By ring, we mean commutative ring with unit. The case 1 = 0 is allowed.
A ring is local iff x_1 + \cdots + x_n \operatorname{inv}\quad\Longrightarrow\quad x_1 \operatorname{inv}\vee \cdots \vee x_n \operatorname{inv}. In particular (case n = 0), 1 \neq 0.
\mathbb{Z}_{(p)}, k[X]_{(X-a)}, \mathbb{R}, \mathbb{Q}_p.
\mathbb{Z}, k[X].
A ring is an integral domain iff x_1 \cdots x_n = 0 \quad\Longrightarrow\quad x_1 = 0 \vee \cdots \vee x_n = 0. In particular (case n = 0), 1 \neq 0.
\mathbb{Z}, \mathbb{Z}[X], and more generally polynomial rings over integral domains.
In \mathbb{Z}/(6), 2 \cdot 3 = 0 even though 2 \neq 0 and 3 \neq 0.
A ring is a field iff \neg (x_1 = 0 \wedge \cdots \wedge x_n = 0) \quad\Longrightarrow\quad x_1 \operatorname{inv}\vee \cdots \vee x_n \operatorname{inv}. In particular (case n = 0), 1 \neq 0; and (case n = 1), x \neq 0 \Rightarrow x \operatorname{inv}.
\mathbb{Q}, \mathbb{Q}(T), \mathbb{F}_q, \bar{\mathbb{Q}}.
\mathbb{R}, \mathbb{C}, \mathbb{F}_p((T)), \mathbb{Q}_p.
The examples in the second block are only fields when disregarding issues of computability. As there are no time bounds on the runtime of the inversion procedure, constructively we only have: “For every x \neq 0, anonymously there is an inverse x^{-1}.”
In general, proof by contradiction and the axiom of choice preclude a computational interpretation of our arguments. For instance, a proof that it is impossible for a given equation to have no solution does not give us any indication how to find one.
In this course, we will always be explicit when using these principles of classical mathematics, by highlighting them with the qualifier “anonymously”:
Let x,y \in \mathbb{R} such that xy = 0. Then anonymously x = 0 or y = 0 (but there is no algorithm which would correctly tell us which alternative holds).
Let a,b be elements of a set X. Then anonymously a = b or a \neq b (but there might not be an algorithm telling us which).
Let f : \mathbb{N}\to \mathbb{N} be a function. Then anonymously f attains a global minimum (but there is no algorithm locating this minimum).
Let X be a set which is not empty. Then anonymously there is an element of X (but there might not be a procedure for finding one).
Anonymously, A or \neg A (but there might not be an algorithm deciding the matter).
By avoiding nonconstructive proofs, we open up three possibilities:
Our proofs give rise to algorithms.
Our results are “stable in continuous families” (this will be explored later on).
Our results are also applicable to R, the central protagonist of synthetic algebraic geometry. This emerging subject allows us to use a simple element-based language to secretly work with Grothendieck-style modern algebraic geometry (with its support for arbitrary base rings which are not fields and for “generic points”).
In the cases where we need an additional assumption to ensure that an argument works constructively, we will highlight such an assumption like this.
A set X has decidable equality iff for every a,b \in X, either a = b or a \neq b.
In classical mathematics—disregarding algorithmic implementability and stability in families—every set has decidable equality.
The rings \mathbb{Z}, \mathbb{Q}, \mathbb{F}_q, \bar{\mathbb{Q}} all have decidable equality. For \bar{\mathbb{Q}}, this is a nontrivial result. The polynomials witnessing algebraicity provide the required additional information to detect equality or inequality by just inspecting a finite number of rational approximations (Mines, Richman, and Ruitenburg 1988, chap. VI.1).
Let A be a ring. The ideal generated by a set S \subseteq A is the set (S) = \Bigl\{ \sum_{i=1}^n a_i s_i \,\Big|\, n \geq 0, a_1,\ldots,a_n \in A, s_1,\ldots,s_n \in S \Bigr\}. An ideal is finitely generated iff it is equal to an ideal of the form (x_1,\ldots,x_n). An ideal is principal iff it is equal to an ideal of the form (x).
A Bézout ring is a ring for which every finitely generated ideal is principal.
