1
$\begingroup$

Let $F$ be a field, $F[X]$ the polynomial ring in one variable and $I$ an ideal of $F[X]$. Then does the quotient map $\pi:F[T]\longrightarrow F[T]/I$ map prime ideals to prime ideals?

2 Answers 2

0

Well if $F$ is a field then $F[x]$ will be what it called a Principal ideal domain (P.I.D). But in a PID all non-zero prime ideals are maximal, so your ideal I will in fact be maximal, so $F[X]/I$ will be a field and this has only two ideals $0$ and itselt, and $0$ is the only prime ideal in a field.

  • 0
    Oh yes sorry i completely missed that :)2012-11-06
1

No.
If $I=\langle T^2\rangle $ , then the image of the prime ideal $\mathfrak p=\langle 0\rangle\subset F[T]$ under $\pi:F[T]\longrightarrow F[T]/I=F[T]/\langle T^2\rangle$ is of course the zero ideal $\overline {\mathfrak p}=\langle \bar 0\rangle\subset F[T]/\langle T^2\rangle$, which is not prime since $F[T]/\langle T^2\rangle$ is not a domain.