Let $k$ be a field of characteristic other than $2$. Consider the ring $$A:= k[X,Y]/(X(Y+1),X(Y+X^2)).$$ Describe all the irreducible components of $\operatorname{Spec}(A)$.

It does not look that hard, but I could not solve this on my exam on commutative algebra.

I know that if $\mathfrak{p}\in A$ is a minimal prime ideal, then $V(\mathfrak{p}) = \{\mathfrak{q} \text{ prime ideal } ∣ \mathfrak{p}\subseteq\mathfrak{q} \}$ is an irreducible component of $\operatorname{Spec}(A)$. But how can I use this on a quotient ring?

  • $\begingroup$ The ideals of $k[X,Y]/I$ correspond to the ideals of $k[X,Y]$ containing $I$... $\endgroup$
    – Tuvasbien
    Mar 18 at 14:47

1 Answer 1


The minimal primes of $A$ correspond to the minimal primes over $I$. If $P$ is a minimal prime containing $I$, then $X(Y+1)\in P$, so either $X\in P$ or $Y+1\in P$. If $X\in P$, then $I\subseteq (X)\subseteq P$, so by minimality, $P=(X)$.

If $X\notin P$, then $Y+1, Y+X^2\in P$, so $I\subseteq (Y+1, Y+X^2)=(X^2-1, Y+1)\subseteq P$. Since $X^2-1\in P$, either $X-1\in P$ or $X+1\in P$, and once again by minimality, we conclude that either $P=(X-1,Y+1)$ or $P=(X+1, Y+1)$.

  • $\begingroup$ (X) is minimal prime containing I, so (X)/I is minimal prime of A and V ((X)/I) is irreducible component of Spec A? $\endgroup$
    – Nik1987
    Mar 22 at 18:36
  • $\begingroup$ Yes, that's correct. $\endgroup$
    – cqfd
    Mar 22 at 23:34

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