Why does K⇝K(X) preserve the degree of field extensions?

Daniell Phillips

Daniell Phillips

Answered

2022-01-13

Why does KK(X) preserve the degree of field extensions?

Answer & Explanation

Neil Dismukes

Neil Dismukes

Expert

2022-01-14Added 37 answers

Step 1
Let Kk be a field extension.
Then
Kkk(X)
is a localization of the integral domain
Kkk[X]=K[X]
thus the natural map
Kkk(X)K(X)
is injective. Now we have the following equivalence:
a) Kk is algebraic
b) Every polynomial over K divides some polynomial over k.
c) Kkk(X)=K(X)
ab: 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: xa. But they divide the minimal polynomial over k.
bc:Kkk(X)
is the localization of K[X] at all nontrivial polynomials over k. Since we have b, we localize at all nontrivial polynomials over K, i.e. we get K(X).
ca:
Take uK.
Then 1(Xu)=pq
for
pK[X]
qk[X]
i.e. p(Xu)
is a polynomial over k with root u. QED
In particular, if Kk is algebraic, we have [K(X):k(X)]=[K:k]

alkaholikd9

alkaholikd9

Expert

2022-01-15Added 37 answers

Step 1
There is also a more elementary argument explaining why
v1, , vn spans K(X)k(X). To understand it one doesn't need to know separable field extensions nor integral ring extensions, only a little bit of linear algebra:
Let E=spank(X){v1, , vn}. Note that K[X]E implies K[X]EE, 1E, as v1, , vn is a basis of Kk
Take an arbitrary hK[X]
Then
h:K(X)aahK(X)
is an isomorphism of linear spaces over k(X) (h1 defines inverse linear mapping h1). Fuethermore h(E)E as K[X]E
Hence
hE:EE
is a monomorphism of finite dimensional linear spaces over k(X), so in fact an isomorphism. Therefore h1E:EE
In particular (h1)(1)=h1E, and gh1E for every gK[X]

alenahelenash

alenahelenash

Expert

2022-01-24Added 366 answers

Let Kk be an extension and X an indeterminate. We identify k(X)kK with k(X)k[X]K[X], and denote this domain by A. Everything will follow from the Observation. Any element of A can be written as 1p(X)P(X) with p(X) in k[X], p(X)0, P(X) in K[X]. Its image in K(X) is 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 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 1(Xa) is in A, then a is algebraic over k. So, A is not a field if Kk is transcendental. If Kk 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 Kk is algebraic.

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