Every finitely generated ideal of \mathbb{Z} is principal. For instance, (a,b) = (\operatorname{gcd}(a,b)).
For every ideal of \mathbb{Z}, anonymously there is a finite set generating it. In particular, there are no ideals which would not be finitely generated.
Let k be a field with decidable equality. Then the Euclidean algorithm is available to ensure that every finitely generated ideal of k[X] is principal.
The radical of an ideal \mathfrak{a}\subseteq A is the ideal \sqrt{\mathfrak{a}} = \{ x \in A \,|\, \exists n \in \mathbb{N}\mathpunct{.} x^n \in \mathfrak{a}\} \subseteq A. An ideal \mathfrak{a} is a radical ideal iff \sqrt{\mathfrak{a}} \subseteq \mathfrak{a} (we always have \mathfrak{a}\subseteq \sqrt{\mathfrak{a}}). An ideal \mathfrak{a} is radically finitely generated iff there exists a finitely generated ideal \mathfrak{b} such that \sqrt{\mathfrak{a}} = \sqrt{\mathfrak{b}}.
A gcd ring is a ring in which every finite family of elements has a greatest common divisor, i.e. a largest element in the quasiorder of divisors.
Every Bézout ring is a gcd ring, a greatest common divisor of a family x_1,\ldots,x_n of ring elements is given by any principal generator of the ideal (x_1,\ldots,x_n).
In the following, we fix a ring k which we presume to be local and a field. Some results will also require k to be algebraically closed or to be an integral domain, but we will not put these as blanket assumptions. Some of the following definitions will also appear to make sense if k is an arbitrary ring. However this will not last beyond the basic notions. The generalization to arbitrary rings requires Grothendieck-style algebraic geometry or synthetic algebraic geometry.
Affine n-space over k is \mathbb{A}^n_k \vcentcolon=k^n regarded as a plain set (not a vector space) equipped with a topology introduced below.
Let f \in k[X_1,\ldots,X_n] be a polynomial. Its nonzero locus is the subset D(f) \vcentcolon=\{ p \in \mathbb{A}^n_k \,|\, f(p) \operatorname{inv}\} \subseteq \mathbb{A}^n_k. Its zero locus or vanishing set is the subset V(f) \vcentcolon=\{ p \in \mathbb{A}^n_k \,|\, f(p) = 0 \} \subseteq \mathbb{A}^n_k.
Let S \subseteq k[X_1,\ldots,X_n] be a set of polynomials. Its nonzero locus is the subset D(S) \vcentcolon=\bigcup_{f \in S} D(f) \subseteq \mathbb{A}^n_k. Its zero locus or vanishing set is the subset V(S) \vcentcolon=\bigcap_{f \in S} D(f) \subseteq \mathbb{A}^n_k.
A subset M \subseteq \mathbb{A}^n_k is algebraic iff it is of the form V(S) for some finitely enumerable set S = \{ f_1,\ldots,f_r \}.
For every polynomial f, D(f) = V(f)^c, and indeed even D(S) = V(S)^c for every finitely enumerable set S of polynomials.
We have the following relations.
We have the following relations.
The vanishing loci have perhaps a greater geometric signifance than the nonzero loci, but require more assumptions for the basic observations about them. In view of Items (2) and (3) there is hope that \sqrt{(S)} can be reconstructed from D(S) or V(S), and for (see below) this is indeed true. In general, however, we are losing information when passing to the geometric side: For instance V(X^2+1) = \emptyset = V(1) \subseteq \mathbb{A}^1_\mathbb{Q}.
Let S \subseteq k[X_1,\ldots,X_n]. Then Hilbert’s basis theorem implies that anonymously, the ideal (S) is finitely generated. Hence V(S) is anonymously an algebraic set even if S is not finitely enumerable.
The ideal I(M) of a subset M \subseteq \mathbb{A}^n_k is the ideal I(M) = \{ f \in k[X_1,\ldots,X_n] \,|\, \forall p \in M\mathpunct{.} f(p) = 0 \} \subseteq k[X_1,\ldots,X_n] of those polynomials which vanish on M.
I(\mathbb{A}^n_k) = (0), if k is infinite.
