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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?

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  • $\begingroup$ 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
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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 $$ [K':F]=\frac{[K:F][F':F]}{[(K\cap F'):F]}. $$
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  • $\begingroup$ 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

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