0
$\begingroup$

Let $L_1$ and $L_2$ be extensions of $K$. Show that

a) If $[L_1:K] = p, [L_2:K] = q, \mbox { $p,q$ prime numbers}$, then $L_1\cap L_2 = K$ or $L_1=L_2$

b) If $[L_2:K]=2, L_2 = K(\alpha)$ and $L_1\cup L_2 = K$, then $[L_1(\alpha):L_1]=2$

For a), I tried to see $L_1$ and $L_2$, of course, as extensions of the same $K$, but 'childs' of a common extension that contains both. I think it has some relation with the multiplicative formula of degrees, but I couldn't find it.

For b), what's so special about the degree being $2$? For this I truly couldn't find any relation

$\endgroup$
  • $\begingroup$ $L_1\cup L_2=K$? But $\alpha\in L_2$ and $\alpha\notin K$, so this does not seem possible. Perhaps you mean $L_1\cap L_2=K$? $\endgroup$ – ajotatxe Oct 2 '17 at 18:19
  • $\begingroup$ @ajotatxe it's possible, I don't know $\endgroup$ – Guerlando OCs Oct 2 '17 at 18:23
  • $\begingroup$ The conditions $[L_2:K] = 2$ and $L_1 \cup L_2$ are incompatible; they can't both be true. For $L_1 \cup L_2 = K \Longrightarrow L_2 \subset K$, so it's not even clear how $[L_2:K]$ would be defined. It looks to me at this point that the best we could have is $[L_2:K] = 1$, i.e., $L_2 = K$. $\endgroup$ – Robert Lewis Oct 2 '17 at 18:38
  • $\begingroup$ I meant to write "The conditions $[L_2: K] = 2$ and $L_1 \cup L_2 = K$ . . . " in the above. $\endgroup$ – Robert Lewis Oct 3 '17 at 15:51
1
$\begingroup$

For (a), suppose $L_1 \cap L_2 \ne K$; then since $K \subset L_1 \cap L_2$, there is some $\alpha \in L_1 \cap L_2$, $\alpha \notin K$. Then $K(\alpha) \subset L_1 \cap L_2 \subset L_1$ and we have

$[L_1:K(\alpha)][K(\alpha):K] = [L_1:K] = p; \tag 1$

thus $[K(\alpha): K]$ is either $1$ or $p$; but we can rule out the case $[K(\alpha): K] = 1$ since it implies $\alpha \in K$, contrary to our assumption $\alpha \notin K$; thus $[K(\alpha): K] = p$, $[L_1: K(\alpha)] = 1$ whence $L_1 = K(\alpha)$. The same argument applied to $L_2$ shows $L_2 = K(\alpha)$ as well, so $q = p$ and $L_1 = L_2 = K(\alpha)$.

For (b), suppose for the moment that $[L_2: K] = n > 1$, a considerable relaxation of the condition $[L_2:K] = 2$; then with $L_2 = K(\alpha)$ we have $[K(\alpha):K] = n$ and $L_1 \cap K(\alpha) = K$. Now if $\alpha \in L_1$, we see that $\alpha \in L_1 \cap K(\alpha) = K$ in contradiction to $[K(\alpha):K] > 1$; therefore, $\alpha \notin L_1$, and we may affirm $[L_1(\alpha): L_1] > 1$. Furthermore, since $[K(\alpha):K] = n < \infty$, $\alpha$ satisfies some irreducible polynomial $p(x) \in K[x]$ with $\deg p(x) = n$:

$p(\alpha) = 0, \tag 2$

and since $K \subset L_1$, we conclude that $\alpha$ is algebraic over $L_1$ as well, and that

$1 < [L_1(\alpha): L_1] \le n = \deg p(x); \tag 3$

note we cannot affirm that $[L_1(\alpha):L_1] = n$ in general, since $p(x)$ may be reducible in $L_1(x)$, though it is not so in $K[x]$; but in the case $n = 2$, we have only the choice $[L_1(\alpha):L_1] =2$, and this establishes our result.

$\endgroup$
0
$\begingroup$

$[L_1:K]=[L_1:L_1\cap L_2][L_1\cap L_2:K]$. This implies that $h=[L_1\cap L_2:K]$ divides $p$ and the same argument shows that it divides $q$. If $p$ and $q$ are distinct primes, $h=1$, otherwise we can have $h=p=q$ and $L_1=L_2$.

$\endgroup$
0
$\begingroup$

For the first question, just write

  • $p = [L_1:K] = [L_1:L_1 \cap L_2] \times [L_1 \cap L_2 : K]$
  • $q = [L_2:K] = [L_2:L_1 \cap L_2] \times [L_1 \cap L_2 : K]$.

Then, as $p$ is a prime number, we must have either :

  • $[L_1 : L_1 \cap L_2] = 1$ and $[L_1 \cap L_2 : K] = p$, which implies that $L_1 = L_1 \cap L_2$, so $L_1 \subset L_2$. Then, you also have $q = [L_2:L_1 \cap L_2] \times p$, but as $q$ is prime, it implies $p = q$ and $[L_2:L_1 \cap L_2] = 1$, so in the same way as before, $L_2 \subset L_1$ : we have proven $L_1 = L_2$.

  • $[L_1 : L_1 \cap L_2] = p$ and $[L_1 \cap L_2 : K] = 1$, which directly implies $L_1 \cap L_2 = K$.

For the second question, I will assume you meant $L_1 \cap L_2 = K$ (intersection instead of union). Define $P$ the minimal polynomial of $\alpha$ over $K$ (which is of degree $2$). The same polynomial will cancel $\alpha$ over $L_1$, so $[L_1(\alpha) : L_1]$ equals either $1$ or $2$.

But if it was $1$, $L_1$ would contain both $K$ and $\alpha$, so it would contain $K(\alpha) = L_2$. You would then have $L_1 \cap L_2 = L_2 \not = K$, a contradiction.

I hope it helps. Next time, try thinking in terms of arithmetic with help of the degree formula. But I agree question b) has little to do with $2$, any prime would have worked here.

$\endgroup$

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.