I(\{(a_1,\ldots,a_n)\}) = (X_1-a_1,\ldots,X_n-a_n).
We have the following relations.
Let M \subseteq \mathbb{A}^n_k be an algebraic subset. A polynomial function on M is a map f : M \to k such that there exists a polynomial g \in k[X_1,\ldots,X_n] with f(p) = g(p) for all p \in M.
In case that k is not infinite, the polynomial g might not be uniquely determined by f. The set of all polynomial functions on M is denoted \mathrm{Hom}_k(M,\mathbb{A}^1_k). There is a canonical surjective homomorphism k[X_1,\ldots,X_n] \longrightarrow \mathrm{Hom}_k(M,\mathbb{A}^1_k) with kernel I(M). So \mathrm{Hom}_k(M,\mathbb{A}^1_k) \cong k[X_1,\ldots,X_n]/I(M). This observation motivates the following definition.
The coordinate algebra of an algebraic subset M \subseteq \mathbb{A}^n_k is the k-algebra k[X_1,\ldots,X_n]/I(M).
In a similar fashion, we define the notion of a polynomial map f : M \to N between algebraic subsets M \subseteq \mathbb{A}^m_k, N \subseteq \mathbb{A}^n_k. The set of such maps is denoted \mathrm{Hom}_k(M,N). We have a canonical isomorphism \mathrm{Hom}_k(\Gamma(N),\Gamma(M)) \longrightarrow \mathrm{Hom}_k(M,N) which maps an k-algebra homomorphism \varphi : \Gamma(N) \to \Gamma(M) to the polynomial map p \mapsto (\varphi([Y_1])(p), \ldots, \varphi([Y_n])(p))).
The fiber of a map f : M \to N over a point p \in N is the subset f^{-1}[\{p\}] = \{ q \in M \,|\, f(q) = p \} \subseteq M.
In this section, we assume that k is field which is also local and an integral domain.
An algebraic subset M \subseteq \mathbb{A}^n_k is irreducible iff for every number r \geq 0 and all algebraic subsets M_1,\ldots,M_r \subseteq \mathbb{A}^n_k, M = M_1 \cup \ldots \cup M_r \quad\Longrightarrow\quad \exists i\mathpunct{.} M = M_i; equivalently, if M \subseteq M_1 \cup \ldots \cup M_r \quad\Longrightarrow\quad \exists i\mathpunct{.} M \subseteq M_i; equivalently still, if for all r \geq 0 and all finitely generated ideals \mathfrak{a}_1,\ldots,\mathfrak{a}_r \subseteq k[X_1,\ldots,X_n], \mathfrak{a}_1 \cdot \ldots \cdot \mathfrak{a}_r \subseteq I(M) \quad\Longrightarrow\quad \exists i\mathpunct{.} \mathfrak{a}_i \subseteq I(M); in other words, iff I(M) is a prime ideal.
By the r = 0 case, the empty set is not irreducible (“too simple to be simple”). If k is, contrary to what we have supposed, not an integral domain, the definition needs to be modified to reflect the paucity of prime ideals in k[X_1,\ldots,X_n].
The algebraic subset V(XY) \subseteq \mathbb{A}^2_k is not irreducible, as V(XY) = V(X) \cup V(Y) but V(X) \neq V(XY) and V(Y) \neq V(XY).
The algebraic subset V(Y-X^2) \subseteq \mathbb{A}^2_k is irreducible, if k is infinite.
Irreducibility of algebraic subsets is related to irreducibility of polynomials, but if k is not algebraically closed and has decidable equality, the correspondence is not perfect: For instance the polynomial X^2+1 \in \mathbb{Q}[X] is irreducible and the ideal (X^2+1) \subseteq \mathbb{Q}[X] is prime, but V(X^2+1) = \emptyset is not. A slightly more involved example is V(Y^2+X^2(X-1)^2) = \{ (0,0), (1,0) \} \subseteq \mathbb{A}^2_\mathbb{Q}.
An affine variety (over k) is an irreducible algebraic subset of some \mathbb{A}^n_k.
A ring A has a primality test for finitely generated ideals if and only if for every ideal \mathfrak{a}\subseteq A, either \mathfrak{a} is prime or alternatively 1 \in \mathfrak{a} or there exist f,g \in A such that fg \in \mathfrak{a} but f,g \not\in \mathfrak{a}. For instance \mathbb{Q}[X_1,\ldots,X_n] has a primality test for finitely generated ideals.
