construct a ring homomorphism

Question

This is an assignment question, please give hints only thank you!

Construct a ring homomorphism $f: R \rightarrow S$ such that $\ker(f)$ is a maximal ideal, and $S$ is not a field.

I know that $I\subseteq R$ is a maximal ideal iff $R/I$ is a field.

I thought of $R$ being $\mathbb{Z}/n\mathbb{Z}$, but then $\ker(f)$ would have to be $n\mathbb{Z}$ and n would have to be a prime number. Then, what would be a suitable $S$ in this case?

Answer 1

The hint you ask for: the map doesn't have to be surjective.

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The solution (for others looking at the question): Just let $R=\Bbb Z$ and $S=\Bbb Z\oplus \Bbb Z/p$ and map

$$\phi:\begin{cases}\Bbb Z\to \Bbb Z\oplus \Bbb Z/p\\ x\mapsto (0, x\mod p)\end{cases}$$

Clearly $S$ is not a field, and clearly the kernel is $(p)$ which is a maximal ideal of $R$.

The main trick of it: don't demand that $S=\text{Im}( \phi)$ is the full image, surjective homomorphisms won't work since the first isomorphism theorem tells you the image is a field. But the image doesn't have to be the full co-domain, so we just slap on an extra $\Bbb Z$ to make it not a field.

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Edit (addendum): In the case this is demanded to be a unital homomorphism, you just tweak things and let $R=\Bbb Z/p$ and $S=\Bbb Z/p\oplus\Bbb Z/p$. Then let

$$\phi:\begin{cases} R\to S \\ x\mapsto (x,x)\end{cases}$$

Here the kernel is $\{0\}$ because the map is injective, and again $S$ is not a field since $(0,1)\cdot (1,0)=(0,0)$.