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Let $A \subset B$ be a finite extension of integral domains. Let $(a)$ be a principal ideal in $A$. If $(a)^e = aB$ is a prime ideal of $B$, does it follow that $(a)$ is prime in $A$ ?

I have a related question, but the given example doesn't work.

user26857
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Alphonse
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    Another example is given in this comment: https://math.stackexchange.com/questions/2980727/contraction-of-non-prime-ideals-in-integral-extensions/2980788#comment6152419_2980788 – Alphonse Nov 02 '18 at 07:55

1 Answers1

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I don't think so. Take $A = k[x,xy,xy^2,y^3] \subset B = k[x,y]$. The ideal $xA$ is not prime, but $xB$ is; $x(xy^2) = (xy)^2$, but $xy \notin xA$ in $A$.

user26857
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Youngsu
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