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On $\Bbb Z_3$, we typically define addition and multiplication as follows:

$$[a]+[b]=[a+b]$$ $$ [a]\cdot[b]=[a\cdot b]$$

Consider defining addition as $[a]+[b]=[0]$ for all $a,b\in \Bbb Z$. This addition is well defined:

Let $a\equiv c \pmod 3$ and $b\equiv d \pmod 3$ for $a,b,c,d \in \Bbb Z$. Then $[a]=[c]$ and $[b]=[d]$, which implies: $$[a]+[b]=[c]+[d]=[0]$$ However, $f([a],[b])=[a]+[b]$ is not an onto mapping from $\Bbb Z_3 \times \Bbb Z_3$ to $\Bbb Z_3$.

EDIT (clarification of question): Is the above definition for addition valid?

That is, is it possible to define addition on $\Bbb Z_3$ in such a way that the function $f:\Bbb Z_3 \times \Bbb Z_3 \rightarrow \Bbb Z_3$ such that $f([a],[b])=[a]+[b]$ is not onto? And likewise for multiplication?

Thank you for your help. I appreciate an explanation suitable for an undergraduate like myself.

monkey king
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  • You ask "Is it possible to define addition on $\Bbb Z_3$ in such a way that the function $f:\Bbb Z_3 \times \Bbb Z_3 \rightarrow \Bbb Z_3$ such that $f([a],[b])=[a]+[b]$ is not onto?" It seems that the addition operation you have defined yourself isn't onto, no? The equivalence class $[1]$ is not in the range of $f$. Does your question have a typo in it? – Joe Sep 03 '23 at 13:48
  • I meant to ask, is defining addition as a non-onto operation valid? – monkey king Sep 03 '23 at 13:52
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    Yes, it is valid, in the sense that the definition makes sense. However, it is not the way we usually define addition in $\mathbb Z_3$. Usually, in $\mathbb Z_3$ we define addition and multiplication so that it is the quotient ring of $\mathbb Z$ modulo $3\mathbb Z$. Are you familiar with quotient rings? – Joe Sep 03 '23 at 13:56
  • No, I would appreciate a simple explanation please! – monkey king Sep 03 '23 at 13:59
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    "Addition" operations are usually supposed to have an additive identity - let's call it $z$. In particular, we should have $[1]+z=[1]$. What is your $z$? – JonathanZ Sep 03 '23 at 14:00
  • Unfortunately, many students find the quotient construction rather difficult to understand. If you want a full explanation, I would suggest turning to your lecture notes, or to the algebra book you are studying from. In the specific case of $\mathbb Z_n$ (also denoted as $\mathbb Z/n\mathbb Z$), we typically define $[a]+[b]$ as $[a+b]$, and $[a]\cdot[b]=[a\cdot b]$ (there is a little work involved in making sure that these definitions make sense). – Joe Sep 03 '23 at 14:06
  • Note that in the expression $[a+b]$, the symbol $+$ refers to addition of integers, whereas in the expression $[a]+[b]$, the symbol $+$ is a new operation $\mathbb Z_n\times \mathbb Z_n\to\mathbb Z_n$. Similar remarks apply to $\cdot.$ – Joe Sep 03 '23 at 14:07
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    Addendum: As JonathanZ remarks, there is a "cultural" expectation that if you wish to denote a binary operation by $+$ and call it "addition", then that operation ought to give you an abelian group. My addition operation does indeed fulfil this expectation (check this!), whereas the operation you defined in the original post does not. In fact, $\mathbb Z_n$, together with $+$ and $\cdot$ as I have defined them, is a (commutative) ring. – Joe Sep 03 '23 at 14:10
  • If you want a quick introduction to what groups, rings, and fields are, then you might be interested in the following answer I wrote. – Joe Sep 03 '23 at 14:27

1 Answers1

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"Addition" operations are usually supposed to have an additive identity - let's call it z. In particular, we should have [1]+z=[1]. What is your $z$?

As to your more general question, the very standard expectations of an additive identity and additive inverses always imply that addition is onto.

What you're doing is abusing nomenclature. You can call your shoes "a car", and while they are a well defined method of transportation and your 4-year old might enjoy it, you can't sit in them and they don't have a steering wheel or gas or brake pedal. You shouldn't really be calling them "a car".

(And the situation for multiplication is similar - assuming you want to define a field and not just a ring. If you set the zero element aside, the expectation is that a multiplicative identity exists, and multiplicative inverses, so multiplication is also onto.

It's hard to tell what level you're at only knowing that you're an undergrad, but these questions are answerable in ones first abstract algebra course, and follow pretty quickly from the group and field axioms.)

JonathanZ
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  • Nitpicking, it is standard to call the second binary operation in a ring "multiplication". So I would not say that there is an expectation that multiplicative inverses exist for non-zero elements. – Joe Sep 03 '23 at 14:18
  • I think the edit removes this problem, but you might want to add that in a field, multiplication distributes over addition. – Joe Sep 03 '23 at 14:24
  • @Joe - Great point. I was a little concerned about what level of detail to go in to, so I silently assumed the OP wanted a field. But I've updated the question to reflect your comment. And while I don't think I'll add it to the main answer, I believe $a\cdot b=0$ for all $a,b$ does define a valid ring. – JonathanZ Sep 03 '23 at 14:24
  • @Joe - I think I'll leave distributivity out of the main answer. It's a requirement to have a fully valid field, but isn't necessary for discussong the onto-ness of the multiplication map. You can edit it in to the question if you feel strongly that it is necessary, and I won't roll back, but like I said, I think fewer details are better. – JonathanZ Sep 03 '23 at 14:31
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    That's okay. It is always difficult to know how much detail to include, and what level of simplification is appropriate, when teaching newcomers to a subject. – Joe Sep 03 '23 at 14:32