Consider the scheme $X:=\mathrm{Spec}(k[X,Y]/(X^2,XY))$. According to Qing Liu's "Algebraic geometry and arithmetic curves", the irreducible components are in $1-1$ correspondence with subschemes of the form $V(\mathcal{P})$ where $\mathcal{P}$ is a minimal prime ideal of $k[X,Y]/(X^2,XY)$. If I understand correctly, this means $X$ has exactly one irreducible component, in particular the subscheme $V(X)$ (as every prime contains the ideal generated by $X$). Is this correct?

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    $\begingroup$ I think you are right. If I recall correctly $V(f,g) = V(f) \cap V(g)$ and $V(fg) = V(f) \cup V(g)$. Then $\text{Spec}(k[X,Y]/(X^2,XY)) \leftrightarrow V(X^2, XY) = V(X^2) \cap (V(X) \cup V(Y)) = V(X)$. $\endgroup$
    – A.P.
    Mar 27, 2015 at 16:21

1 Answer 1


Yes, your scheme $X$ has exactly one irreducible component - i.e., it is an irreducible scheme. Note that an affine scheme $\operatorname{Spec} R$ is irreducible if and only if the nilradical of $R$ is prime. Here, the nilradical of $k[X,Y]/(X^2,XY)$ is $(\overline{X})$ (where $\overline{X}$ denotes the image of $X$ in $k[X,Y]/(X^2,XY)$) and this is a prime ideal since $(k[X,Y]/(X^2,XY))/(\overline{X}) \cong k[Y]$. However, you should be careful to understand your claim that "the subscheme $V(X)$" is the irreducible component of $X$ correctly.

Let me explain. There is a natural way to view $X$ as a closed subscheme of $\mathbb{A}_k^2 = \operatorname{Spec} k[X,Y]$. The underlying set of this subscheme contains precisely those prime ideals $\mathfrak{p}$ of $k[X,Y]$ such that $\mathfrak{p} \supset (X^2,XY)$. As A.P.'s comment to your question shows, we have $$ \mathfrak{p} \supset (X^2,XY) \Leftrightarrow \mathfrak{p} \supset (X) .$$ It is customary to express this by writing $$ V(X) = V(X^2,XY), $$ which is perfectly fine as long as you understand this as an equality of sets only (of course, the subspace topology induced by the Zariski topology on $\mathbb{A}_k^2$ then agrees as well, which is why you could also view the above equality as an equality of topological spaces).

However, sometimes $V(X)$ is not used to denote a subset, but to denote a subscheme (and writing "the subscheme $V(X)$" suggests just that) - here, that would be the scheme $\operatorname{Spec} k[X,Y]/(X)$. But $\operatorname{Spec} k[X,Y]/(X^2,XY)$ and $\operatorname{Spec} k[X,Y]/(X)$ are not isomorphic as schemes. In fact, there is a bijective correspondence between the closed subschemes of any affine scheme $\operatorname{Spec} R$ and the ideals of $R$ - but clearly, $(X^2, XY) \neq (X)$.

(Well, $\operatorname{Spec} k[X,Y]/(X)$ is [naturally isomorphic to] a closed subscheme of $\operatorname{Spec} k[X,Y]/(X^2,XY)$ and, strictly speaking, it is true that $\operatorname{Spec} k[X,Y]/(X)$ is the irreducible component of $\operatorname{Spec} k[X,Y]/(X^2,XY)$ because the underlying topological spaces of those two schemes agree. But, at least to me, this seems to be a rather strange way of asserting that $\operatorname{Spec} k[X,Y]/(X^2,XY)$ is irreducible.)


I now feel a bit silly for writing this pedantic and probably superfluous remark so let me add another comment which might be of more interest to you: While the scheme $X$ is irreducible, it has an embedded prime. For this reason, it might help you to understand the "geometric meaning" of associated primes and primary decompositions - in particular, in what sense they yield more refined information than the decomposition into irreducible components.

  • $\begingroup$ Thank you very much for the detailed answer. It is very helpful. $\endgroup$
    – user45397
    Mar 27, 2015 at 21:27

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