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Let $L/k$ be a field extension. If $A{\otimes }_{k}L$ is a finitely generated algebra over $L$, then $A$ is finitely generated over $k$. Choose finitely many generators of $A{\otimes }_{k}L$ and collect the finitely many coefficients from $A$ occurring in them. The $k$-subalgebra $A_0$ generated by those coefficients satisfies $(A/A_0){\otimes }_{k}L=0$, and faithful flatness of $L/k$ gives $A=A_0$.

Suppose a $k$-algebra $A$ has elements $f_i,g_i$ with ${\sum }_i{g}_i{f}_i=1$, and each $A_{f_i}$ is a finitely generated algebra. Let $B\subseteq A$ contain the $f_i,g_i$ and numerators of finite generating sets of all $A_{f_i}$. Then $B$ is finitely generated and $B_{f_i}=A_{f_i}$. For $a\in A$, clear denominators so that all $f_i^{N}a$ belong to $B$. A unit-ideal identity for powers gives $1={\sum }_i{c}_i{f}_i^N$ in $B$, hence $a={\sum }_i{c}_i{f}_i^{N}a\in B$. Thus $A=B$.

A presentation of an algebra over a field with finitely many generators and relations. The Hilbert basis theorem ensures that every finitely generated algebra over a field admits such a presentation. Its module of Kähler differentials is presented by the Jacobian matrix of the relations.

Let $A$ be a commutative algebra over a field $k$, with a homomorphism $A\to k$ restricting to the identity on $k$, and let $I$ be its kernel. If $A$ is generated by finitely many elements, subtracting their images in $k$ gives finitely many generators in $I$. Every polynomial in those generators, modulo $I^2$, is a linear combination of their classes. Thus $I/I^2$, the algebraic cotangent space at this point, is finite-dimensional. For $A=k+xk[x,y]$ and $I=xk[x,y]$, the classes $x{y}^j$, $j\ge 0$, form an infinite basis modulo $I^2=x^{2}k[x,y]$. This is a subalgebra of a polynomial ring which is not a finitely generated algebra.

Consider the subalgebraIt is closed under multiplication because every nonconstant term of its elements is divisible by $x$. Evaluation at $x=0$ defines a homomorphism $A\to \mathbb{C}$ with kernel $I=x\mathbb{C}[x,y]$. Its square is $I^2=x^2\mathbb{C}[x,y]$, soas a complex vector space. In particular, the classes of $x{y}^j$, $j\ge 0$, are linearly independent and span it.

If $A$ were a finitely generated algebra, take a finite list of generators and subtract their scalar images under evaluation. The resulting generators $z_1,\dots ,z_r$ lie in $I$ and still generate $A$ over $\mathbb{C}$. Every element of $I$ is a polynomial in the $z_i$ with zero constant term. Modulo $I^2$, all terms of degree at least two vanish, leaving a linear combination of the classes of $z_1,\dots ,z_r$. This would make $I/I^2$ finite-dimensional, contradicting the displayed infinite basis. Thus this subalgebra is not finitely generated. This is the infinite cotangent space obstructs finite algebra generation argument; it proves failure of finite generation even when the proposed generators are arbitrary polynomials, rather than just monomials.

Being constant on the two-element orbits is equivalent to $f(-x,-y)=f(x,y)$. In characteristic zero, comparing monomial coefficients shows that precisely the terms of even total degree survive. Thus a vector space basis isSet $u=x^2$, $v=xy$, $w=y^2$. The two basis types are $u^a{w}^b$ and $v{u}^a{w}^b$, so they prove $R=k[u,v,w]$. Hence $R$ is a finitely generated algebra. It is an integral domain as a subring of the integral domain $k[x,y]$.

The generators satisfy $uw-v^2=0$. To prove this is the whole relation ideal, reduce any polynomial in $U,V,W$ modulo the monic relation $V^2-UW$ toIts image in $k[x,y]$ is $A(x^2,y^2)+xyB(x^2,y^2)$. The first summand has both exponents even and the second both odd; their distinct monomials are linearly independent. Thus the image vanishes only when both $A,B$ are zero. ConsequentlyThis gives the requested closed embedding and the quadratic cone invariant ring presentation.

