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

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