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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?

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    $\begingroup$ If $K/F$ is separable, then the Primitive Element Theorem further claims that $K = F(\alpha)$ for some $\alpha\in K$. $\endgroup$
    – anomaly
    Jun 19, 2014 at 2:47

4 Answers 4

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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 $F$.

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  • $\begingroup$ I like this one a lot. $\endgroup$
    – Kaj Hansen
    Apr 29, 2014 at 18:12
  • $\begingroup$ c'est élégant ! $\endgroup$ Apr 1, 2019 at 22:46
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    $\begingroup$ Maybe a bit more should be added. It could happen that some of those powers of $\alpha$ are the same. In that case we can't say those are $n+1$ . However it still works considering that the equality $\alpha^i=\alpha^j$ gives us $\alpha$ is a root of $x^i-x^j$ over $F$ $\endgroup$
    – Wastaken
    Sep 22, 2022 at 15:10
  • $\begingroup$ Very nice one, short but precise. $\endgroup$
    – Aelx
    Nov 10, 2022 at 11:13
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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 such that $g(x) \mapsto g(a)$. 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.

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    $\begingroup$ A pleasing and efficient answer. $\endgroup$
    – Lubin
    Aug 19, 2019 at 16:07
  • $\begingroup$ To me, this answer, as well as the answer from @fretty omit a bit part of what should be in the proof. You crucially state that the map you describe cannot be injective because $K$ is finitely generated whereas $F[x]$ is not. This is not obvious; except in the case where the underlying field $F$ is assumed to be finite (because then we can directly argue via cardinality). If $F$ is not finite, the cardinality of the domain and codomain can be equal. proofwiki.org/wiki/Definition:Linearly_Independent/Set provides some very concrete proofs ("proof 3" is especially awesome). $\endgroup$ May 8, 2022 at 13:37
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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 $.

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As other have said, by definition, a field extension $K$ over $F$ is of finite degree if and only if it is finitely generated, because its degree $n$, by definition, is the minimum number of vectors $v_1,v_2,...v_n \in K$ (from the field extension $K$) that are sufficient to form a spanning set over the underlying field $F$. I.e. if the degree is finite, then a finite number of vectors are enough, and so our extension is finitely generated and conversely, if we can finitely generate our extension, then by definition, the field extension / vector space $K$ has a finite degree equal to the number of vectors we need.

However, regarding the second part of your question about all finite extensions being algebraic:

I believe a lot of common proofs I have seen for this statement online rest on unjustified premises. Therefore, I will now share what I think is an elegant, elementary and self-sufficient actual proof from:

https://proofwiki.org/wiki/Size_of_Linearly_Independent_Subset_is_at_Most_Size_of_Finite_Generator

Let $K$ be a finitely generated vector space / extension over $F$. This means that we can find a finite number of vectors in $K$ that will span $K$ by drawing coefficients over $F$.

Let $ v_1, v_2, ..., v_n$ be such a spanning set for $K$ over $F$ (called a basis for $K$ over $F$). This means, we define $n$ to be the lowest positive integer for the number of vectors that are required to span $K$ over $F$.

Let $x_1, x_2, ..., x_r \in K $ be a sequence non-zero vectors s.t. $r>n$. We will prove that these vectors must be linearly dependent by contradiction! So let us begin by assuming that they are indeed an independent collection of vectors.

We know that there exist $a_1, a_2, ..., a_n \in F$ s.t.

$$ x_1 = a_1v_1 + a_2v_2 + ... + a_nv_n$$

Therefore the elements in the finite sequence $\{x_1, v_1, v_2, v_3, ..., v_n\}$ are linearly dependent. Now consider this beautiful idea: starting from the left-most term and going right-wards, let "$a_i$" be the first term from this ordered sequence that can be expressed as a linear combination of some terms strictly to the left of itself! This term can not be $x_1$ as we have positioned it as the first term in the sequence and so by definition, it cannot be composed of any elements preceding it (as no entries precede it).

Note: Since $x_1$ can, however be expressed as a linear combination of some of the subsequent terms of our sequence (which span $K$), we know that in any case, we will eventually encounter some $i \in \{1,2,3,...,n\}$ s.t. $a_i$ will satisfy our criterion. If $1 \leq j_1 < j_2 < ... j_f \leq n$ are the integer indices s.t. $a_{j_1},a_{j_2},...,a_{j_f} \in F$ are all non-zero and: $$ x_1 = a_{j_1}v_{j_1} + a_{j_2}v_{j_2} + ... + a_{j_f}v_{j_f}$$

holds. Then exactly $i = j_f$ will be the index that we are looking for, as we can write $v_{j_f}$ as a linear combination of the following preceding terms:

$$ a_{j_f}^{-1}(x_1 - a_{j_1}v_{j_1} - a_{j_2}v_{j_2} - ... - a_{j_{f-1}}v_{j_{f-1}})= v_{j_f} = v_i$$

We will throw this now superfluous term "$v_i$" out of our sequence and we will still have a spanning collection of $n$ vectors (you can easily show that this new sequence of vectors yields a basis). So now we have $S_1 = \{x_1, v_1, v_2, v_3, ..., v_n\}$.

Now, inductively, we repeat the same procedure. Consider the sequence of $n+1$ terms: $\{x_2,x_1, v_1, v_2, ...,v_{i-1},v_{i+1},..., v_n\}$. We repeat our argument: $x_2$ can not be a linear combination of any vectors to the left of it as there are no preceding terms in our sequence. On a different note, now: by our assumption of independence for the vectors $x_1,x_2,...,x_n$, $x_1$ cannot possibly be a linear combination of the vectors coming before it. Eventually we must, however come across the term we are searching for as $x_2$ can be written as a linear combination of terms from $S_1$. So we drop the redundant term and we get $S_2 = \{x_2,x_1, v_1, ..., v_n\}$ which is, again, $n$ terms long, as we have added $x_1$ and $x_2$ (2 terms) and have dropped 2 of our original basis vectors.

Inductively, we then repeat this argument until we get our final sequence of basis vectors $S_n = \{x_n,x_{n-1},...,x_2,x_1\}$. Now, recall that we have assumed that $r>n$, so we still have some vectors $x_{n+1},x_{n+2},...,x_{r}$ left. However, since our vectors from the final sequence $S_n$ span $K$, we can express any of these remaining extra vectors as a linear combination of the vectors $x_1,...,x_n$. Therefore, we have directly contradicted our assumption that $x_1,...,x_r$ are linearly independent. Therefore, we have shown that for a finite extension of degree $n$, any collection of $r$ vectors where $r>n$ are going to be linearly dependent. Now letting $\alpha \in K$, we see that a non-trivial solution to:

$0 = a_0 + a_1\alpha + a_2\alpha^2 + ... + a_n\alpha^n$

must exist by the linear dependence of $1,\alpha, \alpha^2, ..., \alpha^n$ (as here we have $n+1 > n$ terms). Hence every term of a field extension of finite degree is algebraic; i.e. a finitely generated extension / an extension of finite degree is algebraic.

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