D(ap)+D(qb)+D(qp)+D(ab)]$
$\displaystyle=(m+n)[D(ap)+D(qb)].$

Using condition $(iv)$ of Theorem 4.1 and the above identity (4.9), we get the
desired result. ∎

######  Lemma 4.4.

Let $b\in\mathcal{T}_{22}$ and $p,q\in\mathcal{T}_{12}$. Then

(4.10) $D(pb+qb)=D(pb)+D(qb).$

###### Proof.

Let $a\in\mathcal{T}_{11}$. Then

(4.11) $\displaystyle(m+n)D[a((p+q)b)]=2mD(a)(p+q)b+2naD((p+q)b).$

Also,

(4.12) $\displaystyle(m+n)D[a((p+q)b)]$ $\displaystyle=(m+n)D[a(pb)+(aq)b]$
$\displaystyle=(m+n)[D(apb)+D(aqb)]~{}(\text{By Lemma \ref{lem4.3}})$
$\displaystyle=2mD(a)(pb)+2naD(pb)+2mD(a)(qb)+2naD(qb).$

Comparing (4.11) and (4.12), we have

(4.13) $\displaystyle 2na[D((p+q)b)-D(pb)-D(qb)]=0$ $\displaystyle\implies
a[D((p+q)b)-D(pb)-D(qb)]=0~{}(\text{By condition (iv) of Theorem \ref{der3}})$
$\displaystyle\implies a[D(pb+qb)-D(pb)-D(qb)]_{11}=0$
$\displaystyle\&~{}a[D((pb+qb)-D(pb)-D(qb)]_{12}=0$
$\displaystyle\implies[D(pb+qb)-D(pb)-D(qb)]_{11}=0~{}(\text{By condition
(iii) of Theorem \ref{der3}})$
$\displaystyle\&~{}[D((pb+qb)-D(pb)-D(qb)]_{12}=0~{}(\text{By condition (b) of
Definition \ref{def1}}).$

Let $s\in\mathcal{T}_{12}$. Then

(4.14) $\displaystyle(m+n)D[s(pb+qb)]=2mD(s)(pb+qb)+2nsD(pb+qb).$

Also,

(4.15) $\displaystyle(m+n)D[s(pb+qb)]$ $\displaystyle=(m+n)D(0)$
$\displaystyle=(m+n)D(spb)+(m+n)D(sqb)$
$\displaystyle=2mD(s)pb+2nsD(pb)+2mD(s)qb+2nsD(qb).$

Comparing (4.14) and (4.15),

(4.16) $\displaystyle 2ns[D(pb+qb)-D(pb)-D(qb)]=0$ $\displaystyle\implies
s[D(pb+qb)-D(pb)-D(qb)]=0~{}(\text{By condition (iv) of Theorem \ref{der3}})$
$\displaystyle\implies s[D(pb+qb)-D(pb)-D(qb)]_{22}=0$
$\displaystyle\implies[D(pb+qb)-D(pb)-D(qb)]_{22}=0~{}(\text{By condition (a)
of Definition \ref{def1}}).$

By (4.13) and (4.16), we have the desired result. ∎

######  Lemma 4.5.

Let $p,q\in\mathcal{T}_{12}$. Then

(4.17) $D(p+q)=D(p)+D(q).$

###### Proof.

Let $b\in\mathcal{T}_{22}$. Then

(4.18) $\displaystyle(m+n)D((p+q)b)=2mD(p+q)b+2n(p+q)D(b).$

Also,

(4.19) $\displaystyle(m+n)D((p+q)b)$
$\displaystyle=(m+n)(D(pb)+D(qb))~{}(\text{By Lemma \ref{lem4.4}})$
$\displaystyle=2mD(p)b+2npD(b)+2mD(q)b+2nqD(b).$