The associated morphism of algebraic varieties is explicitlyIt is surjective. For $(u,v,w)\in Y$, if $u\ne 0$, choose $x$ with $x^2=u$ using algebraic closedness and put $y=v/x$. Then $y^2=v^2/u=w$. If $u=0$, the relation forces $v=0$; take $x=0$ and choose $y^2=w$. This covers every point of $Y$. Except over the origin, the two preimages differ by the simultaneous sign change, consistent with the orbit description.

Finally the gradient of $UW-V^2$ is $(W,-2V,U)$, so the Jacobian criterion makes every nonzero point smooth. At the origin, the maximal ideal modulo its square has the three independent classes of $U,V,W$, since the defining relation is quadratic. The tangent space there has dimension three, while $Y$ has dimension two. Therefore

The spectrum of a ring $A$ is the set of its proper prime ideals. For an ideal $I$, set $V(I)=\{\mathfrak{p}:I\subseteq \mathfrak{p}\}$. These are the closed sets of the Zariski topology. They satisfyThe union identity follows because a prime ideal containing $IJ$ must contain $I$ or $J$: otherwise choose $a\in I\setminus \mathfrak{p}$ and $b\in J\setminus \mathfrak{p}$, contradicting $ab\in \mathfrak{p}$. These identities verify the closed-set axioms. The maximal spectrum has the subspace topology, whose closed sets are $V(I)\cap \mathrm{MaxSpec}A$.

For a finitely generated $\mathbb{C}$-algebra, the needed form of the Hilbert Nullstellensatz isHere an empty intersection means $A$. We prove this formula rather than assume the desired correspondence.

First establish the residue-field fact. If a field $L$ is a finitely generated algebra over $\mathbb{C}$, choose a maximal algebraically independent subset $t_1,\dots ,t_r$ of a finite generating set, with remaining generators $u_1,\dots ,u_s$. Each $u_j$ is algebraic over $\mathbb{C}(t_1,\dots ,t_r)$. Clearing the finitely many denominators in their monic equations gives a nonzero polynomial $d$ such that each $u_j$ is integral overSince $L$ is a field, it contains $1/d$, and $L=R[u_1,\dots ,u_s]$ is an integral extension of $R$. If a field is integral over a domain, the domain is itself a field: for $0\ne a\in R$, an integral equation for $a^{-1}$, multiplied by $a^{n-1}$, expresses $a^{-1}$ in $R$.

But if $r>0$, choose $c\in \mathbb{C}$ for which $t_1-c$ does not divide $d$. Such a choice exists because a nonzero polynomial has only finitely many distinct factors $t_1-c$. The element $t_1-c$ is a nonunit even after inverting $d$: an inverse would imply that $t_1-c$ divides a power of $d$. Thus $R$ would not be a field, a contradiction. Hence $r=0$, and $L$ is algebraic over the algebraically closed field $\mathbb{C}$, forcing $L=\mathbb{C}$. This proves the Zariski lemma in the form used here.

Now if $f\notin \sqrt{I}$, the localization of a ring $B=(A/I)[1/f]$ is nonzero: vanishing would mean $f^n\in I$ for some $n$. It is still a finitely generated algebra over $\mathbb{C}$, since one need only adjoin an inverse of $f$. Choose a maximal ideal $\mathfrak{n}$ of $B$. By the preceding argument, $B/\mathfrak{n}=\mathbb{C}$. The composite $A\to B\to \mathbb{C}$ is surjective because it contains the constant scalars. Its kernel is consequently a maximal ideal $\mathfrak{m}$ of $A$, contains $I$, and does not contain $f$, whose image is invertible. This proves that the displayed intersection is contained in $\sqrt{I}$. The reverse inclusion follows because every maximal ideal is prime and contains the radical of an ideal whenever it contains that ideal.

For a closed subset $Z\subseteq \mathrm{MaxSpec}A$, put $I(Z)={\bigcap }_{\mathfrak{m}\in Z}\mathfrak{m}$. It is a radical ideal. The formulas now showare inverse maps on radical ideals and closed sets. Indeed the first followed by the second gives $\sqrt{I}=I$, while if $Z=V(J)\cap \mathrm{MaxSpec}A$, the second gives $\sqrt{J}$ and $V(\sqrt{J})=V(J)$. Thus the correspondence is a bijection reversing inclusion. This is the relation between radical ideals and closed subsets of the maximal spectrum.

