# Verifying Easy Properties, or Nowhere, Going Nowhere

Whenever I decide to learn something – and especially when it’s learning for learning’s sake – I make sure to be meticulous with things. In particular, whenever I see propositions stated without proof, I break out the old pen and paper and start verifying.

The purpose of this post is to examine a few of the properties on page 661 of Dummit and Foote. Some of the background notation needed is discussed in this previous entry.

Claim. The following properties of the map $\mathcal{I}$ are very easy exercises. Let $A$ and $B$ be subsets of $\mathbb{A}^n$.
(7) $\mathcal{I}(A\cup B)=\mathcal{I}(A)\cap\mathcal{I}(B)$.
(9) If $A$ is any subset of $\mathbb{A}^n$, then $A\subseteq\mathcal{Z}(\mathcal{I}(A))$, and if $I$ is any ideal, then $I\subseteq\mathcal{I}(\mathcal{Z}(I))$.
(10) If $V=\mathcal{Z}(I)$ is an affine algebraic set then $V=\mathcal{Z}(\mathcal{I}(V))$, and if $I=\mathcal{I}(A)$ then $\mathcal{I}(\mathcal{Z}(I))=I$, i.e. $\mathcal{Z}(\mathcal{I}(\mathcal{Z}(I)))=\mathcal{Z}(I)$ and $\mathcal{I}(\mathcal{Z}(\mathcal{I}(A)))=\mathcal{I}(A)$.

Proof. (7) Note that $A\cup B= (A\setminus B)\sqcup (B\setminus A)\sqcup (A\cap B)$. In particular, a polynomial $f$ vanishes on $A\cup B$ if and only if it vanishes on each disjoint component of $(A\setminus B)\sqcup (B\setminus A)\sqcup (A\cap B)$ separately, which happens if and only if it vanishes on the entirety of $A$ and on the entirety of $B$.

(9) Suppose $A\subset\mathbb{A}^n$. Clearly, any polynomial $f$ which vanishes on $A$ is in $\mathcal{I}(A)$, and because $f\in\mathcal{I}(A)$ if and only if $f\equiv 0$ on $A$, $A$ is certainly contained in the zero set of $f$. Therefore, $A\subseteq\mathcal{Z}(\mathcal{I}(A))$. Note that the inclusion doesn’t necessarily reverse since $f(x_1,\ldots,x_n)$ may vanish for some element $(x_1,\ldots,x_n)\not\in A$.

Next, suppose that $I$ is an ideal of the ring $k[x_1,\ldots,x_n]$ and that $f\in I$. By definition, the locus $\mathcal{Z}(I)\subset\mathbb{A}^n$ is the collection of all points $(a_1,\ldots,a_n)\in\mathbb{A}^n$ for which $f(a_1,\ldots,a_n)=0$ for all $f\in \mathcal{I}$. Then, certainly, for all $a=(a_1,\ldots,a_n)\in \mathcal{Z}(I)$, $f(a)=0$, i.e., $f$ is an element in the ideal $\mathcal{I}(\mathcal{Z}(I))$ that vanishes on $\mathcal{Z}(I)$. Hence, $f\in I$ implies that $I\subseteq \mathcal{I}(\mathcal{Z}(I))$.

(10) First, suppose that $I$ is an ideal and that $V=\mathcal{Z}(I)$ is an affine algebraic set. It suffices to show that $V=\mathcal{Z}(\mathcal{I}(V))$ by way of two-sided inclusion. To that end, let $a=(a_1,\ldots,a_n)\in V$. Then $a$ is in the zero-set of the ideal $I$, whereby it follows that $f(a)=0$ for all $f\in I$. But for any $f$ satisfying $f(a)=0$ for arbitrary $a\in V$, $f\in\mathcal{I}(V)$ and $a\in\mathcal{Z}(\mathcal{I}(V))$ by definition. Hence, $V\subset\mathcal{Z}(\mathcal{I}(V))$.

Conversely, if $a\in\mathcal{Z}(\mathcal{I}(V))$, then $f(a)=0$ for all $f\in\mathcal{I}(V)$. But all such functions $f$ disappear for all values $v\in V=\mathcal{Z}(I)$ by definition of $\mathcal{I}(V)$. This means that the polynomials $f\in\mathcal{I}(V)$ for which $f(a)=0$ are precisely the functions which satisfy $f(v)=0$ for all $v\in{Z}(I)$. Hence, $a$ itself must be an element of $\mathcal{Z}(I)=V$, whereby the equality is proved.

The other expression is proved similarly and is omitted for brevity. Therefore, as claimed,

$\mathcal{Z}(\mathcal{I}(\mathcal{Z}(I)))=\mathcal{Z}(I)$ and $\mathcal{I}(\mathcal{Z}(\mathcal{I}(A)))=\mathcal{I}(A)$,

from which it follows that the maps $\mathcal{Z}$ and $\mathcal{I}$ are inverses of one another under the construction given here.    $\square$