For non-constant polynomial $f(x) \in F[x]$ have a zero in $E$ which containing $F$.

Question

I want to prove following theorem

Let $F$ be a field. Then for each non-constant polynomial $f(x) \in F[x]$, there exists a field $E$ containing $F$ such that $f(x)$ has a zero in $E$.

How one can prove this? This seems related to the existence of splitting fields...

Answer 1

Let $f(x) \in F[x]$ be a monic irreducible polynomial. Then $(f(x))$ is an maximal ideal so $F[x]/(f(x))$ is a field. Construct ring homomorphism : $\phi : F \rightarrow F[x]/(f(x))$ by $\phi(a) = a + (f(x))$ for $a\in F$, then $\phi$ is injective [$\because$ if $a+(f(x)) = b+ (f(x))$, then $a-b \in (f(x))$, so $f(x)| (a-b)$ and since $f(x)$ is non-constant, $a=b$. ] So from the first isomorphism theorem $F \simeq \phi(F) = \{a + (f(x)) | a\in F\}$. Let $I = (f(x))$, then $E= F[x]/I$ have a subfield $F$, so $E$ is an extension of $F$. For $\alpha = x+I$, $f(\alpha) = f(x) + I = f(x) + (f(x)) = 0 + I$ so at $E$, $f(\alpha) =0$. i.e., $\alpha \in E$ is a root in $f(x)$.