# How to prove every radical ideal is a finite intersection of prime ideals?

Let $$k$$ be an algebraically closed field, $$\mathbb{A}^n(k)$$ the affine space corresponding to $$k[x_1, \dots, x_n]$$, which is Noetherian by the Hilbert Basis Theorem.

We know that any algebraic set $$V$$ can be written as the finite union of irreducible algebraic varieties, $$V_1 \cup \dots \cup V_k$$, no one containing the other (See Hartshorne, Algebraic Geometry, Corollary I.1.6, p. 5). By a variant of the Strong Nullstellensatz, there is a one-to-one correspondence between radical ideals in $$k[x_1, \dots, x_n]$$ and algebraic sets in $$\mathbb{A}^n(k)$$. Likewise, corresponding to the fact that every prime ideal is radical, there is a one-to-one correspondence between prime ideals in $$k[x_1, \dots, x_n]$$ and algebraic varieties in $$\mathbb{A}^n(k)$$.

Question: Thus we know that for any algebraic set $$V$$, $$Z(I) = V = V_1 \cup \dots \cup V_k = Z(I_1) \cup \dots \cup Z(I_k) \\ \implies Z(I) = Z(I_1) \cup \dots \cup Z(I_k)$$ where $$Z(S)$$ denotes the zero set of a collection of polynomials, $$I$$ is a radical ideal in $$k[x_1, \dots, x_n]$$ and $$I_1, \dots, I_k$$ are prime ideals in $$k[x_1, \dots, x_n]$$.

How do we use this to show that $$I = I_1 \cap \dots \cap I_k$$ which is a special case of the Lasker-Noether theorem when all ideals involved are radical? (Since every prime ideal is the radical of a primary ideal, and again $$k[x_1, \dots, x_n]$$ is Noetherian.)

My attempt: See my community wiki "answer" for what I have tried so far -- this question is already too long due to context.

• This statement is pure commutative algebra, maybe thinking about varieties can cause confusion. If you have a radical ideal $I$ that fails to be prime, then for some $xy \in I$ you have $x\notin I$ and $y\notin I$. Write $I$ as an intersection of two radical ideals $J \cap J'$ where $x\in J$ and $y\in J$. Continue this procedure if some of these ideals fails to be prime. This process should terminate, since otherwise you would get an infinite ascending chain of ideals.
– user144221
Commented Oct 12, 2016 at 13:08
• @Alejo This explanation is helpful and much clearer than any other proof of the Lasker-Noether theorem I have seen online. Plus the result is more general (for all Noetherian rings), as opposed to just polynomial rings over algebraically closed fields. See if you want if you think my attempted proof is correct: math.stackexchange.com/a/1968707/327486 Commented Oct 14, 2016 at 19:30

This is actually fairly simple. Let $$I$$ be a radical ideal in $$k[x_1, \dots, x_n]$$, $$k$$ algebraically closed, then it corresponds to a unique algebraic set $$V \subseteq \mathbb{A}^n(k)$$, $$V=Z(I)$$.

So then $$V = V_1 \cup \dots \cup V_k$$, where $$V_k$$ are algebraic varieties corresponding uniquely to prime ideals $$I_1, \dots, I_k$$. Thus $$Z(I) = Z(I_1) \cup \dots \cup Z(I_k)$$. Now one can show that when taking the union of algebraic sets corresponding to arbitrary ideals, that $$Z(I_1)\cup \dots \cup Z(I_k) = Z(I_1 \cdots I_k)$$ where the product is the product of ideals. Also one has in general that $$Z(J)= Z(\sqrt{J})$$, so that $$Z(I_1)\cup \dots \cup Z(I_k) = Z(I_1 \cdots I_k)=Z\left(\sqrt{I_1 \cdots I_k}\right)$$ Now in general one only has that $$I_1\cdots I_k \subseteq I_1 \cap \dots \cap I_k$$, but taking the radical of the product allows us to conclude (see Lemma 1.7 here and then use induction): $$Z\left(\sqrt{I_1 \cdots I_k}\right)= Z\left( \sqrt{I_1 \cap \dots \cap I_k} \right) = Z\left( \sqrt{I_1} \cap \dots \cap \sqrt{I_k} \right).$$ However, since the $$V_i$$ are algebraic varieties, all of the $$I_i$$ were prime, thus radical, i.e. for all $$i$$ we have $$\sqrt{I_i}=I_i$$, therefore we have actually shown that: $$Z(I_1) \cup \dots \cup Z(I_k) =Z\left(\sqrt{I_1 \cdots I_k}\right) = Z(I_1 \cap \dots \cap I_k)$$ Now since $$\sqrt{I_1 \cdots I_k} = I_1 \cap \dots \cap I_k$$, the former obviously being radical since taking the radical of an ideal is idempotent operation, the latter must also be a radical ideal, and therefore our one-to-one correspondence between zero sets and radical ideals (from the Strong Nullstellensatz) applies, in particular we have that $$Z(I)=Z(I_1)\cup \dots \cup Z(I_k)=Z(I_1 \cap \dots \cap I_k) \iff I = I_1 \cap \dots \cap I_k.$$ Thus starting with an arbitrary radical ideal in $$k[x_1,\dots,x_n]$$, we have returned a (in fact unique) intersection of prime ideals equaling it, thereby proving this special case of the Lasker-Noether theorem. $$\square$$

