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).

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