Daniell Phillips

Answered

2022-01-13

Why does

Answer & Explanation

Neil Dismukes

Expert

2022-01-14Added 37 answers

Step 1

Let

Then

is a localization of the integral domain

thus the natural map

is injective. Now we have the following equivalence:

a)

b) Every polynomial over K divides some polynomial over k.

c)

is the localization of

Take

Then

for

i.e.

is a polynomial over k with root u. QED

In particular, if

alkaholikd9

Expert

2022-01-15Added 37 answers

Step 1

There is also a more elementary argument explaining why

Let

Take an arbitrary

Then

is an isomorphism of linear spaces over

Hence

is a monomorphism of finite dimensional linear spaces over

In particular

alenahelenash

Expert

2022-01-24Added 366 answers

Let $\frac{K}{k}$ be an extension and X an indeterminate. We identify $k(X){\otimes}_{k}K$ with $k(X){\otimes}_{k[X]}K[X]$ , and denote this domain by A. Everything will follow from the
Observation. Any element of A can be written as
$\frac{1}{p(X)}\otimes P(X)$
with $p(X)$ in $k[X],\text{}p(X)\ne 0,\text{}P(X)$ in $K[X]$ . Its image in $K(X)$ is $\frac{P(X)}{p(X)}$ . In particular, the natural map from A to $K(X)$ is injective. Thus we can (and will) view A as the subdomain of $K(X)$ formed by the elements which can be written, in the above notation, as $\frac{P(X)}{p(X)}$ .
The fraction field of A, regarded as a subfield of $K(X)$ , is called the compositum of $k(X)$ and K. It is denoted $k(X)K$ , and coincides with $K(X)$ .
If a is in K and $\frac{1}{(X-a)}$ 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 $k(X)$ , is a field, and is thus equal to $K(X)$ . The same conclusion holds if K is a union of finite degree extensions of k, that is, if $\frac{K}{k}$ is algebraic.

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