Corollary 15.3.8 from Artin (degrees of field extensions)

Question

Here are a few corollaries of the multiplicativity of towers of field extensions.

Corollary 15.3.6(b):

First of all, why is this $\mathcal K$ introduced, what's its purpose?

Second, where does $\beta$ live?

Third, how is 15.3.6(b) applied? So $\beta$ is algebraic over $F$ because by assumption $N$ is finite, right? Then the corollary says that $\beta\in K'$ is algebraic over $K$ and its degree (namely $[K':K]$) is at most equal the degree of $\beta$ over $F$. But I thought this last degree is $[K':F]=N$, isn't it?

Finally, does this proof implicitly use the fact that any finite extension is gotten by adjoining several elements? If so, are these elements just the basis elements? How does the induction work here?

Answer 1

First - $\mathcal K$ is introduced to ensure that the compositum $K'$ of $K$ and $F'$ is a field. An example showing why it's needed is given by certain 2 by 2 matrices over $\Bbb{Q}$. The set of matrices $\begin{pmatrix} a & -b \\ b & a\end{pmatrix}$ for $a,b$ in $\Bbb{Q}$ form a field isomorphic to $\Bbb{Q}[\alpha]$ with $\alpha^2+1=0$. Similarly the set of matrices $\begin{pmatrix} a & -b \\ b & a-b \end{pmatrix}$ for $a,b$ in $\Bbb{Q}$ form a field isomorphic to $\Bbb{Q}[\omega]$ with $\omega^2+\omega+1=0$. Taken together these 2 fields generate $\operatorname{M}_2(\Bbb{Q})$, the full ring of 2 by 2 matrices over $\Bbb{Q}$, and not a field thus the need for $\mathcal K$.

Second - $\beta$ generates $F'$ over $F$, so $\beta$ must live in $F'$.

Third - Yes. Since $F'$ has finite dimension over $F$, it must finitely generated and algebraic. The degree of $\beta$ over $F$ is $[F':F]=n$

Finally - A field extension of finite degree is finitely generated because the degree of a field extension is equal to the minimum number of vectors needed to generate the field as a vector space. See the answers here for instance.

Answer 2

The problem

Artin begins to write the proof by induction. Quickly puts the base case, where there is only one element added to the base field F, shows that the statement of theorem is valid, then suddenly drops the proof. It might be hard to understand the inductive step.

Dummit & Foote

In Algebra by Dummit and Foote, pages 528 and 529, Chapter Field Theory (another widely used standard algebra text), there is an identical Proposition 21. It clearly explains what's going on.

Important note

Dummit and Foote write $F(\alpha_1, \alpha_2)$ when they mean that there is a base field $F$, over which there is another field $F'$ with a basis $(\alpha_1, \alpha_2)$. That is, the number of alphas in braces exactly matches the basis of the extension field. So if the dimension of the extension field is five, i.e. the basis is $(\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5)$, then Dummit will designate this F-field $$F( \alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5)$$

Artin, in its turn, designates an extension field via generators. So if $\alpha_1$ has an irreducible polynomial of degree 5, Artin may designate the same extension field as $$F(\alpha_1)$$ and note $[F(\alpha_1):F] = 5$.

Basic task

The theorem gives us some bigger field $\mathcal{K}$, and two its subfields $K$ and $F'$ and $F$, such that there is another composite field $K'$ constructed as a union of the two. The union assumption is important because element $\beta$ of $F'$ is also $$\beta \in K'$$

so the first part of the theorem is easy to prove by 15.3.5 Multiplicative property of the degree. That is, m divides N because $F \subset K \subset K'$. And n divides N because $F \subset F' \subset K'$.

The slightly more difficult task is to prove the inequality $N \leq mn$

The inequality

What we are doing is this. We take field $F$, add m-basis elements of K, then add to it n-basis elements of F'. This broth lets us span the whole of $K'= K \bigcup F$. We just merge two bases. We can do this as shown in theorem 15.3.5 page 447, and the inequality comes from the fact that while merging two bases, we might encounter dependencies between the bases. Some linear combination may collapse.

Conclusion

It is possible to understand induction on the degree of $F'$, not on the number of additional generators in $F'$ as [incorrectly] done by Artin.

We prove the basis case, with $n=2$, i.e $F'$ is a quadratic field. Then we assume we have proved the theorem for some case n, i.e. $[F':F]=n$. Then if we add one additional basis element $F'$. Not just any element as in Artin, but we instead increase the degree of $F'$ by one as a linear (=vector) space over $F$.

Then while merging $K$ and $F'$ into $K'$ we will have $$N \leq mn \leq m(n+1)$$

Artin's fault is that we must write induction proof on the degree of $F'$ as a vector space, not on the number of generators of $F'$. In Artin's view it is not the same. He picked one generator $\beta$, but the basis for $F'$ in his case would look something like this $$(1,\beta_1, \dots \beta_{n-1})$$