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Suppose $K/F$ is an extension field, and $E_1$ and $E_2$ are two intermediate subfields of finite degree. Let $E$ be the subfields of $K$ generated by $E_1$ and $E_2$. I'm trying to prove that $$ [E:F]\leq[E_1:F][E_2:F].$$

Since $E_1$ and $E_2$ are finite extensions, I suppose they have bases $\{a_1,\dots,a_n\}$ and $\{b_1,\dots,b_m\}$, respectively. If $E_1=F$ or $E_2=F$, then the inequality is actually equality, so I suppose both are proper extension fields. I think $E=F(a_1,\dots,a_n,b_1,\dots,b_m)$. Since $\{a_1,\dots,a_n,b_1,\dots,b_m\}$ is a spanning set for $E$ over $F$, $$[E:F]\leq n+m\leq nm=[E_1:F][E_2:F]$$ since $m,n>1$.

Is this sufficient? I'm weirded out since the problem did not ask to show $[E:F]\leq [E_1:F]+[E_2:F]$ which I feel will generally be a better upper bound.

Chelsea Dirks
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2 Answers2

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The space spanned by the union of the two bases is usually not closed under muliplication, so won't span a field.

Instead you should show that $$ \{a_ib_j\mid 1\le i\le n, 1\le j\le m\} $$ is a spanning set.

Show that the $F$-space $L$ spanned by that set (inside $K$) is closed under multiplication, and that any field containing both $E_1$ and $E_2$ must contain $L$. Then show that $L$ also contains the inverse of all its non-zero elements. There the assumption about finiteness of $n$ and $m$ is crucial. Therefore $L$ is the smallest field containing both $E_1$ and $E_2$.

Note that this spanning set is not always linearly independent over $F$. Anyway, the dimension of $L$ is at most $mn$, which is what you wanted to show.

Jyrki Lahtonen
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    Good luck, Chels! If you get stuck in proving the existence of inverses, you may peek at this. – Jyrki Lahtonen Mar 21 '13 at 07:53
  • How can I show that $\operatorname{span}{ a_i b_j }$ is closed under multiplication? It should suffice to show that $a_i b_j \cdot a_k b_l$ lies in the span for every $1 \leq i,k \leq n$, $1 \leq j,l \leq m$. But how can I show this, can you give me a hint? –  May 31 '18 at 09:39
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    @Brahadeesh: That's the idea! Because $E_1$ and $E_2$ are fields (hence also rings) you can write $a_ia_k$ as an $F$-linear combination $a_ia_k=\sum_tc_{ikt}a_t,c_{ikt}\in F$, The same thing with $b_jb_l$, and then take the product of the two linear combinations. – Jyrki Lahtonen May 31 '18 at 12:24
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The sum of the degrees in general is not going to be an upper bound. Consider $K = \Bbb{Q}(\sqrt[3]{2},e^{2\pi i/3})$. This is a degree $6$ extension of $\Bbb{Q}$. Take $E_1 = \Bbb{Q}(\sqrt[3]{2})$ and $E_2 = \Bbb{Q}(e^{2 \pi i/3})$. Then

$$[E_1 : \Bbb{Q}] + [E_2:\Bbb{Q} ] = 3 + 2 = 5$$

but $E = K$ that has degree $6$ over $\Bbb{Q}$.