Daniell Phillips

2022-01-13

Why does $K⇝K\left(X\right)$ preserve the degree of field extensions?

Neil Dismukes

Expert

Step 1
Let $\frac{K}{k}$ be a field extension.
Then
$K{\otimes }_{k}k\left(X\right)$
is a localization of the integral domain
$K{\otimes }_{k}k\left[X\right]=K\left[X\right]$
thus the natural map
$K{\otimes }_{k}k\left(X\right)\to K\left(X\right)$
is injective. Now we have the following equivalence:
a) $\frac{K}{k}$ is algebraic
b) Every polynomial over K divides some polynomial over k.
c) $K{\otimes }_{k}k\left(X\right)=K\left(X\right)$
$a⇒b$: Because of uniqueness of the division algorithm, it suffices to prove divisibilty in an algebraic extension. So take a splitting field of the polynomial and reduce to the case of linear factors: $x-a$. But they divide the minimal polynomial over k.
$b⇒c:K{\otimes }_{k}k\left(X\right)$
is the localization of $K\left[X\right]$ at all nontrivial polynomials over k. Since we have b, we localize at all nontrivial polynomials over K, i.e. we get $K\left(X\right)$.
$c⇒a:$
Take $u\in K$.
Then $\frac{1}{\left(X-u\right)}=\frac{p}{q}$
for
$p\in K\left[X\right]$
$q\in k\left[X\right]$
i.e. $p\left(X-u\right)$
is a polynomial over k with root u. QED
In particular, if $\frac{K}{k}$ is algebraic, we have $\left[K\left(X\right):k\left(X\right)\right]=\left[K:k\right]$

alkaholikd9

Expert

Step 1
There is also a more elementary argument explaining why
spans $\frac{K\left(X\right)}{k\left(X\right)}$. To understand it one doesn't need to know separable field extensions nor integral ring extensions, only a little bit of linear algebra:
Let . Note that $K\left[X\right]\subset E$ implies , as is a basis of $\frac{K}{k}$
Take an arbitrary $h\in K\left[X\right]$
Then
$\cdot h:K\left(X\right)\ni a↦a\cdot h\in K\left(X\right)$
is an isomorphism of linear spaces over $k\left(X\right)$ (${h}^{-1}$ defines inverse linear mapping $\cdot {h}^{-1}$). Fuethermore $\cdot h\left(E\right)\subset E$ as $K\left[X\right]\subset E$
Hence
$\cdot h{\mid }_{E}:E\to E$
is a monomorphism of finite dimensional linear spaces over $k\left(X\right)$, so in fact an isomorphism. Therefore $\cdot {h}^{-1}{\mid }_{E}:E\to E$
In particular $\left(\cdot {h}^{-1}\right)\left(1\right)={h}^{-1}\in E$, and $g{h}^{-1}\in E$ for every $g\in K\left[X\right]$

alenahelenash

Expert

Let $\frac{K}{k}$ be an extension and X an indeterminate. We identify $k\left(X\right){\otimes }_{k}K$ with $k\left(X\right){\otimes }_{k\left[X\right]}K\left[X\right]$, and denote this domain by A. Everything will follow from the Observation. Any element of A can be written as $\frac{1}{p\left(X\right)}\otimes P\left(X\right)$ with $p\left(X\right)$ in in $K\left[X\right]$. Its image in $K\left(X\right)$ is $\frac{P\left(X\right)}{p\left(X\right)}$. In particular, the natural map from A to $K\left(X\right)$ is injective. Thus we can (and will) view A as the subdomain of $K\left(X\right)$ formed by the elements which can be written, in the above notation, as $\frac{P\left(X\right)}{p\left(X\right)}$. The fraction field of A, regarded as a subfield of $K\left(X\right)$, is called the compositum of $k\left(X\right)$ and K. It is denoted $k\left(X\right)K$, and coincides with $K\left(X\right)$. If a is in K and $\frac{1}{\left(X-a\right)}$ is in A, then a is algebraic over k. So, A is not a field if $\frac{K}{k}$ is transcendental. If $\frac{K}{k}$ has finite degree, the domain A, being finite dimensional over $k\left(X\right)$, is a field, and is thus equal to $K\left(X\right)$. The same conclusion holds if K is a union of finite degree extensions of k, that is, if $\frac{K}{k}$ is algebraic.