Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

2 Answers 2

up vote 8 down vote accepted

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?

share|improve this answer
    
@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 $K$.

share|improve this answer
2  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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