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Let $F<E$ and $F<K$ be finite extensions and assume that $EK$ is defined as composite of two fields. I need to show that $[EK:F] \leq [E:F][K:F]$, with equality if $[E:F]$ and $[K:F]$ are relatively prime.

I am stuck with even the less than or equal to part, and also I couldn't see the thing about the case being these two relatively prime.

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  • $\begingroup$ Welcome to math.SE! For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – frabala Mar 12 '14 at 13:19
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    $\begingroup$ $E \otimes_F K = \{ \sum_j a_j \otimes_F b_j, a_j \in E, b_j \in K\}$ is a ring containing both $E,K$ and we have a surjective morphism $E \otimes_F K \to EK$ which is $\sum_j a_j \otimes_F b_j \mapsto \sum_j a_jb_j$. Thus it reduces to understand the construction of $E \otimes_F K$ and that it is a $[E:F][K:F]$ dimensional $F$-vector space $\endgroup$ – reuns Jul 10 '19 at 20:06
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Since $E$ and $K$ are finite extensions of $F$, they are algebraic and finitely generated. So, we can write $E = F(a_1,\dots,a_n)$ and $K = F(b_1,\dots,b_m)$ for some algebraic elements $a_1,\dots,a_n,b_1,\dots,b_m$. The composite field $EK$ is then equal to $F(a_1,\dots,a_n,b_1,\dots,b_m)$.

Now, $[EK : F] = [EK : E] [E : F]$, so it suffices to show that $[EK : E] \leq [K : F]$. Let $E_0 = E$ and $E_i = E_{i-1}(b_i)$ for each $1 \leq i \leq m$. Note that $E_m = EK$. Let $F_0 = F$ and $F_i = F_{i-1}(b_i)$ for each $1 \leq i \leq m$. In particular, $F_i \subseteq E_i$ for each $i$. Also note that $K = F_m$. Hence, $$ \begin{align} [EK : E] = [E_m : E_0] &= [E_m : E_{m-1}] \cdots [E_1 : E_0], \quad \text{and}\\ [K : F] = [F_m : F_0] &= [F_m : F_{m-1}] \cdots [F_1 : F_0]. \end{align} $$ For each $1 \leq i \leq m$, $$ \begin{align} [E_i : E_{i-1}] = [E_{i-1}(b_i): E_{i-1}] &= \deg \min(E_{i-1},b_i)\\ &\leq \deg \min(F_{i-1},b_i)\\ &= [F_{i-1}(b_i) : F_{i-1}]\\ &= [F_i : F_{i-1}] \end{align} $$ Hence, $$ [EK : E] = \prod_{i=1}^m [E_i : E_{i-1}] \leq \prod_{i=1}^m [F_i : F_{i-1}] = [K : F]. $$ Thus, $[EK:F] \leq [E:F] [K:F]$.


$$ [EK:F] = [EK:E][E:F] \implies [E:F] \mid [EK:F],\\ [EK:F] = [EK:K][K:F] \implies [K:F] \mid [EK:F]. $$ $$ \gcd([E:F],[K:F]) = 1 \implies [E:F][K:F] \mid [EK:F]. $$ $$ \therefore [E:F][K:F] \leq [EK:F]. $$ By the previous part, $$ [EK:F] \leq [E:F] [K:F]. $$ Hence, $[EK:F] = [E:F] [K:F]$ if $[E:F]$ and $[K:F]$ are relatively prime.

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Divide the inequality by one of the factors on the right; you should be able to subsequently simplify the left-hand side by using the product formula for towers. Use the PET for the numerator field on the right-hand side, so that both sides of the inequality turn into the degree of an algebraic element only the one on the left side is considered over a larger base field...

For the second part, remember your knowledge of arithmetic and elementary number theory: in particular recall that $a,b\mid n$ and $\gcd(a,b)=1$ imply $ab\mid n$. What are $a,b,n$ in this problem?

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