Prove $S^{-1}\mathbb{Z_6}\simeq \mathbb{Z}_3$ when $S=\{\overline{1},\overline{2},\overline{4}\}$.

Note: $S^{-1}\mathbb{Z_6}= \frac{S\times \mathbb{Z}_6}{\sim }$ where $(x,y)\sim (u,v) \iff \exists t\in S : t(xv-uy)=0.$

Proving the existence of a homomorphism from $\phi:S^{-1}\mathbb{Z}_6\to \mathbb{Z_3}$ is possible since there are homomorphisms $h:\mathbb{Z}_6\to \mathbb{Z_3}$ and the existence of $\phi$ follows from the universal property for the ring of fractions.

In order to prove the existence of an isomorphism, I would look to the possible homomorphisms and see which of those defines a biyeccion. The possible homomorphisms are given once $h(1)$ is defined, here for $h_1(1)=0, h_2(1)=1, h_3(1)=2$ I have

$$h_1([x])=0\\h_2([x])=[x]\mod 3\\ h_3([x])=[2x] \mod 3$$

It seems that $h_2$ may help to define the isomorphism.

I have to find $\psi:S^{-1}\mathbb{Z}_6 \to \mathbb{Z_6}$ in order to make $\psi\circ h_2$ is an isomorphism. How can I define $\psi$?, it would be correct define $\psi (\frac{x}{a})=x$ where $a:\overline{1},\overline{2},\overline{4}$?

I'm not sure how to fill the details to prove $\psi$ would define an isomorphism.

  • 1
    $\begingroup$ One more thing: the existence of a homomorphism $h:\mathbb{Z}_6\to \mathbb{Z_3}$ is not enough to ensure us that one can use the universal property for the ring of fractions. There is something more we need here. $\endgroup$ – user26857 Nov 30 '14 at 10:05

If you know how to define a homomorphism $\phi:S^{-1}\mathbb{Z}_6\to \mathbb{Z_3}$, why don't check that this is bijective?

In fact, $\phi(\bar x/\bar a)=[x][a]^{-1}$. This is clearly surjective. From $\phi(\bar x/\bar a)=[0]$ you get $[x]=[0]$, that is, $3\mid x$. This means that $\bar x=\bar 0$ or $\bar 3$ (in $\mathbb Z_6$). In the first case it's clear that $\bar x/\bar a=\bar 0/\bar 1$; in the second case you also have $\bar x/\bar a=\bar 0/\bar 1$ (since $\bar 2\cdot\bar 3=\bar 0$ and $\bar 2\in S$), so $\phi$ is injective.

Edit. Your try has no chance at least for the following reason:

There is no (unitary) ring homomorphism from $S^{-1}\mathbb Z_6$ to $\mathbb Z_6$.

Let $\psi:S^{-1}\mathbb Z_6\to \mathbb Z_6$ be a (unitary) ring homomorphism. Then $\psi(\bar 1/\bar 1)=\bar 1$. Say $\psi(\bar 1/\bar 2)=\bar a$. Then $\bar 1=\psi(\bar 1/\bar 1)=\psi(\bar 1/\bar 2+\bar 1/\bar 2)=2\bar a$. But there is no $\bar a\in\mathbb Z_6$ such that $2\bar a=\bar 1$.


There is no need to fiddle around with elements or even equivalence classes. Use the universal property.

$\mathbb{Z}/6$ is the universal ring in which $6=0$ holds. If we localize the element $2$, we get the universal ring in which $6=0$ holds and $2$ is invertible. But then $0=6 = 2 \cdot 3$ simplifies to $0=3$, i.e. $2=-1$ and $2$ is invertible anyway. So we get the universal ring in which $3=0$ holds, which is $\mathbb{Z}/3$.

Here is how to complete your approach (notice that these are essentially the same arguments as above): The natural ring hom. $\mathbb{Z}/6 \to \mathbb{Z}/3$ sends $2$ to a unit, hence extends to a hom. $S^{-1} \mathbb{Z}/6 \to \mathbb{Z}/3$. Conversely, there is a unique hom. $\mathbb{Z} \to S^{-1} \mathbb{Z}/6$ and this maps $3$ to zero, because $2 \cdot 3=0$ in $\mathbb{Z}/6$ and hence $3=0$ in $S^{-1} \mathbb{Z}/6$. Therefore, we get a hom. $\mathbb{Z}/3 \to S^{-1} \mathbb{Z}/6$. The two constructed hom. are inverese to each other (because of uniqueness in the used universal properties).

  • $\begingroup$ Are you saying that my proof is incomplete? Of course "universal ring with ..." means "initial object in the category of rings satisfying ...". Basically what I say is that the category of rings satisfying 6=0 and 2 inv. equals the category of rings satisfying 3=0. Therefore, the initial objects agree. $\endgroup$ – Martin Brandenburg Nov 30 '14 at 0:00
  • $\begingroup$ Cure's approach was to use $\phi$ and show that it is an isomorphism. $\endgroup$ – Martin Brandenburg Nov 30 '14 at 10:34
  • $\begingroup$ Are you sure that you read (the last part of) the question? $\endgroup$ – user26857 Nov 30 '14 at 10:44

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