# By applying the Fundamental Theorem of Homomorphisms, show that there is a ring isomorphism $g: \mathbb{Z}_{4} \to \operatorname{im}(f)$

Theorem: If $R$ and $S$ are rings and $\phi: R \to S$ a ring homomorphism defined by $g(n+\ker(\phi)) = f(n)$, then $R/Ker(\phi) \cong \operatorname{im}(\phi).\\$ Consider the rings $\mathbb{Z}$, $\mathbb{Z}_{4} = \{\bar{0},\bar{1},\bar{2},\bar{3}\}$ and $\mathbb{Z}_{12} = \{[0],[1],[2],\ldots,[11]\}$. Define $\phi: \mathbb{Z} \to \mathbb{Z}_{12}$ by $\phi(x) = 9x$.

By applying the theorem, show that there is a ring isomorphism $g: \mathbb{Z}_{4} \cong \operatorname{im}(\phi))$

My attempt: This is what I did:

Since, the $\ker(\phi) = \{4k| k \in \mathbb{Z}\}$ and the $\operatorname{im}(\phi) = \{[0],[3],[6],[9]\}$,

$g(n+4\mathbb{Z}) = \phi(n)= 9n$

Thus, $\mathbb{Z}_{4} \cong \{[0],[3],[6],[9]\} \in \mathbb{Z}_{12}$ Is this correct? Otherwise, any hints would be much appreciated.

• It's an immediate consequence of the theorem... – Qudit Mar 9 '15 at 18:37
• If it doesn't make sense, I would suggest reviewing your understanding of the concepts involved or asking more specific questions. It's an immediate consequence as I said. – Qudit Mar 9 '15 at 18:43

I'm sure there are plenty of places to find out about this, but anyway. It should be clear that we may restrict to $\operatorname{im}(f)$ as the codomain, which makes the map surjective.
Now, the fundamental theorem says that as far as $f$ is concerned, we might as well identify the elements in the kernel with $0$, since they are mapped to $0$ anyway. Indeed, the induced map $\mathbb Z/\ker(f) \to \operatorname{im}(f)$ is well-defined, and moreover becomes injective, since everything that's mapped to $0$ is identified with $0$, by definition.
The fact that it is still a ring homomorphism is clear (we haven't really done anything), at least supposing you know that $\mathbb Z/\ker(f)$ is in fact a ring.