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I was trying some examples on ring homomorphisms. Got stuck in an intermediate step in which I have to show that:

A map $f$: $\Bbb{Z}_{10}$ $\rightarrow$ $\Bbb{Z}_{20}$ defined as $f(x)$ = $\bar {16}x$ is a ring homomorphism.

Please help!

Gitika
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  • One thing that you have to be careful about is making sure that $f$ is well defined. First, define $g: \mathbb{Z} \to \mathbb{Z}_{20}$ by $g(x) = 16x$. Check that $g(a+b) = g(a) + g(b)$ and $g(ab) = g(a)g(b)$. This will show that $g$ is a ring homomorphism. Next, check to make sure that $10\mathbb{Z} \subset \ker g$. This amounts to showing that $g(10x) = 0$. By the first isomorphism theorem, $g$ factors through $\mathbb{Z}_10$. That is, the map $f$ is indeed a ring homomorphism. – Mike Jul 30 '20 at 20:08
  • What do you mean by g factors through $\Bbb{Z}_{10}$..how to use the first isomorphism theorem..Since you didn't find the kernel..you have just shown that 10Z is contained in kernel of g. – Gitika Jul 30 '20 at 20:35
  • See https://math.stackexchange.com/questions/21932/what-does-it-mean-to-say-a-map-factors-through-a-set. – Mike Jul 30 '20 at 22:45
  • what is $16^-$? – Charlie Chang Jul 30 '20 at 23:59
  • @CharlieChang I think $\bar {16}$ refers to the equivalence class represented by $16$ – Noel Lundström Jul 31 '20 at 01:24
  • Hey how can we verify that g is a ring homomorphism? – Gitika Jan 20 '21 at 13:20
  • @ Mike..Also by using that theorem, we get that g = fk where k will be canonical projection..How to find f from here? – Gitika Jan 20 '21 at 13:21

1 Answers1

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Hint:

Recall ring homs $f : \mathbb{Z}/(n) \to R$ are in bijective correspondence with ring homs $\tilde{f} : \mathbb{Z} \to R$ with the bonus property that $\tilde{f}(n) = 0$. This is sometimes called a Universal Property, and if your book didn't prove it, it's a good exercise to do yourself.

In light of this fact, to show that $f$ is a ring hom, it suffices to show that the map $\tilde{f} : \mathbb{Z} \to \mathbb{Z}/(20)$ which is defined by the same formula ($\tilde{f}(x) = \overline{16}\overline{x}$) is a ring hom with $\tilde{f}(10) = 0$.

Is this problem more approachable? If not, notice $\tilde{f}$ is really a composite of two functions: first we have $\pi : \mathbb{Z} \to \mathbb{Z}/(20)$ the natural quotient map (i.e. $\pi(x) = \overline{x}$). Then we have $m : \mathbb{Z}/(20) \to \mathbb{Z}/(20)$ with $m(\overline{x}) = \overline{16} \overline{x}$.

Can you show that $\pi$ and $m$ are both ring homs? Then, since the composition of ring homs is a ring hom, you will have that $\tilde{f} = m \circ \pi$ is a ring hom too, as needed.

I hope this helps ^_^

HallaSurvivor
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