All statements are easy to check.

See  [ GW1 ] Section (1.3), or  [ Mu ] Ch. I §1, or  [ Alg2 ] 4.3 for proofs based on Noether Normalization, or, for instance, [ AM ] Ch. 5, Ch. 7 for (somewhat) different proofs.

Let \(\mathfrak m\subset k[T_1,\dots , T_n]\) be a maximal ideal. By the theorem, the inclusion \(k\to k[T_1,\dots , T_n]/\mathfrak m\) is a finite field extension, hence – since \(k\) is algebraically closed by assumption – an isomorphism. We define \(t_i\) as the image of (the residue class of) \(T_i\) under its inverse. Then clearly \((T_1-t_1, \dots , T_n-t_n) \subset \mathfrak m\), and since the left hand side is a maximal ideal, equality follows.

Assertions (1), (2) and (3) are easy to check. For (4) note that

and likewise for \({\mathfrak a}_2\), so that we have

Now let \({\mathfrak p}\in V({\mathfrak a}_1{\mathfrak a}_2)\) and assume that \({\mathfrak p}\not \in V({\mathfrak a}_1)\), say \(f\in {\mathfrak a}_1\setminus {\mathfrak p}\). But then for every \(g\in {\mathfrak a}_2\), we have \(fg\in {\mathfrak a}_1{\mathfrak a}_2 \subseteq {\mathfrak p}\), and since \({\mathfrak p}\) is a prime ideal and \(f\not \in {\mathfrak p}\), we get \(g\in {\mathfrak p}\). We have shown that \({\mathfrak a}_2 \subseteq {\mathfrak p}\), i.e., that \({\mathfrak p}\in V({\mathfrak a}_2)\).

The second statement implies that for every closed subset of \(\operatorname{Spec}(R)\) the inverse image under \({}^a \varphi \) is again closed, and hence that the map \({}^a \varphi \) is continuous. To prove it, note that for \({\mathfrak q}\in \operatorname{Spec}(S)\) we have

Since (prime) ideals in the quotient correspond bijectively to (prime) ideals in \(R\) which contain \({\mathfrak a}\), the map \(\operatorname{Spec}(R/{\mathfrak a})\to \operatorname{Spec}(R)\) has image \(V({\mathfrak a})\) and is injective. Since we know that it is continuous, it only remains to show that the inverse map is continuous, as well. In other words, we need to check that \({}^a\pi \) maps closed sets to closed sets. But one checks easily that

which is closed in \(V({\mathfrak a})\), as desired.

It is clear that both \(V(-)\) and \(I(-)\) are inclusion reversing. Furthermore, \(I(Y)\), being an intersection of radical ideals, is itself a radical ideal for every \(Y\). Since \(V({\mathfrak a})\) is closed for every \({\mathfrak a}\), the final statement follows from (1) and (2).

Let us show that \(V(I(Y)) = \overline{Y}\) for every subset \(Y \subseteq \operatorname{Spec}(R)\). Clearly the left hand side is closed and contains \(Y\), so we have \(\supseteq \). To show \(\subseteq \) we need to show that \(V(I(Y))\) is the smallest closed subset containing \(Y\), i.e., that whenever \(Y \subseteq V({\mathfrak a})\), then \(V(I(Y)) \subseteq V({\mathfrak a})\). But if \(Y \subseteq V({\mathfrak a})\), then \({\mathfrak a}\subseteq {\mathfrak p}\) for all \({\mathfrak p}\in Y\), so \({\mathfrak a}\subseteq I(Y)\), and hence \(V(I(Y)) \subset V({\mathfrak a})\) as desired.

Now we show that \(I(V({\mathfrak a})) = \sqrt{{\mathfrak a}}\) for every ideal \({\mathfrak a}\subset R\). But the radical of \({\mathfrak a}\) can be described as

and this is precisely \(I(V({\mathfrak a}))\).

For (1) take \(U \subseteq \operatorname{Spec}(R)\) open, say \(U = \operatorname{Spec}(R)\setminus V({\mathfrak a})\). Then \(V({\mathfrak a}) = \bigcap _{f\in {\mathfrak a}} V(f)\), hence \(U=\bigcup _{f\in {\mathfrak a}} D(f)\). For (2) note that \(D(f)\cap D(g) = D(fg)\) and that the intersection with empty index set, \(\operatorname{Spec}(R)\) equals \(D(f)\).

The set of prime ideals in a localization \(S^{-1}R\) of \(R\) is in bijection to the set of prime ideals \({\mathfrak p}\) in \(R\) with \({\mathfrak p}\cap S=\emptyset \), via \({\mathfrak P}\mapsto \tau ^{-1}({\mathfrak P})\) and \({\mathfrak p}\mapsto {\mathfrak p}S^{-1}R = \left\{ \frac{a}{s};\ a\in {\mathfrak p}, s\in S \right\} \). This implies that the continuous map \(\operatorname{Spec}(R_f)\to \operatorname{Spec}(R)\) restricts to a bijective (and, of course, still continuous) map \(\operatorname{Spec}(R_f)\to D(f)\). To check that it is a homeomorphism, we need to check that open sets in \(\operatorname{Spec}(R_f)\) have opens in \(D(f)\) as their image. It is enough to check this for principal open subsets \(D(g/f^i)\), because those are a basis of the topology, and such a set has image \(D(fg)=D(f)\cap D(g)\) in \(\operatorname{Spec}(R)\), which is open in \(D(f)\). (Equivalently, one could check that closed sets have closed image in \(D(f)\). This is also easy: for an ideal \({\mathfrak a}\subseteq R_f\), the image of \(V({\mathfrak a})\) in \(\operatorname{Spec}(R)\) is \(D(f)\cap V(\tau ^{-1}({\mathfrak a}))\cap D(f)\) which is closed in \(D(f)\).)

Exercise (Problem Sheet 3).

Let \({\mathfrak a}= I(Z)\), so \({\mathfrak a}\subseteq R\) is a radical ideal with \(Z= V({\mathfrak a})\). If \({\mathfrak a}={\mathfrak p}\) is a prime ideal, then \(Z = V({\mathfrak p}) = \overline{ \left\{ {\mathfrak p}\right\} }\) as we have seen above, and in particular \(Z\) is irreducible with generic point \({\mathfrak p}\). Since for prime ideals, \(V({\mathfrak p})=V({\mathfrak q})\) if and only if \({\mathfrak p}= {\mathfrak q}\) (because prime ideals are radical ideals) this argument also shows the uniqueness statement of (2).

Thus it only remains to show that whenever \(Z\) is irreducible, then \(I(Z)\) is prime. First note that \(Z\ne \emptyset \) implies \(I(Z)\ne R\). Now for \(f, g\in R\) with \(fg\in I(Z)\), we have \(Z \subseteq V(fg) = V(f)\cup V(g)\), and if \(Z\) is irreducible, it follows that \(Z \subseteq V(f)\) or \(Z \subseteq V(g)\). Without loss of generality, we have \(Z \subseteq V(f)\), say. But then \(f \in \sqrt{(f)} = I(V(f)) \subseteq I(Z)\).

Use that for any open \(U \subseteq X\), we have \(Y\cap U = \emptyset \) if and only if \(\overline{Y}\cap U=\emptyset \).