# Is the compositum of $L_1$ and $L_2$ equal to $L_1[L_2]$?

In a course about Galois theory, there is the following definition :

Let $L_1$ and $L_2$ be two subfields of the field $L$. We define the compositum $L_1L_2$ of $L_1$ and $L_2$ as the smallest subfield of $L$ containing $L_1$ and $L_2$, that is : $$L_1L_2 = L_2[L_1] = L_1[L_2]$$

So does that mean that $L_1[L_2] = L_1(L_2)$ ? I can't see why.

• How do you define $L_1(L_2)$? – Tobias Kildetoft May 11 '15 at 9:31
• $L_1(L_2)$ is the smallest subfield of $L$ containing $L_1$ and $L_2$, and $L_1[L_2]$ is the smallest $L_1$-subalgebra of $L$ containing $L_2$ – Segipp May 11 '15 at 9:33

In general, for a fields $K \subset L$, and $A \subset L$ the structures $K[A]$ and $K(A)$ are not the same, the former being the smallest ring, or also algebra if you prefer, containing $K$ and $A$ while the latter is the smallest field containing $K$ and $A$. These two can be different.
However, if $K \subset L$ is an extension of finite degree, more generally if it is algebraic, then in fact $K[A]=K(A)$. (As the inverse of $\alpha$ an algebraic element over $K$ is a polynomial expression in $\alpha$.)
• But is it still true when $A$ is a field and $K \subset L$ is not algebraic ? – Segipp May 11 '15 at 10:19
• No, in general this is not true. I did not check the details but $L_1 = Q(x)$ and $L_2 = Q(y)$ for "variables" $x,y$ should work as counter example. – quid May 11 '15 at 10:37
• Ok, so I think this is a counter-example : for $L_1 = K(X)$ and $L_2 = K(Y)$, $L_1L_2 = K(X,Y)$ but $L_1[L_2] = \lbrace P/(QR), P \in K(X,Y), Q \in K(X), R \in K(Y) \rbrace$ which does not contain $1/(X+Y)$. Is it correct ? – Segipp May 11 '15 at 10:37
• I think that $1/(x+y)$ is in $k(x)\bigl(k(y)\bigr)$, but not in $k(x)\bigl[k(y)\bigr]$. – Lubin May 12 '15 at 3:30