Assume that k is a Noetherian domain (that is the case, for instance, if k has decidable equality) and that k[X_1,\ldots,X_n] has a primality test for finitely generated ideals. Assume that k is a Nullstellensatz ring (see below). Let M \subseteq \mathbb{A}^n_k be an algebraic set. Then there are irreducible subsets M_1,\ldots,M_r \subseteq \mathbb{A}^n_k such that M = M_1 \cup \ldots \cup M_r. If we furthermore require M_i \subseteq M_j \Rightarrow i = j, this decomposition is unique up to reordering.
Under these assumptions it is a result of commutative algebra that for every finitely generated ideal \mathfrak{a} there are finitely generated prime ideals \mathfrak{p}_1,\ldots,\mathfrak{p}_r such that \sqrt{\mathfrak{a}} = \sqrt{\mathfrak{p}_1 \cap \ldots \cap \mathfrak{p}_r}. Indeed: By the primality test, the ideal is \mathfrak{a} either prime or not. If it is, we are done. Else we find that 1 \in \mathfrak{a}, in which case we are done by taking r = 0, or we find f,g \not\in \mathfrak{a} such that fg \in \mathfrak{a}; in this case we continue by Noetherian induction with the ideals \mathfrak{a}+ (f) \supsetneq \mathfrak{a} and \mathfrak{a}+ (g) \supsetneq \mathfrak{a} and fold up the two branches of the computation by the rule \sqrt{\mathfrak{a}} = \sqrt{(\mathfrak{a}+(f)) \cap (\mathfrak{a}+(g))}.
Hence V(\mathfrak{a}) = V(\mathfrak{p}_1) \cup \cdots \cup V(\mathfrak{p}_r). The summands V(\mathfrak{p}_i) are irreducible as the Nullstellensatz condition implies I(V(\mathfrak{p}_i)) = \sqrt{\mathfrak{p}_i} = \mathfrak{p}_i, so the defining ideals are prime.
In this subsection, we assume that k has decidable equality. As a consequence, polynomials have well-defined degrees and the Euclidean algorithm is available so that k[X] is a Bézout ring.
Let f, g \in k[X,Y] be polynomials of positive degree with gcd(f,g) = 1. Then V(f,g) is finite in the sense that there is a number N \in \mathbb{N} such that it’s not the case that V(f,g) contains more than N distinct points.
Exercise 3.2.
Let f \in k[X,Y] be an irreducible polynomial such that V(f) is infinite. Then I(V(f)) = (f) and V(f) is irreducible.
Let g \in I(V(f)). Then V(f,g) = V(f) is infinite by assumption. Hence \operatorname{gcd}(f,g) \neq 1 by the previous lemma. As f is irreducible, we must have \operatorname{gcd}(f,g) = f and hence g \in (f).
The claim that (f) is a prime ideal is a standard result about ideals generated by irreducible elements in unique factorization domains.
Additionally assuming that k is infinite, we hence have the following examples for irreducible subsets of \mathbb{A}^2_k:
Conversely, every irreducible subset is of these form, at least anonymously. More precisely, we have the following.
If an irreducible set M is finite in the sense that it is of the form M = \{p_1,\ldots,p_N\}, then by irreducibility already M = \{p_i\} for some i.
If an irreducible set M = V(f_1,\ldots,f_r) is infinite, then (after removing all zero polynomials f_i from this description—if r = 0 after these removals, then M = \mathbb{A}^2_k) we can then consider a partial factorization of the polynomials f_i, so factorizations f_i = p_1^{e_{i1}} \cdots p_s^{e_{is}} where the p_j are pairwise coprime. With an irreducibility test we could even fully factor the f_i into irreducible polynomials. But a partial factorization is much more straightforward to compute.
