# Field Extension

Let $F$ be an extension field of $K$. let $L$ and $M$ be intermediate fields, with both finite algebraic extensions of $K$. Suppose {$a_1,...,a_n$} is a basis for $L$ over $K$ and {$b_1, ...,b_m$} is a basis for $M$ over $K$. Show that {$a_ib_j$} is a spanning set for the field $LM$ ($LM$ is the smallest field between $K$ and $F$ containing both $L$ and $M$) as a vector space over $K$.

What I have done so far is this: Let $x$ $\in$ $L$. Then $x$ can be written as a linear combination of $a_i$. Also $y$ $\in$ $M$ implies that $y$ can be written as a linear combination of $b_j$. This is where I'm stuck.

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What is your definition of $LM$? – Mariano Suárez-Alvarez Feb 8 '11 at 5:29
$LM$ is the smallest subfield containing both $L$ and $M$. – Nana Feb 8 '11 at 5:34
The smallest subfield of what? Presumably, some fixed algebraic closure of $K$ (this is standard). – Arturo Magidin Feb 8 '11 at 5:37
@Arturo: I meant to say the smallest field between $K$ and $F$ containing both $L$ and $M$. – Nana Feb 8 '11 at 6:00
Oh, sorry; somehow, I missed the $F$... – Arturo Magidin Feb 8 '11 at 6:00

Consider the field $L(b_1,\ldots,b_m)=L[b_1,\ldots,b_m]$ (the equality because we are dealing with finite algebraic extensions). Suppose you can prove that it is equal to $LM$.

Then every element of $LM$ can be written as an $L$-linear combination of $b_1,\ldots,b_m$ (there may be a bit of work to be done here if it is not clear; certainly, you can write it as an $L$-linear combination of products of powers of the $b_i$, but since each of those lies in $M$, you can write them as $K$-linear combinations of the $b_i$; take it from there).

And every coefficient in that linear combination can be written as a $K$-linear combination of $a_1,\ldots,a_n$. See where that leads you (assuming you can prove $L[b_1,\ldots,b_m]=LM$, of course).

Corrected. In comments you ask about showing that if $[LM:K]=[L:K][M:K]$, then $L\cap M=K$. I messed up my first attempt in a rather silly way (feel free to look at the edit history to see the screw-up!). Sorry about that.

Again, let $E=L\cap M$ for simplicity. Then: \begin{align*} [L:K][M:K] = [LM:K] &= [LM:E][E:K]\\ &\leq [L:E][M:E][E:K]\\ &= [L:E][M:K]. \end{align*} Can you take it from here?

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@Arturo:This solution is based on being able to show that $L[b_1,...,b_m]$ = $LM$. Is there another approach which doesn't make use of that fact? – Nana Feb 8 '11 at 7:12
@Nana, It should not be very hard to show. One inclusion follows from the fact that $L[b_1, ..., b_m]$ is a field contained in $F$ that contains both $L$ and $M$. For the other inclusion, notice that any field that contains $L$ and $M$, must in particular contain $b_1, ..., b_m$ and all $L$-linear combinations of them. – Vitaly Lorman Feb 8 '11 at 13:44
@Nana: I don't know if there is another approach, but as Vitaly says, this should follow simply from the definition of $LM$ as the "smallest" field that contains both $L$ and $M$, the fact that $b_1,\ldots,b_n$ are in $M$, and that $K[b_1,\ldots,b_n]\subseteq L[b_1,\ldots,b_n]$. – Arturo Magidin Feb 8 '11 at 14:08
@Arturo: Thanks. Does the answer to the question also proves the statement: $[L:K]$ , $[M:K]$ are finite implies that $[LM:K]$ is finite? – Nana Feb 8 '11 at 16:13
@Nina: Yes: you know that $[LM:L]\leq m$ (since $b_1,\ldots,b_m$ are a spanning set, so the dimension is at most $m$), so $[LM:L]\leq [M:K]$ (you also get $[LM:L]\leq[L:K]$ by a symmetric argument). So $[LM:K]=[LM:L][L:K]\leq [M:K][L:K]$. Now try proving that if $\gcd(n,m)=1$, then $[LM:K] = mn$. – Arturo Magidin Feb 8 '11 at 16:45

Comme l'a signalé Arturo Magidin, $LM$ est en fait l'anneau engendré par $L$ et $b_1, \ldots, b_m$. Pour montrer que tout élément de cet anneau est combinaison linéaire des $b_i$, il suffit de voir que $A = \sum Lb_i$ est un anneau. On se ramène donc à prouver que chaque produit $b_i b_j$ (et 1) est dans $A$. Mais en fait, $b_i b_j \in \sum Kb_i$ (et 1 aussi) par l'hypothèse que les $b_i$ forment une base de $M$ sur $K$.

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R.: If you can read English, one may suppose that you can also write English, then can you save me from the French which I cannot understand? Thanks. – awllower Feb 25 '11 at 16:11
@awllower that not always holds true actually, I can read and understand French (this answer in particular) though I have very limited abilities in answering in French. – shabunc Feb 23 at 21:55