I am getting very confused by the definition of a minimal prime ideal

Question

I am trying to prove that in an affine scheme $\operatorname{Spec}A$ that an irreducible component can be written as the vanishing locus $V(\mathfrak p)$ for a minimal prime ideal $\mathfrak p$, but I am getting very confused.

So if $I$ is a minimal prime ideal over an ideal $J$ then it is the smallest prime ideal such that $J\subset I$, i.e. if $\mathfrak p$ is a prime ideal containing $J$ then $I\subset \mathfrak p$.

It is easy to see that every irreducible closed subset of $\operatorname{Spec}A$ corresponds to $V(I)$ with $I$ prime, but isn't every prime minimal over itself? So what does this even mean? Because it seems to imply to me that every prime ideal corresponds to an irreducible component of $\operatorname{Spec}A$, but this can be true because something like $A=\mathbb C[x]$ has infinitely many primes, but in this case $\operatorname{Spec}A$ is a Noetherian scheme so has finitely many irreducible components.

Wikipedia says that a prime ideal is minimal when it is minimal over the zero ideal, but then if $I$ is prime doesn't this mean that $I\subset \mathfrak p$ for every prime ideal of $A$ so $V(I)=\operatorname{Spec}A$?

I suspect I am messing something up here...

Answer 1

i.e. if $\mathfrak p$ is a prime ideal containing $J$ then $I⊂\mathfrak p$.

That is not correct. $I$ being a minimal prime over $J$ means that there is no prime $\mathfrak q$ with $$J \subset \mathfrak q \subsetneq I.$$ But it doesn't mean that $I$ is a minimum in the set $V(I)$ of all prime ideals which contain $I$.

Because it seems to imply to me that every prime ideal corresponds to an irreducible component of $\operatorname{Spec}A$

You always have $V(0) = \operatorname{Spec} A$. So the minimal prime ideals over $0$ correspond to the irreducible components of $\operatorname{Spec} A$. Since every prime ideal contains $0$, this means the irreducible components correspond to the minimal primes of $A$.

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Here is an example. If you take $A = \mathbb C[x,y] / (xy)$, then the minimal primes of $A$ are $(x)$ and $(y)$.