Consider any noetherian ring $A$ and the polynomial ring $A[x]$. Consider the quotient ring $A[x]/\langle x^2+1\rangle$. Is the dimension of this quotient ring equal to dimension of $A$ (i.e. dimension of $A[x] - 1$)?

If there is a chain of ideals in $A[x]$ given by $P_0[x] \subseteq P_1[x] \cdots, P_r[x] \subseteq P_r + \langle x \rangle$. This chain will still be prime in the quotient ring?


1 Answer 1


The ring extension $A\subset A[x]/(x^2+1)$ is integral ($x$ is a root of the monic polynomial $t^2+1\in A[t]$), so $\dim A=\dim A[x]/(x^2+1)$.

Remark. You can replace $x^2+1$ by any monic polynomial.

  • $\begingroup$ thank you! and if there are two variables $A[x,y]/<x^2+1, y^2 + 1>$ then can the same result be applied? $\endgroup$
    – Zoey
    Commented Jun 17, 2015 at 9:56
  • 1
    $\begingroup$ @Anup Yes, of course! ($x$ and $y$ are integral over $A$.) $\endgroup$
    – user26857
    Commented Jun 17, 2015 at 9:58

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