For $A=\mathbb{C}[x]$, the maximal ideals are ${\mathfrak{m}}_c=(x-c)$, for $c\in \mathbb{C}$. Every nonzero proper radical ideal is generated by a squarefree polynomial, and hence has the formfor a finite nonempty subset $S\subset \mathbb{C}$. The associated closed set is $\{{\mathfrak{m}}_c:c\in S\}$. A repeated factor of a polynomial does not change its zero set, but passing to its radical ideal removes that repeated factor. The two endpoint cases areThe empty-product convention makes $I_{\varnothing }=(1)$. Under $c\mapsto {\mathfrak{m}}_c$, the closed sets are exactly the finite subsets and the whole affine line.

The characteristic of $K$ is either a prime $p$ or zero. In characteristic $p$, it is a finitely generated algebra over ${\mathbb{F}}_p$. The course's Zariski lemma says that a field finitely generated as an algebra over a field is a finite algebraic extension of that field. ThereforeIt remains to exclude characteristic zero. Write $K=\mathbb{Z}[a_1,\dots ,a_m]$. Because $K$ is then a field containing $\mathbb{Q}$, also $K=\mathbb{Q}[a_1,\dots ,a_m]$. Zariski lemma makes $K/\mathbb{Q}$ finite algebraic. Each $a_i$ has a monic equation with rational coefficients. Choose a positive integer $N$ clearing all their coefficient denominators. Each $a_i$ is then integral over $A=\mathbb{Z}[1/N]$, so $A[a_1,\dots ,a_m]$ is integral over $A$. This algebra equals $K$: it contains the given integer algebra $K$, and $1/N$ already lies in the field $K$.

An integral field extension forces the base domain to be a field. To see it here, for a nonzero $a\in A$ its inverse in $K$ satisfiesMultiplying by $a^{r-1}$ givesThus $A$ would be a field. But choose a prime $\ell \nmid N$. Reduction modulo $\ell$ is a well-defined map $\mathbb{Z}[1/N]\to {\mathbb{F}}_{\ell }$, so the nonzero element $\ell$ is not invertible in $A$. This contradiction rules out characteristic zero. Therefore the finite-field theorem for finitely generated integer algebras yieldsThis uses Zariski lemma and elementary integrality, not a claim that a field merely finitely generated as a field extension must be algebraic.

The Noether normalization lemma is valid over every field $k$, including finite fields: a nonzero finitely generated algebra $C$ contains elements $y_1,\dots ,y_d$ with algebraic independence over $k$ such that $C$ is finite as a module over the polynomial ring $k[y_1,\dots ,y_d]$. We prove it without an infinite-field hypothesis.

Write $C=k[a_1,\dots ,a_n]$ and induct on $n$. If the generators have algebraic independence, they already give a polynomial ring and the assertion is immediate. Otherwise choose a nonzero relation $F(a_1,\dots ,a_n)=0$. Choose an integer $M$ larger than every exponent in the monomials of $F$, and setIn $F(Y_1+T^M,Y_2+T^{M^2},\dots ,Y_{n-1}+T^{M^{n-1}},T)$, the largest power of $T$ has a nonzero coefficient in $k$. Indeed, a monomial with exponents $(e_1,\dots ,e_n)$ contributes top weight $e_n+{\sum }_{i<n}{e}_i{M}^i$. These weights are distinct by uniqueness of base-$M$ expansion, so only one monomial contributes the highest power. After rescaling, the substituted relation is monic in $T$.

Consequently $a_n$ is an integral element over $C^{\prime }=k[b_1,\dots ,b_{n-1}]$, and $C=C^{\prime }[a_n]$ is finite over $C^{\prime }$. Apply the induction hypothesis to $C^{\prime }$ and compose the finite module extensions. This proves Noether normalization by weighted substitutions. The induction reaches $n=0$, where $C=k$. For an integral domain $C$, taking fraction fields makes the resulting extension finite algebraic, so $d$ is the transcendence degree of $\mathrm{Frac}(C)/k$. More generally $d=\dim C$: integral extensions preserve Krull dimension, and a polynomial algebra in $d$ variables has Krull dimension $d$.

A useful bridge to the Hilbert Nullstellensatz is the Zariski lemma. If a field $E$ is a finitely generated algebra over $k$, normalization makes it finite and integral over a polynomial ring $P=k[y_1,\dots ,y_d]$. A subring over which a field is integral is a field: for nonzero $a\in P$, an integral equation for $a^{-1}\in E$, multiplied by $a^{r-1}$, expresses $a^{-1}$ as an element of $P$. Therefore $P$ must be a field. A polynomial ring in a positive number of variables is not a field, since a variable has no polynomial inverse. Hence $d=0$ and $E/k$ is a finite field extension. This proves the Zariski lemma.

