coquelicot said:
Let R be an integral ring (eventually can be supposed integrally closed), and R' an integral extension of R. Assume that M is a maximal ideal of R'. Must the intersection of M with R be a maximal ideal of R ?
Thx.
This follows from the following two lemma's:
Lemma 1: If ##R\subseteq R^\prime## are rings such that ##R^\prime## is integral over ##R##, and if ##J## is an ideal of ##R^\prime## and if ##I=R\cap J##, then ##R^\prime/J## is integral over ##R/I##.
Proof: Take ##x\in R^\prime##, then we have some equation
x^n + a_1x^{n-1} + ... + a_n = 0
with ##a_i \in R##. Reducing this modulo ##J## then yields that ##x+J## is integral over ##R/I##.
Lemma 2: Let ##R\subseteq R^\prime## be integral domains such that ##R^\prime## is integral over ##R##. Then ##R## is a field if and only if ##R^\prime## is a field.
Proof: If ##R## is a field, then let ##y\in R^\prime## be nonzero. Then there is some equation
y^n + a_1y^{n-1} + ... + a_n = 0
with ##a_i \in R##. We can take this equation of smallest possible degree. But then
y^{-1} = -a_n^{-1}(y^{n-1} + a_1 y^{n-2} + ... + a_{n-1})
note that ##a_n\neq 0## because of the integral domain requirement. Thus ##R^\prime## is a field.
Conversely, if ##R^\prime## is a field and if ##x\in R## is nonzero, then ##x^{-1}## is integral over ##R## and thus satisfies an equation
x^{-n} + a_1x^{-n+1} + ... + a_n=0
It follows that ##x^{-1} = - (a_1 + a_2x + ... + a_nx^{n-1})\in R##.
