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Prove that $f:\Bbb{Z}/12\Bbb{Z}\to\Bbb{Z}/4\Bbb{Z}:[x]_{12}\mapsto[x]_4$ is a homomorphism.

I think you can prove this very simply using the following reasoning: \begin{align*} [x]_{12},[y]_{12}\in\Bbb{Z}/12\Bbb{Z}\implies f([x]_{12}+[y]_{12}) &= f([x+y]_{12})\\ &=[x+y]_4 \\ &=[x]_4+[y]_4 \\ &=f([x]_{12})+f([x]_{12}) \end{align*}

But I was thinking that I may need to prove that $f$ is a well-defined function, otherwise this (homework) exercise seems a little bit too easy. Should I prove this ? And any hints how to do this ?

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Should I prove this? Not, for example, if your sister is having a baby or the house is burning down... in the absence of other catastrophes I see no reason not to! :) –  rschwieb Feb 22 '13 at 21:14
@rschwieb Okay rephrasing: "Do I get full-points for my homework even if I don't prove this?" ;) –  Kasper Feb 22 '13 at 21:21

4 Answers 4

up vote 2 down vote accepted

The idea is that if $x$ and $y$ are such that $[x]_{12}=[y]_{12}$, then $f([x]_{12})=f([y]_{12})$ because $4$ divides $12$.

Indeed, there exist $k \in \mathbb{Z}$ such that $y=x+12k$, so $f([y]_{12})=[y]_4=[x+12k]_4=[x]_4=f([x]_{12})$.

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I don't see directly how this proves that $f$ is well defined. –  Kasper Feb 22 '13 at 21:09
@Kasper: I added an explanation. –  Seirios Feb 22 '13 at 21:11
So in general if I want to prove it is well defined, I only need to check that if $x=y\implies f(x)=f(y)$ ? I thought my teacher once told me I need to check for something different also, but maybe I remember wrong. –  Kasper Feb 22 '13 at 21:19
@Kasper: By definition, a point has exactly one image by a function, and it is exactly what you check. –  Seirios Feb 22 '13 at 21:21
okay thanks for your explanation ! –  Kasper Feb 22 '13 at 21:22

It might appear a bit finnicky, but until you have shown that the function is well defined, you cannot use the function symbol, because it might be ambiguous.

So you have to prove that if $[x]_{12} = [y]_{12}$ for two integers $x, y$, then $[x]_{4} = [y]_{4}$. The first equality means $12 \mid x -y$, the second one $4 \mid x -y$. Since $4 \mid 12$, and divisibility is transitive, you get the first equality implies the second one.

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Suppose $x\equiv y \mod 12$. Then $x-y=12k=4(3k)=4m$ so $x\equiv y\mod 4$. It is well defined.

In general, $a\equiv b \mod mn\implies a\equiv b\mod m,\mod n$, but not conversely.

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The key point is that $4$ divides $12$ or, equivalently $12\mathbb{Z}\subseteq 4\mathbb{Z}$.

Also, note that this is a ring homomorphism.

I hear already some people tell me that you probably have not seen this yet, but I will still propose you an alternative that could be interesting, now or later.

This has to do with the factorization of homomorphisms through a quotient. The condition is that the kernel contains the ideal, subgroup, subspace, etc... by which you want to mod out the domain of the initial homomorphism. This a general and useful principle.

First note that the canonical surjection $$ x\longmapsto [x]_{4} $$ is a (well-defined, of course) homomorphism from $\mathbb{Z}$ onto $\mathbb{Z}/4\mathbb{Z}$.

Now observe that the kernel of the latter is $4\mathbb{Z}$ and contains $12\mathbb{Z}$.

It follows that the above homomorphism factors through $\mathbb{Z}/12\mathbb{Z}$ into a homomorphism $$ [x]_{12}\longmapsto [x]_{4}. $$

See theorem 2.1 (for group homomorphisms, but the same is true for ring homorphisms through quotients modulo ideals like here) here: http://www.math.rochester.edu/people/faculty/jonpak/N4.pdf

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Indeed, that is the universal property that characterizes quotient rings. –  Math Gems Feb 22 '13 at 23:22
@MathGems I should have mentioned that. Thanks for the link. –  1015 Feb 23 '13 at 3:22

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