[proofplan] Both identities are proved by double containment, working directly from the definitions of the ideal operations $+$ and $\cap$ and of the vanishing locus operator $V$. For (1) we use that a sum $I + J$ contains both $I$ and $J$ and that every element of $I + J$ is a sum $f + g$ with $f \in I$, $g \in J$. For (2) the inclusion $V(I) \cup V(J) \subset V(I \cap J)$ is monotonicity of $V$, and the reverse uses that $I \cdot J \subset I \cap J$ together with the integral domain property of $k$ to argue that if $V(I)$ misses a point $p$ then $V(J)$ must contain it. [/proofplan] [step:Establish monotonicity and reversal of $V$ under inclusion] We first record the only general fact about $V$ that the proof uses. Let $A \subset B$ be subsets of $k[X_1, \ldots, X_n]$. Then \begin{align*} V(B) = \{p \in \mathbb{A}^n_k : f(p) = 0 \text{ for all } f \in B\} \subset \{p \in \mathbb{A}^n_k : f(p) = 0 \text{ for all } f \in A\} = V(A), \end{align*} since requiring vanishing on a larger set $B$ is a stronger condition than vanishing on $A$. Thus $V$ is order-reversing: $A \subset B \implies V(B) \subset V(A)$. We use this throughout. [/step] [step:Prove $V(I + J) = V(I) \cap V(J)$ by double containment] Recall that the sum of ideals is \begin{align*} I + J = \{f + g : f \in I,\ g \in J\}, \end{align*} and that $I \subset I + J$ and $J \subset I + J$ (taking $g = 0$ or $f = 0$ respectively). For the inclusion $V(I + J) \subset V(I) \cap V(J)$: from $I \subset I + J$ and $J \subset I + J$, the order-reversing property of $V$ gives $V(I + J) \subset V(I)$ and $V(I + J) \subset V(J)$, so $V(I + J) \subset V(I) \cap V(J)$. For the reverse inclusion $V(I) \cap V(J) \subset V(I + J)$: let $p \in V(I) \cap V(J)$, so $f(p) = 0$ for every $f \in I$ and $g(p) = 0$ for every $g \in J$. An arbitrary element of $I + J$ has the form $f + g$ with $f \in I$, $g \in J$. Evaluating at $p$, \begin{align*} (f + g)(p) = f(p) + g(p) = 0 + 0 = 0, \end{align*} so $p \in V(I + J)$. Combining the two inclusions, $V(I + J) = V(I) \cap V(J)$. [guided] We are asked to identify the vanishing locus of the sum-ideal $I + J$ with the intersection of the individual vanishing loci. The slogan is: "vanishing on more polynomials = vanishing at fewer points". Sums of ideals enlarge the constraint set on polynomials, so they shrink the constraint set on points to an intersection. We prove $V(I + J) = V(I) \cap V(J)$ by showing each side is contained in the other. \textbf{Forward inclusion} $V(I + J) \subset V(I) \cap V(J)$. We use the order-reversing property of $V$ from the previous step. The sum-ideal contains both summands: $I \subset I + J$ because $f = f + 0$ with $0 \in J$, and similarly $J \subset I + J$. Reversing inclusions under $V$ yields $V(I + J) \subset V(I)$ and $V(I + J) \subset V(J)$, hence \begin{align*} V(I + J) \subset V(I) \cap V(J). \end{align*} Why does this work? A point in $V(I + J)$ kills every element of $I + J$, in particular every element of $I$ and every element of $J$, so it lies in both $V(I)$ and $V(J)$. \textbf{Reverse inclusion} $V(I) \cap V(J) \subset V(I + J)$. This direction is the substantive content. Take $p \in V(I) \cap V(J)$. By definition, every $f \in I$ satisfies $f(p) = 0$, and every $g \in J$ satisfies $g(p) = 0$. We must check that every element of $I + J$ also vanishes at $p$. Recall that an element of the sum-ideal is by definition a sum of an element of $I$ and an element of $J$: \begin{align*} I + J = \{f + g : f \in I,\ g \in J\}. \end{align*} Take any such element $f + g$. Polynomial evaluation at $p$ is a ring homomorphism $\mathrm{ev}_p : k[X_1, \ldots, X_n] \to k$, and hence respects addition: \begin{align*} (f + g)(p) = \mathrm{ev}_p(f + g) = \mathrm{ev}_p(f) + \mathrm{ev}_p(g) = f(p) + g(p) = 0 + 0 = 0. \end{align*} So $p$ kills every element of $I + J$, i.e. $p \in V(I + J)$. Combining: $V(I + J) = V(I) \cap V(J)$, as required. [/guided] [/step] [step:Prove $V(I) \cup V(J) \subset V(I \cap J)$ by monotonicity] The intersection of ideals satisfies $I \cap J \subset I$ and $I \cap J \subset J$. By the order-reversing property of $V$ (Step 1), $V(I) \subset V(I \cap J)$ and $V(J) \subset V(I \cap J)$. Taking the union gives \begin{align*} V(I) \cup V(J) \subset V(I \cap J). \end{align*} [/step] [step:Prove $V(I \cap J) \subset V(I) \cup V(J)$ via the integral-domain argument] We must show: if $p \in V(I \cap J)$, then $p \in V(I)$ or $p \in V(J)$. We argue by cases on whether $p \in V(I)$. If $p \in V(I)$, there is nothing to prove. Otherwise, $p \notin V(I)$, so there exists $g \in I$ with $g(p) \neq 0$. We claim $p \in V(J)$. Take any $f \in J$. The product satisfies $f g \in I \cap J$: indeed, $g \in I$ implies $f g \in I$ (since $I$ is an ideal and absorbs multiplication), and $f \in J$ implies $f g \in J$ (since $J$ is an ideal). Hence $f g \in I \cap J$, and the hypothesis $p \in V(I \cap J)$ gives $(fg)(p) = 0$. Therefore \begin{align*} f(p)\, g(p) = (fg)(p) = 0. \end{align*} Since $g(p) \neq 0$ and $k$ is a field (hence an integral domain), the relation $f(p) g(p) = 0$ forces $f(p) = 0$. As $f \in J$ was arbitrary, $p \in V(J)$. Combining with the previous step, $V(I \cap J) = V(I) \cup V(J)$. [guided] This is the only step where we use a non-formal property — namely that $k$ has no zero divisors — and the proof structure is dichotomy on the negation. We need to show: any point $p \in V(I \cap J)$ lies in $V(I) \cup V(J)$, equivalently, in at least one of $V(I)$, $V(J)$. \textbf{The dichotomy.} If $p \in V(I)$ we are done immediately. Otherwise we must show $p \in V(J)$ — every $f \in J$ kills $p$. \textbf{Setting up the obstruction.} Since $p \notin V(I)$, the definition of the vanishing locus gives a witness: there is at least one polynomial $g \in I$ with $g(p) \neq 0$. Fix this $g$ for the remainder of the argument. Why is having such a witness useful? Because $g$ provides a "non-zero scalar" at the point $p$, and we will use this to cancel. \textbf{Producing elements of $I \cap J$.} How do we leverage the hypothesis $p \in V(I \cap J)$? We need elements of $I \cap J$ to feed into it. Take any $f \in J$. Then $fg \in I \cap J$, because: \begin{itemize} \item $fg \in I$: $g \in I$ and $I$ is an ideal, so $I$ absorbs multiplication on either side — in particular $f \cdot g \in I$. \item $fg \in J$: $f \in J$ and $J$ is an ideal, so $g \cdot f \in J$, and by commutativity $fg \in J$. \end{itemize} This is the ideal-theoretic identity $I \cdot J \subset I \cap J$ in disguise: products of an element of $I$ with an element of $J$ always land in the intersection. \textbf{Using the hypothesis.} Since $fg \in I \cap J$ and $p \in V(I \cap J)$, the polynomial $fg$ vanishes at $p$: \begin{align*} (fg)(p) = f(p)\, g(p) = 0. \end{align*} We have written this as a product of two elements of $k$. \textbf{Cancelling using the integral domain property.} The polynomial ring $k[X_1, \ldots, X_n]$ takes values in $k$, and $k$ is a field. In a field, $a \cdot b = 0$ implies $a = 0$ or $b = 0$. We have $f(p) g(p) = 0$ and $g(p) \neq 0$, so the field axioms force $f(p) = 0$. This held for an arbitrary $f \in J$, so every element of $J$ vanishes at $p$, i.e. $p \in V(J)$. \textbf{Why we needed $k$ to be an integral domain.} Over a ring with zero divisors, $f(p) g(p) = 0$ does not imply $f(p) = 0$ or $g(p) = 0$, so the cancellation step would fail. The result $V(I \cap J) = V(I) \cup V(J)$ uses the integrality of $k$ in an essential way; it is not a purely formal consequence of $V$ being order-reversing. [/guided] [/step]