Comparing (4.18) and (4.19),

(4.20) $\displaystyle 2m[D(p+q)-D(p)-D(q)]b=0$
$\displaystyle\implies[D(p+q)-D(p)-D(q)]b=0~{}(\text{By condition (iv) of
Theorem \ref{der3}})$ $\displaystyle\implies[D(p+q)-D(p)-D(q)]_{12}b=0$
$\displaystyle\&~{}[D(p+q)-D(p)-D(q)]_{22}b=0$
$\displaystyle\implies[D(p+q)-D(p)-D(q)]_{12}=0~{}(\text{By condition (b) of
Definition \ref{def1}})$
$\displaystyle\&~{}[D(p+q)-D(p)-D(q)]_{22}=0~{}(\text{By condition (ii) of
Theorem \ref{der3}}).$

Let $s\in\mathcal{T}_{12}$. Then

(4.21) $\displaystyle(m+n)D((p+q)s)=2mD(p+q)s+2n(p+q)D(s).$

Also,

(4.22) $\displaystyle(m+n)D((p+q)s)$ $\displaystyle=(m+n)D(0)$
$\displaystyle=(m+n)(D(ps)+D(qs))$
$\displaystyle=2mD(p)s+2npD(s)+2mD(q)s+2nqD(s).$

By (4.21) and (4.22),

(4.23) $\displaystyle 2m[D(p+q)-D(p)-D(q)]s=0$
$\displaystyle\implies[D(p+q)-D(p)-D(q)]s=0~{}(\text{By condition (iv) of
Theorem \ref{der3}})$ $\displaystyle\implies[D(p+q)-D(p)-D(q)]_{11}s=0$
$\displaystyle\implies[D(p+q)-D(p)-D(q)]_{11}=0~{}(\text{By condition (a) of
Definition \ref{def1}}).$

By (4.20) and (4.23), we get the desired result. ∎

######  Lemma 4.6.

Let $a_{1},a_{2}\in\mathcal{T}_{11}$ and $b_{1},b_{2}\in\mathcal{T}_{22}$.
Then

(4.24) $\displaystyle(i)~{}D(a_{1}+a_{2})=D(a_{1})+D(a_{2})$
$\displaystyle(ii)~{}D(b_{1}+b_{2})=D(b_{1})+D(b_{2}).$

###### Proof.

Let $t_{2}\in\mathcal{T}_{22}$. Then

(4.25)
$\displaystyle(m+n)D((a_{1}+a_{2})t_{2})=2mD(a_{1}+a_{2})t_{2}+2n(a_{1}+a_{2})D(t_{2}).$

Also,

(4.26) $\displaystyle(m+n)D((a_{1}+a_{2})t_{2})$ $\displaystyle=(m+n)D(0)$
$\displaystyle=(m+n)(D(a_{1}t_{2})+D(a_{2}t_{2}))$
$\displaystyle=2mD(a_{1})t_{2}+2na_{1}D(t_{2})+2mD(a_{2})t_{2}+2na_{2}D(t_{2}).$

Comparing (4.25) and (4.26),

(4.27) $\displaystyle 2m[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]t_{2}=0$
$\displaystyle\implies[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]t_{2}=0~{}(\text{By
condition (iv) of Theorem \ref{der3}})$
$\displaystyle\implies[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]_{12}t_{2}=0$
$\displaystyle\&~{}[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]_{22}t_{2}=0$
$\displaystyle\implies[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]_{12}=0~{}(\text{By
condition (b) of Definition \ref{def1}})$
$\displaystyle\&~{}[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]_{22}=0~{}(\text{By
condition (ii) of Theorem \ref{der3}}).$

Let $p\in\mathcal{T}_{12}$. Then

(4.28)
$\displaystyle(m+n)D((a_{1}+a_{2})p)=2mD(a_{1}+a_{2})p+2n(a_{1}+a_{2})D(p).$

Also,

(4.29) $\displaystyle(m+n)D((a_{1}+a_{2})p)$
$\displaystyle=(m+n)D(a_{1}p+a_{2}p)$
$\displaystyle=(m+n)[D(a_{1}p)+D(a_{2}p)]~{}(\text{By Lemma \ref{lem4.5}})$
$\displaystyle=2mD(a_{1})p+2na_{1}D(p)+2mD(a_{2})p+2na_{2}D(p).$

