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Let $E,F,L$ be fields such that $E,F\subset L$. Define $EF$ be the smallest subfield in $L$ contains both $E,F$.

I don't know whether the following statement is correct: Let $K$ be a field and consider the extensions $K(a)$ and $K(b)$, then $K(a,b)=K(a)K(b)$.

Any ideas?

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Yes, this is true, as long as $a, b$ are contained in some common field. What difficulties are you having in proving it? – Dylan Moreland Dec 3 '11 at 19:49
I'm currently proving a theorem which I believe I found a short cut by applying K(a,b)=K(a)K(b). I just want to double check it is right. Thx – CC_Azusa Dec 3 '11 at 19:54
Right, so we should prove that $K(a, b) = K(a)K(b)$. Did you attempt to do so? Note that $K(a, b)$ is the smallest subfield containing $K$ and $a, b$. Perhaps you think of it as the set of $f(a, b)/g(a, b)$ where $f, g \in K[x, y]$ and $g(a, b) \neq 0$? – Dylan Moreland Dec 3 '11 at 19:56
up vote 3 down vote accepted

If $K\subseteq L$ and $a,b\in L$, then $K(a,b)$ is by definition the smallest subfield of $L$ that contains $K$ and $a$ and $b$. On the other hand, $K(a)K(b)$ is the smallest subfield of $L$ that contains $K(a)$ and $K(b)$. These descriptions are almost identical. It is easy to prove that they are the same by proving that $K(a,b)\subseteq K(a)K(b) \subseteq K(a,b)$.

Here is a slightly more sophisticated approach. For $U\subseteq L$ any subset, $K(U)$ is the smallest subfield of $L$ that contains $K$ and $U$. Then $$ K(a)K(b)=K(K(a)\cup K(b)) = K(K \cup \{a,b\}) = K(\{a,b\}) = K(a,b) $$

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