1
$\begingroup$

This is problem 2.1 of Falko Algebra I Fields and Galois Theory. I have finished the proof but it looks very unsatisfying.

$E/K$ is a field extension. $L_1,L_2$ are intermediate fields. $L_1L_2$ denotes the composite field formed by $L_1$ and $L_2$. Suppose $[L_1:K],[L_2:K]<\infty$. Prove if $[L_1L_2:K]=[L_1:K][L_2:K]$, then $L_1\cap L_2=K$.

I can show by presenting the basis of $L_1L_2/K$. If $K\subset L_1\cap L_2$ is proper, this means $L_1\cap L_2$ has non-trivial overlap(i.e. I can start removing redundant basis elements from either $L_1$ or $L_2$). This will show $[L_1L_2:K]<[L_1:K][L_2:K]$.

Questions:

  1. This proof lacks elegance though simple. Is there a proof based on purely showing algebraic manipulation to complete the proof(i.e. showing either $[L_1\cap L_2:K]\leq 1$, $[L_1L_2: L_1\cap L_2]=[L_1:K][L_2:K]$ or other equivalent relations)?

  2. Why should I expect this is the case without looking at the basis?(i.e. Assume I do not know the proof but I want to know the reason why this algebraic relation leads to this special vector space construction.)

  3. If $[L_1:K],[L_2:K]$ are relatively prime, this becomes trivial. In my view point $L_1L_2=L_1\otimes_K L_2$, relatively primeness implies for simple extensions $L_1,L_2$ over $K$, the $K-$algebra can be generated by the simple tensor of the two generators. How do I generalize this statement for arbitrary case by assuming finite extension say $L_1=K(a_1,\dots, a_m),L_2=K(b_1,\dots b_n)$ where I assume $m,n$ are minimal generator and $[L_1:K]=m_1,[L_2:K]=m_2$ which may not have any straightforward relationship with $m$ and $n$ respectively?(i.e. I want a straightforward proof showing that for $[L_1:K],[L_2:K]$ relatively prime, I can present an explicit construction of basis of $L_1\otimes_K L_2$ This statement does not say $[L_1:K],[L_2:K]$ must be relatively prime here.

$\endgroup$
3
$\begingroup$

This indeed is a bit laboured. Suppose $L_1\cap L_2\ne K$. Then $|L_1\cap L_2:K|=d>1$. Let $n_i=|L_i:K|$. Then $|L_i:L_1\cap L_2| =n_i/d$. Also $|L_1L_2:L_1|\le|L_2:L_1\cap L_2|=n_2/d$, so $$|L_1L_2:L_1\cap L_2|=|L_1L_2:L_1||L_1:L_1\cap L_2|\le n_1n_2/d^2$$ and then $$|L_1L_2:K|=|L_1L_2:L_1\cap L_2||L_1\cap L_2:K|\le n_1n_2/d<n_1n_2.$$

Here I used the principle $|LM:L|\le|M:K|$ where $L$ and $M$ are extensions of $K$ and $LM$ is a compositum of $L$ and $M$ over $K$. This boils down to noting that if $M=K(\alpha)$ then $LM=L(\alpha)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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