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This is part of Exercise 1.2.6(2) on page 4 of Springer's "Linear Algebraic Groups (Second Edition)". The first part is here.

The Details:

Since definitions vary:

A topological space $(X,\tau)$ is a set $\tau$ of subsets of $X$, called closed subsets, such that

  • $\varnothing, X\in\tau$,
  • The intersection $$\bigcap_{i\in I}X_i$$ of any closed subsets $(X_i)_{i\in I}$ is closed, where $I$ is arbitrary, and
  • The union of finitely many closed sets is closed.

Note that $\tau$ is omitted sometimes when the context is clear.

Let $k$ be an algebraically closed field.

We have that the radical $\sqrt{I}$ of an ideal $I$ of $S=k[T_1,\dots, T_n]$ is the ideal of all $f\in S$ with $f^m\in I$ for some $m\in\Bbb N$.

We denote by $\mathcal{V}(I)$ the set of all zeros of $I$.

Let $X\subseteq V=k^n$. Denote by $\mathcal{I}(X)\subseteq S$ the ideal of the $f\in S$ with $f(v)=0$ for all $v\in X$.

Hilbert's Nullstellensatz: For any ideal $I$ of $S$, we have $$\mathcal{I}(\mathcal V(I))=\sqrt{I}.$$

On page 2, 1.2.1 reads,

A topological space $X\neq\varnothing $ is reducible if it is the union of two proper closed subsets. Otherwise $X$ is irreducible. A subset $A\subset X$ is irreducible if it is irreducible for the induced topology. Notice that $X$ is irreducible if and only if any two nonempty open subsets of $X$ have a nonempty intersection.

On page 3, we have

1.2.5 Proposition: A closed subset $X$ of $V$ is irreducible${}^\dagger$ if and only if $\mathcal I(X)$ is a prime ideal.

The Question:

Suppose $I=\sqrt{I}$ is an ideal of $S$ s.t. $$I=P_1\cap \dots\cap P_s$$ for prime ideals $P_i$. (This can be done for all $I$ by the previous part.) Suppose there are no inclusions among the $P_i$. Show that the $P_i$ are uniquely determined up to order.

Thoughts:

Suppose

$$I=Q_1\cap\dots\cap Q_t$$

for prime $Q_j$. Select a $Q_r$ for some $r$. An idea suggested to me is to consider

$$Q_rI=(Q_rP_1)\cap \dots\cap(Q_rP_s);$$

however, I'm not sure where to go with this.

By Hilbert's Nullstellensatz, we have

$$\mathcal{I}(\mathcal V(I))=I.$$

I think we could then use 1.2.5 Proposition somehow.

Please help :)

$\dagger$: . . . with respect to the Zariski topology . . .

Shaun
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  • The idea is that the ideal $P_1$ Is contained in $Q_i$ for some $ i$, and similarly $Q_i \subset P_j$ for some $j$. $P_1 \subset P_j$ implies $P_1=Q_i=P_j$ since there are not inclusions among $P_k$. What you can do for that is, supposing $P_1\nsubseteq Q_i$ for all $i$, to construct an element $f\in P_1$ such that $f\notin Q_i$ to lead to contradiction. – Acrobatic Sep 20 '22 at 14:22
  • Why the downvote? – Shaun Sep 20 '22 at 15:03

1 Answers1

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Suppose $Q$ is any prime ideal that contains $I$. Then $P_1P_2\dots P_n\subseteq P_1\cap P_2\cap\dots\cap P_n= I\subseteq Q$. Since $Q$ is prime, this implies $P_i\subseteq Q$ for some $i$.

In particular, if $Q$ is a minimal prime containing $I$, then $Q$ must be equal to $P_i$ for some $i$, i.e. the $P_i$ include all the minimal primes over $I$. Since every prime containing $I$ contains a minimal such prime and there are no inclusions among the $P_i$, this implies the $P_i$ must be exactly the minimal primes over $I$. So, they are uniquely determined up to order.

Eric Wofsey
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