According to Artin's Algebra, chapter 15, section 3, the mapping property of the degree of field extension is as follows:
Let $F\subset K\subset L$ be fields. Then $[L:F]=[L:K][K:F]$, where $[K:F]$ represents the dimension of $K$, as an $F$-vector space.

My question is based on its Corollary: Let $\mathcal{K}$ be an extension field of a field $F$, let $K$ and $F'$ be subfields of $\mathcal{K}$ that are finite extensions of $F$, and let $K'$ denote the subfield of $\mathcal{K}$ generated by the two fields $K$ and $F'$ together. Let $[K':F]=N$, $[K:F]=m$ and $[F':F]=n$. Then $N\leq mn$.

At this corollary, I guess $N=mn-d$, where $d$ is the degree of the $gcd(f(x),g(x))$. Here $f(x)$, $g(x)$ represent the monic irreducible polynomials that generate $K$ and $F'$. I mean: $K=F[x]/(f)$, $F'=F[x]/(g)$. Am I right? If not, what is $N$ supposed to be?

  • $\begingroup$ Probably $N = mn/[K\cap F':F]$. What terrible notation, by the way.... $\endgroup$ – Greg Martin Jun 1 '14 at 3:45
  • $\begingroup$ Why exists $f$ monic irreducible such that $K\cong F[x]/(f)$? That is true if $K=F(a)$ where $f$ is the minimal polynomial for $a$ over $K$. Finite extensions must be algebraic but not always simple if the field extension is not separable. $\endgroup$ – Gaston Burrull Jun 1 '14 at 4:16

This question is a bit subtle. Because both $K$ and $F'$ are subfields $K'$ we can compute $[K':F]$ in two different ways $$ [K':K][K:F]=[K':F]=[K':F'][F':F] $$ and conclude that $N$ must be divisible by both $m$ and $n$. A common argument here is the case when we know that $m$ and $n$ are coprime. Then we can conclude that we must have $N=mn$.

But that's about all that we can say in the general case.

  • Even if the extensions are simple and we can write $F'=F[x]/\langle g\rangle$, $K'=F[x]/\langle f\rangle$ for some polynomials $f,g\in F[x]$ we should not expect a formula in terms of the gcd of the polynomials $f$ and $g$. After all many polynomials give rise to the same field extension (as a subfield of, say, some fixed algebraic closure of $F$). For example when $F=\Bbb{Q}$ then $f(x)=x^2-2$ and $g(x)=x^2-8$ give the same field extension $K=F'=F(\sqrt2)$, but $\gcd(f,g)=1$.
  • The formula $N=mn/[(K\cap F'):F]$ suggested in the comments holds in some cases (more below), but not always. Consider again the case $F=\Bbb{Q}$. Let $z_1=\root3\of2$ and $z_2=\omega z_1$ be two roots of the irreducible cubic $x^3-2$, where $\omega=(-1+\sqrt{-3})/2$ is a primitive cubic root of unity. Let $K=F(z_1)$ and $F'=F(z_2)$. Here clearly $F'\cap K=F$, $m=n=3$, so this "formula" would give $N=9$. But the correct value in this case is $N=6$. This is because in this case $K'=F(z_1,z_2)=F(z_1,\omega)$ is the splitting field of $x^3-2$ that we know to be sextic (and it also follows from the fact $\omega$ is a root of the quadratic $x^2+x+1$).

A concept arising from this is that of linearly disjoint extension. I once wrote a quick explanation of this for our study group. Basically the equation $N=mn$ holds, iff $K$ and $F'$ are linearly disjoint over $F$. A necessary (but not sufficient - see the second bullet) condition for linear disjointness is that $F'\cap K=F$. It is not hard to show (see the notes) that if both $K$ and $F'$ are Galois over $F$, then this condition is also sufficient. As in that case they are also Galois over their intersection, it follows that

  • If $K/F$ and $F'/F$ are both finite Galois extensions (inside an algebraic closure of $F$), and $K'$ is their compositum, then $$ [K':F]=\frac{[K:F][F':F]}{[(K\cap F'):F]}. $$
  • $\begingroup$ More can be said about linear disjointness, but this gives you the basics. IIRC Pete L. Clark has asked here about more powerful results implying linear disjointness. $\endgroup$ – Jyrki Lahtonen Jun 1 '14 at 8:29
  • $\begingroup$ This is the question I remembered. Pete L. Clark's course notes should give you more to chew. $\endgroup$ – Jyrki Lahtonen Jun 1 '14 at 9:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.