Lost on introductory abstract algebra proof.

Question

I'm fairly lost on where to begin this proof and would like direction to resources that might help me understand what I'm meant to show.

The proof is:

Prove that If $m|n$, then $\pi_{m,n}: \mathbb{Z}_n \to \mathbb{Z}_m$ is well-defined.

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I know that $m|n$ means that $n=m*k$ where $k \in \mathbb{Z}$.

I know that if $m|n$, then $\pi_{m,n} ([a]_n)=[a]_m$

And I know that for a function $f:X \to Y$ to be well-defined I must prove that:

1.) f is a relation from X to Y f⊆X×Y

2.) The domain of f is X, every element in X is related to some element of Y ∀x∈X,∃y∈Y:(x,y)∈f

3.) No element of X is related to more than one element of Y ∀x∈X,∀y1,y2∈Y:(x,y1),(x,y2)∈f⟹y1=y2

I'm not sure how these facts/properties relate to the requested proof and any help that can be offered would be greatly appreciated. I think the Chinese Remainder Theorem may be involved, but that may just be a stab in the dark on my part.

Answer 1

Write $a+n\mathbb{Z}$ for the element of $\mathbb{Z}_n$ corresponding to the equivalence class of integers modulo $n$ which contains $a$. The function $\pi_{n,m}$ in question is defined by $$\pi_{n,m}(a+n\mathbb{Z})=a+m\mathbb{Z}\qquad \mbox{ for all } a\in\mathbb{Z}.$$ Recall that a function associates each member of the domain to precisely one member of the codomain. Therefore, we say that $\pi_{n,m}$ is well-defined if $\pi_{n,m}(a+n\mathbb{Z})=\pi_{n,m}(b+n\mathbb{Z})$ whenever $a,b\in\mathbb{Z}$ satisfy $a+n\mathbb{Z}=b+n\mathbb{Z}$.

In other words, the question is asking you to show that if $m\mid n$ and $a\equiv b\mod n$, then $a\equiv b\mod m$.

Answer 2

What you have stated is definition of a function. To check if $f$ is well-defined you need to ensure that regardless of the representative used for the input, the output will remain the same. Said differently, if $x_1,x_2$ are two representatives of the same object in the domain (so $x_1=x_2$), then for well-definedness you need to ensure that $f(x_1)=f(x_2)$.

To give you an example, let $f:\mathbb{Z}_8 \rightarrow \mathbb{Z}_{12}$ be given by $f([x]_8)=[x]_{12}$. Is this well-defined? Observe that $[0]_8=[8]_8$ but $f([0]_8)=[0]_{12}$ and $f([8]_8)=[8]_{12}$ are not the same outputs. Hence the function is NOT well-defined.

In your question with $\pi_{m,n}([x]_{n})=[x]_{m}$. Let $[x]_n=[y]_n$, then $ x\equiv y \pmod{n}$. Now we are given that $m | n$, so $x \equiv y \pmod{m}$ (by transitivity--$m$ divides $n$ and $n$ divides $x-y$, so $m$ divides $x-y$). But this also implies,that $[x]_m=[y]_m$, which means $$\pi_{m,n}([x]_{n})=\pi_{m,n}[y]_{n}.$$ Hence well-defined.