Extending valuations and linear disjointness of fields

Question

Let $F$ be a field and let $K$ be a field extension of $F$. Suppose that $K$ can be written as the compositum of field extensions $E$ and $L$ of $F$, linearily disjoint over $F$, thus $K$ can be identified with the field of fractions $E \otimes _F L$, an integral domain.

Suppose that we are given a valuation of $E$ over $F$ with residue field $F$, for example, $E$ could be the field of Laurent series over $F$. Can $K$ be given a valuation over $L$ with residue field $L$ which is compatible with this valuation? Is such a valuation unique? Note that the valuation need not be of rank one or archimedean.

This is probably very basic, but I am new to this material.

Answer 1

The answer seems to follow more or less directly from the axioms.

It is enough to show that the valuation on elements of the form $\sum a_i \cdot b_i$ for $a_i \in E$ and $b_i \in L$ is uniquely determined by the constraints. By basic properties of the tensor product and the axioms of the valuation, we may assume that the $a_i$ and the $b_i$ are linearly independent over $F$, and that $v(a_i) + v(b_i)$ is independent of $i$. Since each $b_i$ is in $L$ and nonzero, it has valuation zero, so in fact $v(a_i)$ is independent of $i$. Dividing by $a_1$, say, we may assume that $v(a_i) = 0$ for all $i$.

Let $c_i$ be the image of $a_i$ in $F$: it is nonzero. Therefore $\sum c_i \cdot b_i$ is nonzero, as the $b_i$ are linearly independent over $F$. It follows that the valuation of $\sum a_i \cdot b_i$ is zero.