Now let $k$ be an algebraically closed field. The Weak Hilbert Nullstellensatz says that every maximal ideal of $k[X_1,\dots ,X_n]$ is uniquely of the formTo prove it, the residue field $k[X]/\mathfrak{m}$ is a field generated as a $k$-algebra by the images of the variables. The Zariski lemma makes it finite algebraic over $k$, and algebraic closedness makes it $k$. Thus each $X_i$ has an image $a_i\in k$. The evaluation map has kernel the displayed ideal: subtracting the constant value of a polynomial expresses its difference as a combination of $X_i-a_i$. That kernel is maximal and contained in $\mathfrak{m}$, so equality holds. Conversely every evaluation kernel is maximal because its quotient is $k$. Uniqueness follows from the variable images. Every proper ideal is contained in a maximal ideal, so it has a common zero; equivalently, an ideal with no common zero is the whole ring.

For an ideal $I\subseteq k[X_1,\dots ,X_n]$, let $V(I)$ be its common-zero set and let $I(V(I))$ be all polynomials vanishing on that set. The Strong Hilbert Nullstellensatz statesThe inclusion $\sqrt{I}\subseteq I(V(I))$ follows because a field has no nonzero nilpotent elements. For the other inclusion, take $f$ vanishing on $V(I)$, with $f\ne 0$, and form the Rabinowitsch trick idealIt has no common zero: at a zero of $I$ the second generator has value one. The Weak Hilbert Nullstellensatz in $n+1$ variables gives $J=(1)$. Hence a finite identity has the form $1={\sum }_j{h}_j(X,T)g_j(X)+h(X,T)(1-Tf)$, with $g_j\in I$. Substitute $T=f^{-1}$ in the localization of a ring $k[X{]}_f$. Clearing the finitely many powers of $f$ occurring in denominators gives $f^r\in I$ for some $r$, so $f\in \sqrt{I}$. The case $f=0$ is immediate. This completes all three proofs. The algebraically closed hypothesis belongs to the two forms of the Hilbert Nullstellensatz; it was not needed for the Noether normalization lemma or the Zariski lemma.

The localization of global sections on a principal open gives $A_{f_i}=\Gamma (X_{f_i},{\mathcal{O}}_X)$. Each of these rings is a finitely generated algebra because $X_{f_i}$ is an affine variety. Choose finite generators of every $A_{f_i}$ and write them as $a_{ij}/f_i^{e_{ij}}$ with $a_{ij}\in A$. Let $B$ be the $k$-subalgebra of $A$ generated by all $f_i,g_i,a_{ij}$. It is a finitely generated algebra, and $B_{f_i}=A_{f_i}$ since these localizations contain the chosen generators and $f_i^{-1}$.

For any integer $N\ge 1$, raise ${\sum }_i{g}_i{f}_i=1$ to the power $r(N-1)+1$. Every resulting monomial contains some $f_i^N$, so it gives a unit-ideal identity for powersFor an arbitrary $a\in A$, equality $A_{f_i}=B_{f_i}$ implies $f_i^{N_i}a\in B$ for some $N_i$: multiply by an additional power if equality of localized fractions requires it. Choose a common $N$ and the identity above. Then $a={\sum }_i{c}_i(f_i^{N}a)\in B$. Consequently $A=B$, so $A$ is a finitely generated algebra. It is a reduced ring, since a nilpotent element global regular function has zero germ everywhere on the reduced variety $X$. This is finite generation from finitely many principal localizations.

The Weak Hilbert Nullstellensatz states that if $k$ is an algebraically closed field, every maximal ideal of $k[Y_1,\dots ,Y_s]$ isEquivalently, every proper ideal has a common zero. We prove the more general field statement, the Zariski lemma: a field finitely generated as a $k$-algebra is a finite field extension of $k$.

Write such a field as $L=k[b_1,\dots ,b_s]$. Choose a maximal set of algebraically independent elements $t_1,\dots ,t_r$ among these generators and denote the remaining generators by $c_1,\dots ,c_h$. Each $c_j$ is an algebraic element over $K=k(t_1,\dots ,t_r)$, so $L/K$ is a finite field extension. Clearing the denominators in their monic equations, choose a nonzero $d\in k[t_1,\dots ,t_r]$ such that each $c_j$ is an integral element overBecause $L$ is a field and $d^{-1}\in L$, we have $L=D[c_1,\dots ,c_h]$. Thus $L$ is an integral extension of $D$.

