# How to prove that this ring homomorphism is isomorphic?

Here is the question:

Let F be a field, and R a commutative ring with identity. If φ:F[x]→R is a surjective ring homomorphism such that ker φ = (m(x)) for some m(x) ∈ F [x], prove that φ induces an isomorphism F[x]/(m(x)) →∼ R.

So based on my understanding, to prove isomorphism I must show the transformation is injective and surjective. We already know that it is surjective so I just need to show that it is also one-to-one.

Since the kernal is m(x), we know the kernal is non-trivial. So I'm guessing for that reason φ cannot be injective, but somehow F[x]/(m(x)) →∼ R is injective but I am stuck with that.

I am thinking that I need to show that the kernal of F[x]/m(x)) is 0. But the question says it is m(x). But m(x) cannot be 0 since it is the denominator.

• Remember the notation $F[x]/(m(x))$ is for the space of polynomials $F[x]$ modulo the ideal $(m(x))$. This not a fraction in the usual sense (which would make no sense here), but a space in which $m(x)$ and all its multiples are zero. – Joel Cohen Nov 11 '15 at 23:57
• I never understand why people ask students such conflated questions. The real question is: Let $\phi : A \to B$ be a surjective ring homomorphism show that $\phi$ induces an isomorphism $A/\text{Ker}(\phi)\to B$. Or an even better question: Let $\phi : A \to B$ be a surjective ring homomorphism. Show that for any ring homorphism $\psi : A\to C$ such that $\psi(\text{Ker}(\phi))=\{0\}$ there exists a (unique) homorphism $\bar \psi : B \to C$ such that $\bar\psi \circ \phi = \psi$. – Nex Nov 25 '15 at 8:33

## 1 Answer

You're conflating two separate things here:

• There is the map $\varphi$, from $F[x]$ to $R$. $\varphi$ is surjective but not injective. Certainly, $\varphi$ is not an isomorphism.

• Meanwhile, we can consider the ring $F[x]/(m(x))$, where $(m(x))$ is the kernel of $\varphi$. The question says that $\varphi$ induces a map from $F[x]/(m(x))$ to $R$; this induced map is different from $\varphi$, although closely related to it. It is this map which you want to show is an isomorphism. So, the first step is . . .

What is the induced map?

(If your abstract algebra class looks anything like the one I took, you're probably much more comfortable with groups right now. If so, think about the analogous question for groups: I have a group $G$, and a surjective homomorphism $\psi$ from $G$ to some group $H$. This induces a map $\hat{\psi}: G/ker(\psi)\rightarrow H$ - do you remember how?)

Once you've identified the induced map, you'll need to show that it is:

• a (well-defined) homomorphism,

• surjective, and

• injective.