Algebraic extensions
Idea
An algebraic element is the zero of some polynomial. An algebraic extension is an extension in which every element is algebraic.
Definition
Definition. A field extension $K\subset L$ is called algebraic if for all $x\in L$ there exists $F\in K[t]$ such that $F(x)=0$.
Definition. A field $K$ is called algebraically closed iff $K$ has no algebraic extensions. Equivalently, any nonconstant polynomial $P\in K[T]$ is of degree 1.
Proposition. Suppose $K$ is algebraically closed. Then if $K\subset L$ is an algebraic extension, $K=L$.
Properties
- Every finite field extension is algebraic.
- There exist infinite algebraic extensions
- Every finitely generated algebraic field extension is finite.
- If $K$ is generated by (any finite number of) algebraic elements, then $K/k$ is algebraic.
- Suppose $L/K$ and $K/k$ are both algebraic (not necessary finite). Then $L/k$ is finite.
- Any endomorphism of an algebraic extension is an automorphism.
Proofs
1
Let $F\subset E$ be a finite field extension. Then for $\alpha\in E$, the powers of $\alpha$ are not all linearly independent, for otherwise $E$ would be an infinite dimensional field extension. Then we can write a linear equation in some variables $a^i$ with nonzero coefficients (in $F$) that equals 0. This shows that $\alpha$ is the root of a polynomial in $F[x]$.
An example of an infinite algebraic extension is the subfield of the complex numbers containing all algebraic number over $\mathbb{Q}$ (as an extension of $\mathbb{Q}$)
2
If $K$ is generated by one element, then $K=k(\alpha)$. Since $K$ is algebraic, $\alpha$ is an algebraic element and so we can apply (Property 2) and (Example $k_P$) to see that $K$ is finite. Now assume $K=k(\alpha,\beta)$. Then $[K:k]=[k(\alpha,\beta):k(\alpha)][k(\alpha):k]$ which is finite. We are done by induction.
3
Observe that if $\alpha$ is algebraic over $k$ and $F$ is an extension of $k$, and also $\alpha\in L$ and $F\subset L$ for some field $L$, then certainly $\alpha$ is algebraic over $F$. We can apply this to our situation by observing that there is a tower $$ k\subset k(\alpha_1)\subset k(\alpha_1,\alpha_2)\subset\cdots\subset k(\alpha_1,\dots,\alpha_n). $$ Each extension is algebraic over the previous, and by (Property 2) each extension is finite. Hence $k(\alpha_1,\dots,\alpha_n)$ is finite over $k$ by (Of dimensions, Property 1). By (Property 1) this implies it is an algebraic extension over $k$.
4
Let $\alpha\in L$. It is by assumption algebraic over $K$, hence the root of a polynomial $P_{\alpha,K}\in K[T]$. Let $k_1$ be the subfield of $K$ generated over $k$ by all coefficients of the polynomial $P_{\alpha,K}$. Then $$k\subset k_1\subset k_1(\alpha).$$ By (General, Property 2) the second extension is finite, and by (Property 2) the first extension is also finite. So by (Of dimensions, Property 1) we have that $k_1(\alpha)/k$ is finite, hence algebraic. So $\alpha$ is algebraic over $k$.
5
It suffices to show the endomorphism $\sigma:K/k\to K/k$ is surjective, since any field homomorphism is injective. Let $\alpha\in K$, and $\alpha=\alpha_1,\dots,\alpha_n$ be the roots of $P_\alpha$. By (Property 3) this is a finite extension. We also note that $\sigma(k(\alpha_1,\dots,\alpha_n))\subset k(\alpha_1,\dots,\alpha_n)$ because by (Of morphisms, Property 1) a root of $P_\alpha$ must map to a root of $P_\alpha$, and $k(\alpha_1,\dots,\alpha_n)$ contains all such elements by construction. But $\ker\sigma=0$. So $\sigma$ is injectively mapping between two finite dimensional vector spaces, hence $\sigma$ is an isomorphism when restricted to these vector spaces. In particular $\alpha\in\text{im}\sigma$, so $\sigma$ is surjective.