A subring over which a field is integral must itself be a field. Indeed, for $0\ne a\in D$, an integral equation for $a^{-1}$ multiplied by $a^{m-1}$ expresses $a^{-1}$ as an element of $D$. But if $r>0$, the ring $D$ is not a field: choose an irreducible polynomial $q(t_1)\in k[t_1]$ which does not divide $d$. There are infinitely many monic irreducible polynomials in $k[t_1]$, by the usual product-plus-one argument, whereas only finitely many can divide the nonzero $d$. The prime ideal $(q)$ stays proper after inverting $d$, so $q$ has no inverse in $D$. Therefore $r=0$, proving that $L/k$ is a finite field extension, and in particular an algebraic field extension.

Apply the Zariski lemma to the residue field $k[Y_1,\dots ,Y_s]/\mathfrak{m}$ of a maximal ideal. When $k$ is an algebraically closed field, this field is $k$ itself. The images $a_i$ of $Y_i$ then give $\mathfrak{m}=(Y_1-a_1,\dots ,Y_s-a_s)$, proving the Weak Hilbert Nullstellensatz.

For the real Laurent polynomial ring, the same argument applies because it is the finitely generated algebraEvery residue field at a maximal ideal is consequently a finite field extension of $\mathbb{R}$, and so is $\mathbb{R}$ or $\mathbb{C}$. Here we use the fundamental theorem of algebra, which makes $\mathbb{C}$ the algebraic closure of $\mathbb{R}$ of degree two. The images of $X_1,X_2$ must be nonzero since these elements are units. It follows that the answer iswith $(\alpha ,\beta )$ and $(\overline{\alpha },\overline{\beta })$ defining the same maximal ideal. If both coordinates are real, evaluation has image $\mathbb{R}$; otherwise its image is $\mathbb{C}$. Thus every displayed kernel is maximal, and the Zariski lemma shows that this list is exhaustive.

For explicit generating sets of an ideal, there are three cases. For real nonzero $a,b$,If $\alpha =a\in {\mathbb{R}}^{\times }$ and $\beta =s+it$ with $t\ne 0$,If $\alpha =u+iv$ with $v\ne 0$ and $\beta =s+it$, thenThe last two quotients are $\mathbb{C}$; the nonzero coordinate condition ensures that the Laurent inverses exist in them. There are no further coincidences in the classification: equal kernels induce an $\mathbb{R}$-isomorphism between the evaluation image fields. The only $\mathbb{R}$-automorphisms of $\mathbb{C}$ are the identity and complex conjugation, so the two pairs must agree or be conjugate.

To prove the intersection assertion, fix a prime ideal $P$ and an element $f\notin P$. The integral domain $B=R/P$ has nonzero image $\bar{f}$. Its localization of a ring $B[{\bar{f}}^{-1}]$ is nonzero and still a finitely generated algebra over $\mathbb{R}$, so it has a maximal ideal $\mathfrak{n}$. By the Zariski lemma, the field $L=B[{\bar{f}}^{-1}]/\mathfrak{n}$ is finite over $\mathbb{R}$.

The image of $B$ in $L$ is a subring containing $\mathbb{R}$ and is a finite-dimensional $\mathbb{R}$-vector space. It is an integral domain, hence a field: multiplication by any nonzero element is an injective linear map of a finite-dimensional vector space and therefore surjective. The kernel of $B\to L$ is thus a maximal ideal which avoids $\bar{f}$, since $\bar{f}$ is invertible in $L$. Its inverse image in $R$ is a maximal ideal containing $P$ but avoiding $f$. As this works for every $f\notin P$,This is the Jacobson ring property for this Laurent polynomial ring.

For a nonzero finitely generated module $M$, its set of associated primes isHere ${\mathrm{Ann}}_R(m)=\{a\in R:am=0\}$ is the annihilator of the element. The Laurent polynomial ring $R$ is a Noetherian ring, so the nonempty collection of annihilators of nonzero elements of $M$ has a maximal member, say $P={\mathrm{Ann}}_R(m)$. It is proper because $m\ne 0$.

