Let $K\hookrightarrow K'$ be a regular extension of fields, and $K[x_{1},\cdots,x_{n}]\hookrightarrow K'[x_{1},\cdots,x_{n}]$ the corresponding ring extension. Does every prime ideal of the first ring expand to a prime ideal in the second?
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The answer is Yes.
A field extension $K \to K'$ is regular if and only if for every $K$-algebra $A$ which is an integral domain the $K'$-algebra $A' := A \times_K K'$ is an integral domain (Bourbaki, Algebra, Chap. V, §17, No.3). Applying this to $A = K[x_1,\dots,x_n]/{\mathfrak p}$ for a prime ideal ${\mathfrak p}$ we obtain $A' = K'[x_1,\dots,x_n]/{\mathfrak p}K'[x_1,\dots,x_n]$. Thus ${\mathfrak p}K'[x_1,\dots,x_n]$ is a prime ideal.
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$\begingroup$ Yes, and it is an astonishing thing that this fact doesn't appear in other classical books of algebra (I went to Lang's one, and he shows the result only for finite extensions and certain kind of primes. Wikipedia doesn't make any reference to things like this, and other sources don't even talk about regular extensions). Thanks for the reference. $\endgroup$ Mar 30, 2011 at 16:31