Rings, Modules, Fields

11 Fields

Fields are simple objects in the category of rings. Since they are integral domains, their characteristic is either a prime \(p\) or \(0\). Much of the way we study fields relies on the fact that their module theory is very simple, in particular, all modules are free. Moreover, any morphism of fields is injective since the kernels are proper ideals. Thus our basic notion is that of a field extension \(L/F\), which is just an inclusion \(F \hookrightarrow L\). In particular, the field \(L\) becomes a \(F\)-vector space, so has a dimension which we call the degree of the extension, denoted \([L:F]\). Note that by Lemma 7.3 a finite degree extension (which we can just call finite) is algebraic. We will start by studying algebraic extensions, mostly finite ones.

Lemma 11.1. If \(M/L/K\) are two extensions of fields, then \([M:K] = [M:L][L:K]\).

Proof. Take a basis of \(L/K\), and a basis of \(M/L\), and the product of the two sets will be a basis of \(M/K\). □

We say that an extension \(L/K\) is simple if it is generated over \(K\) by one element, \(\alpha \), which we call the primitive element. The degree of \(\alpha \) is then \([L:K] = [K(\alpha ):K]\).

Lemma 11.2. The degree of an element divides the degree of the field extension.

Proof. Apply Lemma 11.1 to the fact that we have the extensions \(L/K(\alpha )/K\). □

Definition 11.3. The characteristic polynomial \(\chi _L(\alpha )\) of an element \(\alpha \) of a finite extension \(L/K\) is the characteristic polynomial of multiplication by that element as a \(K\)-linear map.

Lemma 11.4. The degree of \(\alpha \) is the degree of its minimal polynomial \(\mu _K(\alpha )\).

Proof. Just note that the span of \(1,\alpha ,\dots , \alpha ^{\deg (\alpha )-1}\) form a field containing \(a\), giving one inequality, and the other inequality comes from applying the Cayley-Hamilton Theorem to \(\chi _L(\alpha )\). □

We already have enough to prove an interesting fact:

Theorem 11.5 (Lüroth). A subfield of \(K(t)\) strictly containing \(K\) is isomorphic to \(K(t)\).

Proof. If \(K(t)/L/K\) are extensions, with \(L \neq K\), \(L\) contains some rational function \(f/g\), where \(f,g\) are relatively prime. \(g(t)f/g-f(t)\) then shows that \(K(t)\) is a simple finite extension of \(L\), and its degree is \(\min _{f/g \in L}(\max (\deg f,\deg g))\). Then a minimizing \(f/g\) and using Lemma 11.1, \(L = K(f/g)\). □

Corollary 11.6. \(\chi _L(\alpha ) = \mu _L(\alpha )^{\frac {[L:K]}{\deg {\alpha }}}\).

Proof. This follows from doing the proof of Lemma \(\ref {degreemult}\) on \(L/K(\alpha )/K\). □

Definition 11.7. A field is algebraically closed if there are no proper algebraic field extensions. An algebraic extension is called the algebraic closure if it is algebraically closed.

Lemma 11.8. A field is algebraically closed iff any polynomial factors into linear terms.

Proof. This is immediate, since nonlinear irreducibles \(f\) correspond to simple nontrivial extensions via \(K[x]/(f(x))\). □

We use \(\bar {K}\) to denote the algebraic closure of \(K\). The algebraic closure behaves similarly to the “universal cover" of a space. In particular, we would like every algebraic extension to be describable in terms of an algebraic closure.

Lemma 11.9 (Embedding Lemma). If \(L/K\) is an algebraic extension and \(\bar {K}/K\) is an algebraic closure, then there is an embedding of \(K\)-algebras \(L \hookrightarrow \bar {K}\).

Proof. Consider the poset of embeddings of subfields use Zorn’s Lemma to choose a maximal one, and note that we can always embed a simple extension, so that the maximal element is everywhere defined. □

Corollary 11.10 (Extending Lemma). Any automorphism of \(K\) extends to one of \(\bar {K}\).

Proof. Apply the embedding lemma, and note that the embedding must be an isomorphism as \(\bar {K}\) is algebraically closed. □

Proposition 11.11. Algebraic closures exist and are unique up to isomorphism.

Proof. Given a field \(F\), write \(F = F_0\). Now inductively given \(F_i\) produce a field \(F_{i+1}\) by adjoining an element \(g_\alpha \) to \(F_i\) for each irreducible polynomial \(f_\alpha \) over \(F_i\), and modding out by a maximal ideal containing \(f_\alpha (g_\alpha )\). Then the union of the \(F_i\) is algebraically closed. For uniqueness, given \(M,L\), two algebraic closures of \(K\), By the embedding lemma \(L\) embeds into \(M\), but this embedding is an isomorphism as \(L\) is algebraically closed. □

Note that fields don’t have many limits, but if we consider the category of fields embedded in some fixed algebraic closure, then we get a lattice, which does have limits and colimits. The product of some fields \(F_i\) is then their intersection, and the coproduct is called their compositum.