Proof of the lemma (in case the link goes dead in the future)

The fact that $$\sqrt{I \cdot J} \subseteq \sqrt{I \cap J}$$ follows from the fact that $$I \cdot J \subseteq I \cap J$$.

If $$a \in \sqrt{I \cap J}$$ then $$a^n \in I \cap J$$ for some $$n\in \mathbb{N}$$, thus both $$a^n \in I$$ and $$a^n \in J$$ are true, and thus $$a \in \sqrt{I} \cap \sqrt{J}$$. Thus $$\sqrt{I \cap J} \subseteq \sqrt{I} \cap \sqrt{J}$$.

Now if $$a \in \sqrt{I} \cap \sqrt{J}$$, then $$a^n \in I$$, $$a^m \in J$$, for some $$m,n \in \mathbb{N}$$. Therefore $$a^{m+n} \in I \cdot J$$, so $$\sqrt{I} \cap \sqrt{J} \subseteq \sqrt{I \cdot J}$$. $$\square$$

– Shaun
Commented Sep 19, 2022 at 14:37
• @Shaun: I replaced the broken link with one that seems to match the context. Also note the short proof given at the bottom of the Answer. Commented Oct 5, 2022 at 16:02

Let's write out the definitions and see if this gives us any insights: $$Z(I) = Z(I_1) \cup \dots \cup Z(I_k) \\\iff \\ \{ x \in \mathbb{A}^n(k): \forall p \in I,\ p(x)=0 \} \\= \{ x \in \mathbb{A}^n(k): \forall p_1 \in I_1, \ p_1(x)=0 \} \cup \dots \cup \{x \in \mathbb{A}^n(k): \forall p_k\in I_k,\ p_k(x)=0 \} \\ \ \\ = \{ x \in \mathbb{A}^n(k): (\forall p_1 \in I_1,\ p_1(x)=0)\lor\dots\lor(\forall p_k \in I_k,\ p_k(x)=0 ) \}$$

One has that $$(\forall p_1 \in I_1,\ p_1(x)=0)\lor(\forall p_12 \in I_2,\ p_2(x)=0) \implies (\forall p \in (I_1 \cap I_2),\ p(x)=0)$$ since obviously $I_1 \cap I_2 \subset I_1, I_1 \cap I_2 \subset I_2$ together imply that $$(\forall p_1\in I_1,\ p_1(x)=0) \implies (\forall p \in (I_1 \cap I_2),\ p(x)=0) \\ (\forall p_2\in I_2,\ p_2(x)=0) \implies (\forall p \in (I_1 \cap I_2),\ p(x)=0)$$

This shows that $$\left[(\forall p_1 \in I_1,\ p_1(x)=0)\lor\dots\lor(\forall p_k \in I_k,\ p_k(x)=0 )\right] \implies (\forall p \in (I_1 \cap \dots \cap I_k),\ p(x)=0)$$ So then by this we should have that $$\{x \in \mathbb{A}^n(k): (\forall p_1 \in I_1,\ p_1(x)=0)\lor\dots\lor(\forall p_k \in I_k,\ p_k(x)=0 ) \} \subseteq \{ x: \mathbb{A}^n(k): (\forall p \in (I_1 \cap \dots \cap I_k),\ p(x)=0) \}$$ In other words (other symbols): $$Z(I) = Z(I_1) \cup \dots \cup Z(I_k) \subseteq Z(I_1 \cap \dots \cap I_k)$$ So, if we can show that $Z(I_1 \cap \dots \cap I_k) \subseteq Z(I)$, and thus further that $Z(I)=Z(I_1 \cap \dots \cap I_k)$, we can parlay our one-to-one correspondence between algebraic sets and radical ideals to reach our desired conclusion: $I = I_1 \cap \dots \cap I_k$.

Update: What I showed above is that $I_1 \cap \dots \cap I_k \subseteq I = I_1 \cdots I_k$, which I thought was the easy direction. However, it is actually the hard direction, since the analogous result is not true in general -- the result which is true in general is that $I_1 \cdots I_k = I \subseteq I_1 \cap \dots \cap I_k$. Thus I had essentially already shown the result previously, despite my not realizing that that was the case.

