Is any finite-dimensional extension of a field, say $F$, algebraic and finitely generated?

As title, Is any finite-dimensional extension of a field, say $F$, algebraic and finitely generated?

Say if $K/F$ is a finite extension when $K$ is a finite-dimensional vector space of $F$. Clearly, this implies that $K$ is finitely generated (as an algebra) over $F$, since a basis is a generating set. So every finite extension is finitely generated.

So indeed they all are, is my logic correct?

• If $K/F$ is separable, then the Primitive Element Theorem further claims that $K = F(\alpha)$ for some $\alpha\in K$. – anomaly Jun 19 '14 at 2:47

By definition, a field extension of finite degree is finitely generated because the degree is the number of linearly independent "vectors" in the extension required to form a spanning set.

Now let $$K/F$$ be a finite extension. Consider an arbitrary $$a \in K$$ and the evaluation homomorphism $$\text{ev}_a:F[x] \rightarrow K$$ defined where $$g(x) \mapsto g(a)$$. This map cannot be injective because $$K$$ is finitely generated whereas $$F[x]$$ is not, so the kernel of this map must be a nontrivial ideal. This is to say: we can find nontrivial polynomials in $$F[x]$$ with $$a$$ as a root for any $$a \in K$$, namely the nonzero elements of $$\ker(\text{ev}_a)$$. In other words, $$K/F$$ is an algebraic extension.

• A pleasing and efficient answer. – Lubin Aug 19 at 16:07

A similar but easier answer to the one given above is as follows.

To show $K/F$ is algebraic if finite we must show that every element of $K$ satisfies a polynomial over $F$.

Suppose $[K : F] = n$ and choose $\alpha\in K$. Then consider the elements $1,\alpha,\alpha^2,...,\alpha^n$.

This is a list of $n+1$ elements in an $n$ dimensional $F$-vector space so must be linearly dependent. Thus there exists $a_0,a_1,...,a_n\in F$ not all zero such that $a_n \alpha^n + ... + a_2\alpha^2 + a_1\alpha + a_0 = 0$.

But then $\alpha$ is a root of the polynomial $a_nx^n + ... + a_2x^2 + a_1x + a_0$ over $F$.

• I like this one a lot. – Kaj Hansen Apr 29 '14 at 18:12
• c'est élégant ! – Malik Apr 1 at 22:46

Other answers provide nice proofs, here is a very short one based on the multiplicativity of the degree over field towers: If $K/F$ is a finite extension and $\alpha \in K$, then $F(\alpha)$ is a subfield of $K$, and we have a tower of fields $F \subseteq F(\alpha) \subseteq K$. The Tower Law then asserts that

$$[F(\alpha):F][K:F(\alpha)] = [K:F]$$

Since $K/F$ is finite, by definition the RHS is finite. On the other hand, it is seen from the above equality that $[F(\alpha):F]$ is less than or equal to $[K:F]$, therefore is also finite. By definition, $\alpha$ is algebraic over $F$.