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

  • $\begingroup$ $L_1\cup L_2=K$? 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

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.


$[L_1:K]=[L_1:L_1\cap L_2][L_1\cap L_2:K]$. 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 $L_1=L_2$.


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.


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