By distributivity of \cap over \cup and by irreducibility, we have M = V(p_{k_1}^{e_{1,k_1}}, \ldots, p_{k_r}^{e_{r,k_r}}) for some indices k_1,\ldots,k_r. All the exponents e_{i,k_i} are positive (else M = V(1,\ldots) = \emptyset), hence we can also write M = V(p_{k_1},\ldots,p_{k_r}). If two different k_i appear in this description, then the lemma above would predict that M is finite, in contradiction to M being infinite. Hence k_1 = \ldots = k_r =\vcentcolon k and M = V(p_k).
Assume that k is algebraically closed. Let f = f_1^{e_1} \cdots f_r^{e_r} \in k[X,Y] be a factorization of a polynomial into irreducible factors. Then V(f) = V(f_1) \cup \cdots \cup V(f_r) is the decomposition of V(f) into irreducible components, and I(V(f)) = (f_1 \cdots f_r).
By Exercise 2.4(b), the subsets V(f_i) are infinite, hence irreducible with I(V(f_i)) = (f_i) by the lemma above. We have I(V(f)) = I(\bigcup_i V(f_i)) = \bigcap_i I(V(f_i)) = \bigcap_i (f_i); we conclude with the standard observation \bigcap_i (f_i) = (f_1 \cdots f_r) valid in unique factorization domains.
A Nullstellensatz ring is a ring A such that for every number n \geq 0 and every finitely generated ideal \mathfrak{a}\subseteq A[X_1,\ldots,X_n], I(V(\mathfrak{a})) \subseteq \sqrt{\mathfrak{a}}.
Explicitly, writing \mathfrak{a}= (f_1,\ldots,f_r), this means that if f_1(p) = \ldots = f_r(p) = 0 \Rightarrow g(p) = 0 for all p \in A^n, then g \in \sqrt{(f_1,\ldots,f_r)}. The converse holds trivially.
Every algebraically closed field K with decidable equality is a Nullstellensatz ring (Lombardi and Quitté 2015 Theorem III.9.7); in fact, for every finitely generated ideal \mathfrak{a}\subseteq K[X_1,\ldots,X_n] and every polynomial g \in K[X_1,\ldots,X_n], we either have g \in \sqrt{\mathfrak{a}} (in which case V(\mathfrak{a}) \subseteq V(g)) or V(\mathfrak{a}) \cap D(g) is inhabited. The proof cleverly employs the resultant to reduce the problem to univariate polynomials.
The ring R of synthetic algebraic geometry is a Nullstellensatz ring; in fact, for every finitely generated ideal \mathfrak{a}\subseteq R[X_1,\ldots,X_n], we even have I(V(\mathfrak{a})) = \mathfrak{a}.
The Nullstellsatz property expresses a kind of completeness, stating that geometric truths have algebraic certificates.
Assume that k is a Nullstellensatz ring. Then the associations \begin{array}{rcl} \mathfrak{a}&\longmapsto& V(\mathfrak{a}) \\ I(M) &⟻& M \end{array} define order-reversing bijective maps between:
A basic fact in linear algebra is the structure theorem regarding solution sets of linear equations: A linear equation Ax = b over a field with decidable equality either has no solution at all, or the solution set M is of the form x_0 + \operatorname{span}(v_1,\ldots,v_r) where (v_1,\ldots,v_r) is a linearly independent family. In this case it makes sense to call r the dimension of the solution set; the solutions are parametrized by the points of \mathbb{A}^r_k, we have a bijection M \to \mathbb{A}^r_k. Noether normalization is the analogous result for algebraic sets. To state it, we will require the notion of a finite map:
(Noether normalization) Let k be an algebraically closed field with decidable equality. Let M \subseteq \mathbb{A}^n_k be an algebraic set. Then either M = \emptyset or there is a finite surjective map M \to \mathbb{A}^r_k for some r \leq n.
Intuitively, as a first approximation, a map is finite if all its fibers are finite sets. For instance, the projection maps V(Y-X^2) \to \mathbb{A}^1_k, V(X-Y^2) \to \mathbb{A}^1_k and V((Y-X)(Y+X)) \to \mathbb{A}^1_k which all map (x,y) \mapsto x are finite. In contrast, the map V(XY) \to \mathbb{A}^1_k, (x,y) \mapsto x is not finite.
A polynomial map f : M \to N between algebraic sets M \subseteq \mathbb{A}^m_k, N \subseteq \mathbb{A}^n_k is finite if the induced ring homomorphism f^* : \Gamma(N) \to \Gamma(M) is finite. A ring homomorphism A \to B is finite iff it exhibits B as a finitely generated A-module.