Using (4.28) and (4.29),

(4.30) $\displaystyle 2m[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]p=0$
$\displaystyle\implies[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]p=0~{}(\text{By
condition (iv) of Theorem \ref{der3}})$
$\displaystyle\implies[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]_{11}p=0$
$\displaystyle\implies[D(a_{1}+a_{2})-D(a_{1})-D(a_{2})]_{11}=0~{}(\text{By
condition (a) of Definition \ref{def1}})$

By (4.27) and (4.30), we have $(i)$ of Lemma 4.6. Similarly, we can prove
$(ii)$ of Lemma 4.6. ∎

######  Lemma 4.7.

Let $a\in\mathcal{T}_{11},p\in\mathcal{T}_{12}$ and $b\in\mathcal{T}_{22}$.
Then

(4.31) $D(a+p+b)=D(a)+D(p)+D(b).$

###### Proof.

Let $a_{1}\in\mathcal{T}_{11}$. Then

(4.32) $\displaystyle(m+n)D((a+p+b)a_{1})=2mD(a+p+b)a_{1}+2n(a+p+b)D(a_{1}).$

Also,

(4.33) $\displaystyle(m+n)D((a+p+b)a_{1})$ $\displaystyle=(m+n)D(aa_{1})$
$\displaystyle=(m+n)[D(aa_{1})+D(pa_{1})+D(ba_{1})]$
$\displaystyle=2mD(a)a_{1}+2naD(a_{1})+2mD(p)a_{1}$
$\displaystyle+2npD(a_{1})+2mD(b)a_{1}+2nbD(a_{1})$

Comparing (4.32) and (4.33), we have

(4.34) $\displaystyle 2m[D(a+p+b)-D(a)-D(p)-D(b)]a_{1}=0$
$\displaystyle\implies[D(a+p+b)-D(a)-D(p)-D(b)]a_{1}=0$
$\displaystyle~{}(\text{By condition (iv) of Theorem \ref{der3}})$
$\displaystyle\implies[D(a+p+b)-D(a)-D(p)-D(b)]_{11}a_{1}=0$
$\displaystyle\implies[D(a+p+b)-D(a)-D(p)-D(b)]_{11}=0$
$\displaystyle~{}(\text{By condition (i) of Theorem \ref{der3}}).$

Let $t_{2}\in\mathcal{T}_{22}$. Then

(4.35) $\displaystyle(m+n)D((a+p+b)t_{2})=2mD(a+p+b)t_{2}+2n(a+p+b)D(t_{2}).$

Also,

(4.36) $\displaystyle(m+n)D((a+p+b)t_{2})$
$\displaystyle=(m+n)D(pt_{2}+bt_{2})$
$\displaystyle=(m+n)[D(pt_{2})+D(bt_{2})]~{}(\text{By Lemma \ref{lem4.2}})$
$\displaystyle=(m+n)[D(at_{2})+D(pt_{2})+D(bt_{2})]$
$\displaystyle=2mD(a)t_{2}+2naD(t_{2})+2mD(p)t_{2}$
$\displaystyle+2npD(t_{2})+2mD(b)t_{2}+2nbD(t_{2}).$

Comparing (4.35) and (4.36), we have

(4.37) $\displaystyle 2m[D(a+p+b)-D(a)-D(p)-D(b)]t_{2}=0$
$\displaystyle\implies[D(a+p+b)-D(a)-D(p)-D(b)]t_{2}=0$
$\displaystyle~{}(\text{By condition (iv) of Theorem \ref{der3}})$
$\displaystyle\implies[D(a+p+b)-D(a)-D(p)-D(b)]_{12}t_{2}=0$
$\displaystyle\&~{}[D(a+p+b)-D(a)-D(p)-D(b)]_{22}t_{2}=0$
$\displaystyle\implies[D(a+p+b)-D(a)-D(p)-D(b)]_{12}=0$
$\displaystyle~{}(\text{By condition (b) of Definition \ref{def1}})$
$\displaystyle\&~{}[D(a+p+b)-D(a)-D(p)-D(b)]_{22}=0$
$\displaystyle~{}(\text{By condition (ii) of Theorem \ref{der3}}).$

By (4.34) and (4.37), we get the desired result. ∎

Now, we are ready to prove Theorem 4.1.

