$K\rightarrow K[x]/(g), a\mapsto a+(g)$ is injective

Question

Let $K$ be a field, $g\in K[x]$ irreducible. Then $\pi: K\rightarrow K[x]/(g), a\mapsto a+(g)$ is injective.

In my book they imply that this is trivial, altough I'm not quite sure whether I understood why.

Answer 1

Assume $a\in K$ with $\pi(a)=0$, i.e., $a\in (g)$. Then in $K[x]$, $a$ is a multiple of $g$, hence either $a=0$ or $\deg a\ge \deg g>0$. But from $a\in K$, we have $\deg a\le 0$. We conclude $a=0$.

Answer 2

Since $K$ is a field, its ideals are $\{0\}$ and $K$. The map $\pi$ is a ring homomorphism (and it preserves the identity). Therefore its kernel is not $K$, hence it is $\{0\}$.

More generally, if $f\colon K\to R$ is a ring homomorphism (preserving the identity) and $R$ is not the trivial ring, then $f$ is injective.

The fact that $g$ is irreducible is irrelevant: it suffices to assume that $g$ is not constant.