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.
Best Answer
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 $\sqrt{I} \cap \sqrt{J} \subseteq \sqrt{I \cdot J}$. $\square$