The fiber of a map f : Y \to X over a point x_0 \in X is the set f^{-1}[x_0] \vcentcolon=\{ y \in Y \,|\, f(y) = x \} \subseteq Y.
In algebraic geometry, we often try to draw the domain of a map in such a way that the map looks like the vertical projection to the codomain.
Let \varphi : A \to B be a ring homomorphism. The following are equivalent:
Let X \subseteq \mathbb{A}^n_k be an algebraic set. Then the following are equivalent:
Let X be a variety over k.
The function field of X is the field k(X) \vcentcolon=\operatorname{Frac}(\Gamma X).
The fraction field of \mathbb{A}^1_k is k(X), if k is infinite (so that I(\mathbb{A}^1_k) = (0)).
Let f \in k(X). (a) f is defined at a point p \in X iff f can be written as f = g/h with g,h \in \Gamma X and h(p) \neq 0. (b) J_f \vcentcolon=\{ g \in \Gamma X \,|\, gf \in \Gamma X \} \subseteq \Gamma X.
Let f \in k(X).
If \Gamma X is a gcd domain, then f can be written as f = g/h where g and h are coprime. In this case Euclid’s lemma implies that J_f = (h).
The ring of germs at a point p \in X is the subring \mathcal{O}_p \vcentcolon=\{ f \in k(X) \,|\, p \in D(J_f) \} of k(X).
Let p \in X.
If k is a Nullstellensatz ring, then \Gamma X = \bigcap_{p \in X} \mathcal{O}_p \cap \{ f \in k(X) \,|\, J_f \text{ is finitely generated} \}.
Exercise 5.2.
Of particular importance is that case that the maximal ideal of a ring of germs is principal. Exercise 5.3 tells us that in this case (under some conditions) the ring is a discrete valuation domain.
Assume that k is infinite with decidable equality. Let X = \mathbb{A}^1_k. Let p \in X. Then \mathcal{O}_p is a discrete valuation domain with uniformizing parameter X - p.
The rational spectrum of a k-algebra A is \mathrm{Specr}(A) \vcentcolon=\mathrm{Hom}_k(A,k), i.e. the set of k-algebra homomorphisms A \to k.
\mathrm{Specr}(k[X]) = \mathbb{A}^1.
\mathrm{Specr}(k[X_1,\ldots,X_n]/(f_1,\ldots,f_m)) = V(f_1,\ldots,f_m).
\mathrm{Specr}(k/(s)) = \{ \star \,|\, s = 0 \mathrel{\text{in}} k \}, i.e. \mathrm{Specr}(k/(s)) contains at most one element, and is inhabited iff s = 0 in k.
\mathrm{Specr}(k[s^{-1}]) = \{ \star \,|\, s \in k^\times \}.
For finitely presented k-algebras A, a chart of \mathrm{Specr}(A) is the bijection P_\varphi : \mathrm{Specr}(A) \to V(I),\ \alpha \mapsto (\alpha(\varphi([X_1])),\ldots,\alpha(\varphi([X_n]))) induced by an isomorphism \varphi : k[X_1,\ldots,X_n]/I \to A with I finitely generated.
Let A and B be finitely presented k-algebras. A map f : \mathrm{Specr}(A) \to \mathrm{Specr}(B) is polynomial if P_\psi \circ f \circ P_\varphi^{-1} is a polynomial map between algebraic subsets for some (equivalently: every) chart P_\varphi : \mathrm{Specr}(A) \to V(I), P_\psi : \mathrm{Specr}(B) \to V(J).
Given a finitely presented k-algebra A, there is a canonical k-algebra homomorphism A \longrightarrow \mathrm{Hom}_\mathrm{poly}(\mathrm{Specr}(A), k) mapping an element s \in A to the polynomial map \mathrm{ev}_s \vcentcolon=(\alpha \mapsto \alpha(s)). (Given a chart P_\varphi of A, the composition \mathrm{ev}_s \circ P_\varphi^{-1} maps a point p to \varphi^{-1}(s)(p) and is hence indeed polynomial.)
