0

Let $A$ be a Noetherian domain of dimension 1 with infinitely many maximal ideals. Let $f \in A[y]$. Show that $(f)$ is not a maximal ideal of $A[y]$.

Since $A$ is noetherian, Krull dimension of $A[y]$ is 2, so not every prime ideal is maximal. For an arbitrary field $k$, also the ideal of $k[x,y]$ generated by $f$, where $f \in$ $k[x,y]$ is any non constant polynomial, is not maximal. Any idea for the proof ? Thanks

user26857
  • 52,094
user737722
  • 17
  • 3
  • If $A[y]/(f(y))$ is a field and $f$ is non-constant, with $a$ its leading coefficient then $A[a^{-1}]$ is a field, then look at the primary decomposition of $aA$ ? When $A=k[x]$ it reduces to that $k[x,h(x)^{-1}]$ is not a field. – reuns Dec 27 '19 at 19:57
  • @reuns If f is monic, then how can we show that A is a field provided that A[y]/(f) is a field? – user26857 Dec 30 '19 at 00:24

0 Answers0