If $ab\in P$ and $b\notin P$, then $bm\ne 0$ andMaximality forces ${\mathrm{Ann}}_R(bm)=P$, hence $a\in P$. Thus $P$ is a prime ideal, proving $\boxed{{\mathrm{Ass}}_R(M)\ne \varnothing }$. This argument in fact only needs that $R$ is Noetherian, not the finite generation of $M$.

A proper ideal $Q$ is $P$-primary when its radical of an ideal is $P$ andEquivalently, $ab\in Q$ and $b\notin P$ imply $a\in Q$; equivalently, every zero divisor of $R/Q$ is nilpotent. These formulations distinguish the actual primary ideal property from the weaker assertion that there is only one minimal prime ideal.

For the final example, set $u=X_1-1$ and $v=X_2-1$. ThenSince $R/P\cong \mathbb{R}[X_2,X_2^{-1}]$ is an integral domain, $P$ is a prime ideal. Moreover $I\subseteq P$ and $u^2\in I$, so $\sqrt{I}=P$. Consequently $P$ is the unique minimal prime ideal over $I$.

Nevertheless $uv\in I$, while $u\notin I$ and no power of $v$ belongs to $I$. For the first assertion, reduce modulo $v$: the image of $I$ is $(u^2)$ in $\mathbb{R}[X_1,X_1^{-1}]$, and $u$ is not a multiple of $u^2$ because $u$ is a nonunit in this integral domain. For the second, $v\notin P$, so $v^j\notin P\supseteq I$ for every $j\ge 1$. Thus $I$ is not a $P$-primary ideal: a unique minimal prime does not imply primary. In $R/I$, the nonzero class of $u$ is killed by $v$, a zero divisor which is not nilpotent.

Zariski lemma says that a field which is a finitely generated algebra over a field $k$ is a finite algebraic extension of $k$. The Strong Hilbert Nullstellensatz says that, for an ideal $I\subseteq k[T_1,\dots ,T_n]$ over an algebraically closed field,

Let the unique point of $V(\mathfrak{a})$ be $x=(x_1,\dots ,x_n)$ and letThe Nullstellensatz gives $\sqrt{\mathfrak{a}}=\mathfrak{m}$, so $\mathfrak{a}\subseteq \mathfrak{m}$. Each of the finitely many generators $u_i=T_i-x_i$ of $\mathfrak{m}$ has some power $u_i^{e_i}\in \mathfrak{a}$. Ifthen every monomial of total degree $r$ in the $u_i$ is divisible by one of the $u_i^{e_i}$. HenceThus the assertion is true; algebraically, the quotient defines a punctual scheme supported at $x$.

True. Choose a maximal ideal $\mathfrak{m}\subset B$ and write $L=B/\mathfrak{m}$, a field extension of $k$. Taking the quotient by $A{\otimes }_k\mathfrak{m}$ givesso $A{\otimes }_{k}L$ is a finitely generated algebra over $k$, and therefore over $L$.

By finite generation descends along a field extension, $A$ is finitely generated over $k$. Indeed, collect the finitely many coefficients from $A$ occurring in a finite set of $L$-algebra generators of $A{\otimes }_{k}L$. If $A_0$ is the $k$-subalgebra generated by those coefficients, then $(A/A_0){\otimes }_{k}L=0$; because a field extension is a faithfully flat module, $A=A_0$. Interchanging $A$ and $B$ proves that $B$ is also finitely generated.

Write the finitely generated algebra asBy the Weak Hilbert Nullstellensatz, $\mathbb{C}$-algebra homomorphisms $A\to \mathbb{C}$ correspond exactly to the points of the affine algebraic set $V(I)$.

If $V(I)$ is finite, its cardinality is finite. If it is infinite, the complex affine algebraic set cardinality dichotomy giveswhich is uncountable. This also covers the zero algebra, whose homomorphism set is empty. Therefore

For a finitely generated algebra over an algebraically closed field $k$, radical ideals correspond, with inclusion reversed, to closed subsets of the maximal spectrum. If $f\notin \sqrt{I}$, the nonzero algebra $(A/I)[1/f]$ is still finitely generated over $k$. Its maximal residue field is $k$ by the Zariski lemma, so contracting a maximal ideal gives a maximal ideal of $A$ containing $I$ and avoiding $f$. Hence the intersection of all maximal ideals containing $I$ is $\sqrt{I}$. This proves the inverse correspondence, including the empty closed set and unit ideal.