Let $L$ be a field extension of $K$, and $\alpha \in L$. In this class I am taking, the following proof that $K[\alpha]$ is a field implies $\alpha$ is algebraic over $K$ is given:
If $\alpha$ is not algebraic, $\not \exists P \in K[x]$ s.t. $P(\alpha) = 0$.
This means $i \colon K[x] \to L$ given by $i(P) = P(\alpha)$ is injective (since if $P(\alpha) = 0$ then $P = 0$ by the above assumption that $\alpha$ is not algebraic). But $K[x]$ is not a field and the image of $i = K[\alpha]$ is a field, which is a contradiction.
I'm fine with everything up to this last statement. My thought process is that, since $\ker(i) = (0)$, then by the first ring isomorphism theorem $K[x]/(0) \cong K[\alpha]$, and since $K[x]/(0) \cong K[x]$ this implies $K[x] \cong K[\alpha]$, which is a contradiction since $K[x]$ is not a field.
Is there a more obvious explanation? Or is this obvious enough that it is ok to leave it out as the instructor did in the original proof.
Best Answer
I think it is way simpler to observe that if $\;K[a]\;$ is a field (and assuming $\;a\neq0\;$ otherwise it is trivial), then
$$\frac1a\in K[a]\implies \exists \;0\neq p(x)\in K[x]\;s.t.\;\;\frac1a=p(a)\implies ap(a)-1=0$$
and certainly the polynomial $\;f(x)=xp(x)-1\in K[a]\;$ is non zero, thus $\;a\in L\;$ is algebraic over $\;K\;$ .