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.

  • $\begingroup$ How do you define $L_1(L_2)$? $\endgroup$ – Tobias Kildetoft May 11 '15 at 9:31
  • $\begingroup$ $L_1(L_2)$ 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 $L_2$ $\endgroup$ – 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 $\alpha$.)

Normally, in your context the extension will be algebraic, so there is no real problem. Still, at least in isolation that definition you recall seems a bit confusing.

  • $\begingroup$ But is it still true when $A$ is a field and $K \subset L$ is not algebraic ? $\endgroup$ – Segipp May 11 '15 at 10:19
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    $\begingroup$ 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. $\endgroup$ – quid May 11 '15 at 10:37
  • $\begingroup$ 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 ? $\endgroup$ – Segipp May 11 '15 at 10:37
  • $\begingroup$ Ok, thanks ! (sorry, didn't see your answer) $\endgroup$ – Segipp May 11 '15 at 10:38
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    $\begingroup$ I think that $1/(x+y)$ is in $k(x)\bigl(k(y)\bigr)$, but not in $k(x)\bigl[k(y)\bigr]$. $\endgroup$ – Lubin May 12 '15 at 3:30

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