contraction of prime ideal is maximal if that prime ideal is maximal but why the ideal is supposed to be prime A theorem from Atiyah says that, if $A \subset B$ is an integral extension, and $P$ is a prime ideal of $B$, then $P$ is maximal if and only if $A \cap P$ is maximal.
I was wondering why do we need $P$ to be prime here.
Consider $A=\mathbb{C}[{x^2}]$ and $B=\mathbb{C}[{x}]$. Then $x$ is integral over $A=\mathbb{C}[{x^2}]$ and hence $B$ is integral over $A$.
Now consider the ideal $P=(x^2) \subset \mathbb{C}[{x}]$ and its contraction in $\mathbb{C}[{x^2}]$ is $A\cap P=(x^2)$, then it is clear that $A \cap P$ is maximal but the ideal $P$ is not even a prime ideal.
Does this look completely correct?
 A: Look at what the claim says:

then $P$ is maximal if and only if $A \cap P$ is maximal.

You've shown that $A\cap P$ is maximal in $A$, but $(x^2)\lhd B$ is not prime and clearly has no chance of being maximal either.  So the hypothesis wasn't satisfied and the conclusion wasn't true either: no surprises here.

As a side note, when I saw the question it looked to me as if an error in logic was unfolding.  What I mean is it looked like this:

"I've read a theorem that if $P$ is prime, then statement Q holds.  Why do we need $P$ to be prime? Let me show you an example where $P$ is not prime but statement Q holds..."

To which one should say "When $P\implies Q$ you don't need $P$ to hold for $Q$ to hold: it might hold anyway whether or not $P$ holds.  If it is a square it is a rectangle, but I don't need it to be a square for it to be a rectangle.
This need or more accurately necessity expresses the opposite implication: $Q\implies P$.  In that case, we say $P$ is necessary for $Q$, because $Q$ can't hold without $P$ holding also.
I'm not sure if that is what you were attempting, but I figured I would mention it.
