# Commutative algebra: Irreducible components of $\operatorname{Spec}(A)$

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?

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

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 $$P=(X+1, Y+1)$$.