Alternative Proof of the First direction:

Because everything involved is algebraic/radical, we have that $I(V)=I(Z(I))=I$ always, where $I$ is the ideal generated by the set (you'll see the definition shortly if you don't know it already).

Anyway so we have that $$I(V) = I ( V_1 \cup \dots \cup V_k ) \\ \iff \\ \{ p \in k[x_1, \dots, x_n]: \forall x \in V,\ p(x)=0 \} = \{ p \in k[x_1, \dots, x_n]: \forall x\in (V_1 \cup \dots \cup V_k),\ p(x)=0 \}$$

The key point turns out to be that $V_1 \cup \dots \cup V_k \subseteq Z(I_1 \cap \dots \cap I_k)$, we have that since $x \in (V_1 \cup \dots \cup V_k) \implies x \in Z(I_1 \cap \dots \cap I_k)$, that $$(\forall x\in Z(I_1 \cap \dots \cap I_k))\ A(x) \implies (\forall x \in (V_1 \cup \dots \cup V_k))\ A(x)$$ and then applying this nameless principle, $$\{p \in k[x_1, \dots, x_n]: \forall x\in Z(I_1 \cap \dots \cap I_k),\ p(x)=0 \} \\ \subseteq \{ p\in k[x_1,\dots,x_n]:\forall x \in (V_1 \cup \dots \cup V_k),\ p(x)=0 \} \\ = \{p\in k[x_1,\dots,x_n]:\forall x \in V,\ p(x)=0 \}$$ in other words $$I_1 \cap \dots \cap I_k = I(Z(I_1 \cap \dots \cap I_k)) \subseteq I(V_1 \cup \dots \cup V_k) = I(V) = I \\ I_1 \cap \dots \cap I_k \subseteq I \iff Z(I) \subseteq Z(I_1 \cap \dots \cap I_k)$$

Update: What I showed above is that $I_1 \cap \dots \cap I_k \subseteq I = I_1 \cdots I_k$, which I thought was the easy direction. However, it is actually the hard direction, since the analogous result is not true in general -- the result which is true in general is that $I_1 \cdots I_k = I \subseteq I_1 \cap \dots \cap I_k$. Thus I had essentially already shown the result previously, despite my not realizing that that was the case.

Furthermore

$I = \langle f_1, \dots, f_r \rangle$ because all ideals are finitely generated. Then so basically we want to show that $$\langle f_1, \dots, f_r \rangle = \bigcap_{i=1}^k \langle f^i_1, \dots, f^i_{r_i} \rangle$$ where $I_i = \langle f^i_1, \dots, f^i_{r_i} \rangle$.

Now given $V_1 = Z(f_1^1,\dots, f_{r_1}^1), V_2=Z(f_1^2,\dots,f_{r_2}^2)$, we have that $$V_1 \cup V_2 = Z(f_1^2 \cdot f_1^1,\dots, f_1^2 \cdot f_{r_1}^1, \dots, f_{r_2}^2 \cdot f_1^1, \dots, f_{r_2}^2 \cdot f_{r_1}^1).$$ So do we have that, and if so why, $$\langle f_1^2 \cdot f_1^1,\dots, f_1^2 \cdot f_{r_1}^1, \dots, f_{r_2}^2 \cdot f_1^1, \dots, f_{r_2}^2 \cdot f_{r_1}^1 \rangle = \langle f_1^1,\dots, f_{r_1}^1 \rangle \cap \langle f_1^2,\dots,f_{r_2}^2 \rangle ???$$

EDIT: FINALLY FOUND A SOLUTION

Product of ideals corresponding to vanishing of points is equal to their intersection

Apparently that complicated formula I wrote is just the product of ideals. Also, all that I showed above is that the product of ideals is a subset of the intersection of ideals, something which is always true. However, the intersection of ideals is a subset of the product only in certain cases, including this one, according to Proposition 1.1.10 in Atiyah-MacDonald, Introduction to Commutative Algebra.

That this holds appears to be a special case of the fact that the intersection of radical ideals is equal to the product of radical ideals, see here -- a proof that this is true can found in Lemma 1.7 here.

However, I am not sure if the radical of the product of ideals is equal to the product of the radicals of ideals, see here.

A bunch of links about the product and intersection of finitely generated ideals, possibly in Noetherian rings:

When is the product of ideals equal to their intersection -- apparently a question that comes up with schemes as well: https://mathoverflow.net/questions/49259/when-is-the-product-of-two-ideals-equal-to-their-intersection