###### Proof of Theorem 4.1.

Let $a,b\in\mathcal{T}$. Then $a=a_{11}+a_{12}+a_{22}$ and
$b=b_{11}+b_{12}+b_{22}$ where $a_{ij}$, $b_{ij}\in\mathcal{T}_{ij}$.

(4.38) $\displaystyle D(a+b)$
$\displaystyle=D(a_{11}+a_{12}+a_{22}+b_{11}+b_{12}+b_{22})$
$\displaystyle=D(a_{11}+b_{11}+a_{12}+b_{12}+a_{22}+b_{22})$
$\displaystyle=D(a_{11}+b_{11})+D(a_{12}+b_{12})+D(a_{22}+b_{22})~{}(\text{By
Lemma \ref{lem4.7}})$
$\displaystyle=D(a_{11})+D(b_{11})+D(a_{12})+D(b_{12})+D(a_{22})+D(b_{22})$
$\displaystyle~{}(\text{By Lemma \ref{lem4.5} and \ref{lem4.6}})$
$\displaystyle=D(a_{11})+D(a_{12})+D(a_{22})+D(b_{11})+D(b_{12})+D(b_{22})$
$\displaystyle=D(a_{11}+a_{12}+a_{22})+D(b_{11}+b_{12}+b_{22})~{}(\text{By
Lemma \ref{lem4.7}})$ $\displaystyle=D(a)+D(b).$

Hence, $D$ is additive. Thus, $D$ is a $(m,n)$-derivation. ∎

## Acknowledgement

The first author is thankful to the University Grants Commission (UGC), Govt.
of India for financial supports under Sr. No-2121540952, Ref. No.
20/12/2015(ii)EU-V dated 31/08/2016 and all authors are thanful to the Indian
Institute of Technology Patna for providing the research facilities.

## References

  * [1] S. Ali and A. Fošner, On generalized (m, n)-derivations and generalized (m, n)-Jordan derivations in rings, Algebra Colloq. 21(03) (2014), 411–420.
  * [2] M. N. Daif, When is a multiplicative derivation additive?,  Int. J. Math. Math. Sci. 14(3) (1991), 615–618.
  * [3] B. L. M. Ferreira, Multiplicative maps on triangular n-matrix rings, Int. J. Math. Game Theory Algebr. 23(2) (2014), 1–14.
  * [4] B. L. M. Ferreira, Jordan derivations on triangular matrix rings, Extracta Math. 30(2) (2015), 181–190.
  * [5] I. N. Herstein, Jordan derivations of prime rings, Proc. Amer. Math. Soc. 8(6) (1957), 1104–1110.
  * [6] W. Jing and S. Lu, Generalized Jordan derivations on prime rings and standard operator algebras,  Taiwan. J. Math. 7(4) (2003), 605–613.
  * [7] R. E. Johnson, Rings with unique addition, Proc. Amer. Math. Soc. 9(1) (1958), 57–61.
  * [8] W. S. Martindale III, When are multiplicative mappings additive?, Proc. Amer. Math. Soc. 21(3) (1969), 695–698.
  * [9] C. E. Rickart, One-to-one mappings of rings and lattices, Bull. Amer. Math. Soc. 54(8) (1948), 758–764.
  * [10] M. S. T. E.-Sayiad, M. N. Daif and V. D. Filippis, Multiplicativity of left centralizers forcing additivity, Bol. Soc. Parana. Mat. 32(1) (2014), 61–69.
  * [11] Y. Wang, The additivity of multiplicative maps on rings, Comm. Algebra 37(7) (2009), 2351–2356.
  * [12] Y. Wang, Additivity of multiplicative maps on triangular rings, Linear Algebra Appl. 434(3) (2011), 625–635.
  * [13] J. G. Wendel, Left centralizers and isomorphisms of group algebras, Pacific J. Math. 2(2) (1952), 251–261.