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$K$ and $L$ are subfields of the field $F$. I think the composite of $K$ and $L$ (smallest subfield of $F$ containing $K$ and $L$) is just their product, namely

$$\{\sum _{i=1}^n k_i l_i |k_i\in K, l_i\in L, i=1, \dots, n, n\in \mathbb N\}. $$

But if this is true, why no book mention it?

user150248
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2 Answers2

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That will be the smallest ring containing both $K$ and $L$. In some cases it will not be a field. The smallest field will be the set $$\{ \frac{ \sum k_i l_i}{\sum k'_j l'_j}\ | \ k_i, k_j' \in K, l_i, l'_j \in L \textrm{ and } \sum k_j' l'_j \ne 0 \}$$

To see that you do not get a field in general, consider $K = \mathbb{Q}(x)$, $L = \mathbb{Q}(y)$, $K,L \subset \mathbb{Q}(x,y)$. You can check that your subset consists of all the fractions $\frac{P(x,y)}{Q(x) R(y)}$. The element $x+y$ has no inverse here. Indeed, assume that we have $$(x+y)\cdot \frac{P(x,y)}{Q(x)R(y)} = 1$$ that is $$(x+y)P(x,y) = Q(x) R(y)$$ an equality of non-zero polynomials in $x$, $y$. Now substitute $y=-x$ and get $Q(x) \cdot R(-x) = 0$, not possible.

Note however, that if $K$, $L$ both contain a sub field $F$ and $K$, $L$ algebraic over $F$, then you will get the composite field with your construction.

orangeskid
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  • Regarding the last statement, why do we have a simpler representation when the two fields are both algebraic extension over the same field? – Juggler Dec 24 '20 at 06:14
  • @Juggler: If both are algebraic over a subfield $F$ (that means all their elements are algebraic) thenm since all the products and all the sums of algebraic are algebraic, any of the sums you wrote call it $x$ is algebraic over $F$. But if $\sum a_i x^i=0$, and $x\ne 0$, then you can write $x^{-1}$ as a polynomial in $x$. – orangeskid Dec 24 '20 at 07:02
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For you to get started you need to have a bigger field $\Omega$ such that $F,K,L\subset\Omega$. For otherwise you have problem defining those products $k_il_i$ as well as their sums.

Anyway, you get the compositum this way if $[K:F]$ and $[L:F]$ are both finite. For then you can check that the set of such sums, call it $M$, is closed under addition and multiplication. Also, you can easily prove that $M$ is a vector space over $F$ of dimension at most $[K:F]\cdot [L:F]$. So $M$ is a finite dimensional integral domain over $F$, and therefore it is a field.

Clearly $M$ is contained in any subfield of $\Omega$ that contains both $K$ and $L$, so it is the compositum.

Jyrki Lahtonen
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  • What about $\displaystyle K \otimes_{K \cap L} L$ ? (It seems an horrible notation, but I interpret as just saying we need basis of $K,L$ as $K \cap L$ algebras (or vector spaces) to construct the compositum explicitly) – reuns Sep 01 '17 at 23:26
  • That's a possibility @reuns. We then need to mod out some maximal ideal of that tensor product to get a field. But without $\Omega$ even the intersection $K\cap L$ is not really defined. Of course, we can use $K\otimes_FL$ and mod out a maximal ideal again. – Jyrki Lahtonen Sep 02 '17 at 00:27
  • But the choice of that maximal ideal may make a difference! Even $K\otimes_FK$ usually has several maximal ideal to choose from. Also, see here. – Jyrki Lahtonen Sep 02 '17 at 00:29
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    Tks, the example $K = \mathbf{Q}(\sqrt[3]{2}),L= \mathbf{Q}(\zeta_3\sqrt[3]{2})$ explains what you meant with modding out some maximal ideal (As $[\displaystyle K \otimes_{\mathbf{Q}} L: \mathbf{Q}] = 9$ while the compositum should be $[\mathbf{Q}(\sqrt[3]{2},\zeta_3):\mathbf{Q}] =6$). And so choosing an intersection $K \cap L$ together with a maximal ideal of $\displaystyle K \otimes_{K \cap L} L$ is the same as embedding $K,L$ into a larger field. – reuns Sep 02 '17 at 01:22