Let A be a finitely presented k-algebra. The canonical homomorphism A \longrightarrow \mathrm{Hom}_\mathrm{poly}(\mathrm{Specr}(A), k) is an isomorphism, if A is reduced and k is a Nullstellensatz ring.
Using the chart P_\varphi : \mathrm{Specr}(A) \to V(I) induced by an isomorphism \varphi : k[X_1,\ldots,X_n]/I \to A with I finitely generated, the canonical homomorphism turns into the canonical map k[X_1,\ldots,X_n]/I \longrightarrow \Gamma V(I). As k is a Nullstellensatzring, the codomain equals k[X_1,\ldots,X_n]/\sqrt{I}. As A is reduced, \sqrt{I} = I.
Summarizing, the \mathrm{Specr} construction and the construction \mathrm{Hom}_\mathrm{poly}(\cdot,k) constitute the two parts of a contravariant equivalence between the category of finitely presented reduced k-algebras and the category of sets of the form \mathrm{Specr}(A) and polynomial maps, if k is a Nullstellensatz ring.
This (amazing!) equivalence raises three follow-up questions:
To accomodate these more general situations, we need to go beyond naive algebraic geometry and turn to Grothendieck-style algebraic geometry.
An integral domain A is integrally closed in \operatorname{Frac}(A) iff every element of Frac(A) which is integral over A (i.e. is the zero of a monic polynomial with coefficients from A) is an element of A (i.e. of the form x/1 for some x \in A).
A variety X over k is normal at a point p \in X iff \mathcal{O}_p = \{ f \in k(X) \,|\, f \text{ is defined at } p \} is integrally closed in its field of fractions. A variety is normal iff its normal at all its points.
GCD domains are integrally closed in their fields of fractions.
Let x \in \operatorname{Frac}(A). Assume that x = p/q, wlog p \perp q (i.e. the unit element is a gcd of p and q), is integral over A: x^n + a_{n-1} x^{n-1} + \ldots + a_1 x^1 + a_0 = 0. Multiplying by q^n and rearranging yields p^n = q (\ldots), so q \mid p^n, so q \mid p by Euclid’s lemma, hence x \in A.
For the rest of this section, we assume that the base field k is algebraically closed (hence in particular infinite) and has decidable equality.
The variety \mathbb{A}^n is normal: The ring \Gamma \mathbb{A}^n = k[X_1,\ldots,X_n] is integrally closed in its field of fractions (as it is a gcd domain (even a Bézout domain)). From this it follows that all the local rings \mathcal{O}_p are integrally closed in their fields of fractions, as detailed in Exercise 6.2(a).
The variety V = V(Y^2 - X^3 - X^2) is not normal at p = (0,0), but is normal at all other points. We have \Gamma V = k[X,Y]/(Y^2 - X^3 - X^2). The rational function t \vcentcolon=Y / X \in k(V) is integral over \mathcal{O}_p, but not contained in \mathcal{O}_p:
Assume that t = Y/X = p/q with p,q \in \Gamma V. Write q = [q_{00} + q_{10} X + q_{01} Y + q_{11} X Y + \ldots]. So qY = pX in \Gamma V = k[X,Y]/(Y^2 - X^3 - X^2). So (qY)(0,Y) = (pX)(0,Y) in k[Y]/(Y^2 - 0^3 - 0^2) = k[Y]/(Y²). So q_{00}Y + Y^2 (...) = 0 in k[Y]/(Y²) (this k-vector space has (1,[Y]) as a basis), so q_{00} = 0 in k.
On the other hand, t^2 = Y^2 / ^2 = (X^3 + X^2) / X^2 = X + 1, so t is a zero of the monic polynomial T^2 - (X + 1) \in (ΓV)[T] and hence integral over \Gamma V.
The normalization of a variety X is X' \vcentcolon=\mathrm{Specr}(A'), where A' is the integral closure of A = \Gamma X in Frac(A), i.e. A' = \{ x \in \operatorname{Frac}(A) \,|\, x \text{ is integral over } A \}.
Let A be an integral domain. Let A' be the integral closure of A in \operatorname{Frac}(A). Then A' is integrally closed in \operatorname{Frac}(A') = \operatorname{Frac}(A).