## Algebraic field extensions exist and are unique

We will now prove some basic facts about algebraic field extensions.

Let $F$ be a field and let $p(x)$ be an irreducible monic polynomial of positive degree over $F$. Define $E = F[x]/(p(x))$. We claim the following.

1. $E$ is an extension of $F$.
2. $E$ contains an element $\eta$ which is algebraic over $F$ and whose minimal polynomial is $p(x)$.
3. If $K$ is an extension of $F$ which contains an element $\xi$ which is algebraic over $F$ with minimal polynomial $p(x)$, then there is an injective ring homomorphism $E \hookrightarrow K$.

Note that $F[x]$ is a Euclidean domain, hence a principal ideal domain. Since $p(x)$ is irreducible, it is also prime, so that $(p(x))$ is a prime ideal. Since $F[x]$ is a principal ideal domain, $(p(x)) \subseteq F[x]$ is maximal. So $F[x]/(p(x))$ is a field. Now let $\iota : F \rightarrow F[x]/(p(x))$ be the map $\iota(\alpha) = \alpha + (p(x))$. If $\iota(\alpha) = \iota(\beta)$, then $\alpha - \beta \in (p(x))$. Since every nonzero element of $(p(x))$ has degree at least 1, $\alpha - \beta = 0$, so that $\alpha = \beta$. Hence $E$ is (isomorphically) an extension of $F$.

Let $\eta$ denote the image of $x$; $\eta = x + (p(x))$. Certainly $p(\eta) = 0$.

Now suppose $K$ exists. By the universal property of polynomial rings, we can define a homomorphism $\varphi : F[x] \rightarrow K$ by fixing $F$ and mapping $x \mapsto \xi$. Note that $\varphi(p(x)) = p(\xi) = 0$, so that $(p(x)) \subseteq \mathsf{ker}\ \varphi$. Conversely, if $q(x) \in \mathsf{ker}\ \varphi$, then $q(\xi) = 0$, so that the minimal polynomial $p(x)$ divides $q(x)$. Thus $\mathsf{ker}\ \varphi = (p(x))$. By the first isomorphism theorem for rings, $\Phi : F[x]/(p(x)) \rightarrow K$ given by $s(x) + (p(x)) \mapsto \varphi(s(x))$ is a well defined ring homomorphism. Suppose $\Phi(\overline{s(x)}) = \Phi(\overline{t(x)})$; then $s(\xi) = t(\xi)$, so that $(s-t)(\xi) = 0$. So $p(x)$ divides $s(x) - t(x)$, and we have $\overline{s(x)} = \overline{t(x)}$. So $\Phi$ is injective.

This allows us to characterize $E$ as the “smallest” extension of $F$ which contains a root of $p(x)$. Moreover, in this previous exercise we showed that the elements of $F[x]/(p(x))$ are uniquely of the form $a(x) + (p(x))$ for some $a(x)$ of degree less than the degree of $p(x)$.