So the question is to prove that if $X = Z(\mathfrak{U})$ is a Zariski-closed subset in $A^n$, then $Y = Z(\mathfrak{P})$ is an irreducible component of $X$ if and only if $\mathfrak{P}$ is a minimal prime ideal containing $\mathfrak{U}$ in $\mathbb{k}[x_1, \ldots, x_n]$

I get that since $A^n$ is Noetherian, there is a finite decomposition into $X = X_1 \cup X_2\ \cup \cdots \ X_r$, with each $X_i = Z(\mathfrak{Q}_i)$ for some prime $\mathfrak{Q}_i$ in $\mathbb{k}[x_1, \ldots, x_n]$, with (I think) $\mathfrak{U} = \mathfrak{Q}_1 \cap \cdots \cap \mathfrak{Q}_r$.

So my job is to show that if $\mathfrak{P}$ is a minimal prime ideal of this kind then $\mathfrak{P} = \mathfrak{Q}_i$ for some $1 \leq I \leq r$, and that if $Y = X_i$ then $\mathfrak{P}$ is a minimal prime ideal &c &c.

And then I'm stuck. I mean, intuitively I can see why it has to be so - if $\mathfrak{P}$ isn't minimal then there's something in $Z(\mathfrak{P})$ which isn't in $Z(\mathfrak{U})$ - or conversely if there's no $\mathfrak{Q}_i$ such that $\mathfrak{P}$ = $\mathfrak{Q}_i$ then you can't have $Y = X_i$, but I can't see a way to do it formally.

YuiTo Cheng
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Matt Bright
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1 Answers1


I recognise this was asked quite a while ago now, but in case it remains of interest to anyone, here is a proof.

I will assume that $\mathbb{K}$ is algebraically closed and use Hilbert's Nullstellensatz. As you did, write $X = \bigcup_{i=1}^{r} X_i$ for some distinct irreducible components $X_i = Z(\mathfrak{Q}_i)$, and we chose $r$ minimally so that $X_i \not\subset X_j$ whenever $i \neq j$.

Suppose now that $\mathfrak{P}$ is a minimal prime of $R = \mathbb{k}[x_1, \dots, x_n]$ such that $Y = Z(\mathfrak{P})$, then

$$ \prod_{i=1}^{r} \mathfrak{Q}_i \subseteq \bigcap_{i=1}^{r} \mathfrak{Q}_i = \bigcap_{i=1}^{r} I(X_i) = I\left(\bigcup_{i=1}^{r} X_i \right) = I(X) \subseteq I(Y) = \mathfrak{P}. $$

Thus at least one $\mathfrak{Q}_i \subseteq \mathfrak{P}$ since $\mathfrak{P}$ is prime. Now $\mathfrak{Q}_i = I(X_i)$ with $X_i$ irreducible, and so $\mathfrak{Q}_i$ is a prime ideal and is thus a prime ideal contained within the minimal prime ideal $\mathfrak{P}$ and so $\mathfrak{Q}_i = \mathfrak{P}$. But then $Y = X_i$ and so in particular $Y$ is an irreducible component of $X$.

Suppose now that $\mathfrak{P}$ is not a minimal prime ideal in $R$, then since $R$ is a Noetherian ring, it has finitely many minimal prime ideals and every prime ideal of $R$ contains a minimal prime ideal. Thus there exists some minimal prime $\mathfrak{P}'$ of $R$ such that $\mathfrak{P}' \subsetneq \mathfrak{P}$. Thus $Y' = Z(\mathfrak{P}') \subsetneq Y \subseteq X$ is a Zariski closed subset of $X$ with $\mathfrak{P}'$ a minimal prime of $R$, and so $Y'$ is in fact an irreducible component of $X$, $Y' = X_i$ say. Now suppose, for the sake of contradiction, that $Y$ is an irreducible component of $X$, then $Y = X_j$ for some $j$. But then $X_j = Y \subsetneq Y' = X_i$ contradicting the supposition on the $X_i$. Thus $Y$ is not an irreducible component of $X$.

Adam Higgins
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    Thanks! Although I elected to leave Algebraic geometry behind a little bit ago (more in the Algebraic topology direction now, with a view to getting into the theory and applications of TDA) I am determined, in the off hours, to get my head around it. The fact that this proof, which might well have confounded me when I asked the question, seems clear now is heartening..: – Matt Bright May 20 '19 at 18:01
  • There seem to be some problems with your argument in the second paragraph. If $\mathfrak p'\subsetneq\mathfrak p$ then $Z(\mathfrak p)\subsetneq Z(\mathfrak p')$ not the reverse order (although the arguments afterward are pretty much the same). So you really need to make it clear that $\mathfrak p$ and $\mathfrak p'$ contain the ideal of $X$ (maybe you assumed this implicitly in both paragraphs). – GoogleME Dec 28 '21 at 16:42