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In my algebra book they define for field extensions $L/K_1/F$ and $L/K_2/F$ the field $K_1K_2 = K_1(K_2)$.

The formal definition I have for this is $$ K_1(K_2) = \bigcap_{K_2 < E < L} E \quad \text{where } E/K_1. $$ So the smallest field extension of $K_1$ containing $K_2$.

For extensions such as $F(\alpha)$ though sometimes it is written, $$ F(\alpha) = \{ a_0 + a_1\alpha + a_2\alpha^2 + \cdots + a_{n - 1}\alpha^{n - 1} : a_i \in F\} $$

Can I extend this definition for $K_1(K_2)$ and write the following? $$ K_1(K_2) = \left\{ \sum_{i = 1}^n \sum_{j = 1}^{m_i} a_{ij}b_i^j : a_{ij} \in K_1, b_i \in K_2, n, m_i \in \mathbb{N} \right\}$$

EDIT: $K_1, K_2$ are algebraic.

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  • $\begingroup$ That expression for $F(\alpha)$ only works for $\alpha$ algebraic over $F$ right? $\endgroup$ – anon May 17 '12 at 2:07
  • $\begingroup$ That is right, It does happen that the $K_1, K_2$ are algebraic in my situation. $\endgroup$ – zrbecker May 17 '12 at 2:18
  • $\begingroup$ You can probably ask Elman. He is in his office a lot of times, and he always welcomes anyone who walks by it. $\endgroup$ – Daniel Montealegre May 17 '12 at 5:10
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There are several different-looking definitions of the compositum $K_1K_2$ of two fields, and it’s a very useful exercise to sit down and prove them equivalent. Here's my suggestion: Let's consider only subfields and subsets of some larger field $\Omega$. Then for a field $K$ and a set $S$, take as definition of $K(S)$ something close to what you wrote, namely $$ K(S)=\bigcap_{L\supset K\cup S}L\,, $$ where the index $L$ runs through fields only. The construct is a field, of course.

Now on the other hand, consider the set $K^{\mathrm{r}}(S)$ whose elements are the finite rational expressions (with coefficients in $K$) of the form $f({\mathbf{s}})=n({\mathbf{s}})/d({\mathbf{s}})$, where in each case, $\mathbf{s}$ represents some finite set of elements of $S$, and the numerator $n$ and the denominator $d$ are multivariable polynomials over $K$. Clearly $K^{\mathrm{r}}(S)$ is also a field. Now you can easily check that $K(S)=K^{\mathrm{r}}(S)$. It follows immediately that for two fields, $K_1(K_2)=K_2(K_1)$, and you also see that in your case, where the $K_i$ are algebraic over some smaller field, you can dispense with rational expressions, and use just polynomial ones.

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