Zariski's Lemma
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algebra Algebra is a branch of mathematics that deals with abstract systems, known as algebraic structures, and the manipulation of expressions within those systems. It is a generalization of arithmetic that introduces variables and algebraic ope ...
, Zariski's lemma, proved by , states that, if a field is finitely generated as an
associative algebra In mathematics, an associative algebra ''A'' over a commutative ring (often a field) ''K'' is a ring ''A'' together with a ring homomorphism from ''K'' into the center of ''A''. This is thus an algebraic structure with an addition, a mult ...
over another field , then is a finite field extension of (that is, it is also finitely generated as a
vector space In mathematics and physics, a vector space (also called a linear space) is a set (mathematics), set whose elements, often called vector (mathematics and physics), ''vectors'', can be added together and multiplied ("scaled") by numbers called sc ...
). An important application of the lemma is a proof of the weak form of
Hilbert's Nullstellensatz In mathematics, Hilbert's Nullstellensatz (German for "theorem of zeros", or more literally, "zero-locus-theorem") is a theorem that establishes a fundamental relationship between geometry and algebra. This relationship is the basis of algebraic ge ...
: if ''I'' is a proper ideal of k _1, ..., t_n/math> (''k'' an
algebraically closed field In mathematics, a field is algebraically closed if every non-constant polynomial in (the univariate polynomial ring with coefficients in ) has a root in . In other words, a field is algebraically closed if the fundamental theorem of algebra ...
), then ''I'' has a zero; i.e., there is a point ''x'' in k^n such that f(x) = 0 for all ''f'' in ''I''. (Proof: replacing ''I'' by a
maximal ideal In mathematics, more specifically in ring theory, a maximal ideal is an ideal that is maximal (with respect to set inclusion) amongst all ''proper'' ideals. In other words, ''I'' is a maximal ideal of a ring ''R'' if there are no other ideals ...
\mathfrak, we can assume I = \mathfrak is maximal. Let A = k _1, ..., t_n/math> and \phi: A \to A / \mathfrak be the natural surjection. By the lemma A / \mathfrak is a finite extension. Since ''k'' is algebraically closed that extension must be ''k''. Then for any f \in \mathfrak, :f(\phi(t_1), \cdots, \phi(t_n)) = \phi(f(t_1, \cdots, t_n)) = 0; that is to say, x = (\phi(t_1), \cdots, \phi(t_n)) is a zero of \mathfrak.) The lemma may also be understood from the following perspective. In general, a ring ''R'' is a
Jacobson ring In algebra, a Hilbert ring or a Jacobson ring is a ring such that every prime ideal is an intersection of primitive ideals. For commutative rings primitive ideals are the same as maximal ideals so in this case a Jacobson ring is one in which ever ...
if and only if every finitely generated ''R''-algebra that is a field is finite over ''R''. Thus, the lemma follows from the fact that a field is a Jacobson ring.


Proofs

Two direct proofs are given in Atiyah–MacDonald; the one is due to Zariski and the other uses the Artin–Tate lemma. For Zariski's original proof, see the original paper. Another direct proof in the language of
Jacobson ring In algebra, a Hilbert ring or a Jacobson ring is a ring such that every prime ideal is an intersection of primitive ideals. For commutative rings primitive ideals are the same as maximal ideals so in this case a Jacobson ring is one in which ever ...
s is given below. The lemma is also a consequence of the
Noether normalization lemma In mathematics, the Noether normalization lemma is a result of commutative algebra, introduced by Emmy Noether in 1926. It states that for any field k, and any finitely generated commutative ''k''-algebra A, there exist elements y_1,y_2,\ldot ...
. Indeed, by the normalization lemma, ''K'' is a finite module over the polynomial ring k _1, \ldots , x_d/math> where x_1, \ldots , x_d are elements of ''K'' that are algebraically independent over ''k''. But since ''K'' has Krull dimension zero and since an integral ring extension (e.g., a finite ring extension) preserves Krull dimensions, the polynomial ring must have dimension zero; i.e., d=0. The following characterization of a Jacobson ring contains Zariski's lemma as a special case. Recall that a ring is a Jacobson ring if every prime ideal is an intersection of maximal ideals. (When ''A'' is a field, ''A'' is a Jacobson ring and the theorem below is precisely Zariski's lemma.) Proof: 2. \Rightarrow 1.: Let \mathfrak be a prime ideal of ''A'' and set B = A/\mathfrak. We need to show the
Jacobson radical In mathematics, more specifically ring theory, the Jacobson radical of a ring R is the ideal consisting of those elements in R that annihilate all simple right R- modules. It happens that substituting "left" in place of "right" in the definitio ...
of ''B'' is zero. For that end, let ''f'' be a nonzero element of ''B''. Let \mathfrak be a maximal ideal of the localization B ^/math>. Then B ^\mathfrak is a field that is a finitely generated ''A''-algebra and so is finite over ''A'' by assumption; thus it is finite over B = A/\mathfrak and so is finite over the subring B/\mathfrak where \mathfrak = \mathfrak \cap B. By integrality, \mathfrak is a maximal ideal not containing ''f''. 1. \Rightarrow 2.: Since a factor ring of a Jacobson ring is Jacobson, we can assume ''B'' contains ''A'' as a subring. Then the assertion is a consequence of the next algebraic fact: :(*) Let B \supseteq A be integral domains such that ''B'' is finitely generated as ''A''-algebra. Then there exists a nonzero ''a'' in ''A'' such that every ring homomorphism \phi: A \to K, ''K'' an algebraically closed field, with \phi(a) \ne 0 extends to \widetilde: B \to K. Indeed, choose a maximal ideal \mathfrak of ''A'' not containing ''a''. Writing ''K'' for some algebraic closure of A/\mathfrak, the canonical map \phi: A \to A/\mathfrak \hookrightarrow K extends to \widetilde: B \to K. Since ''B'' is a field, \widetilde is injective and so ''B'' is algebraic (thus finite algebraic) over A/\mathfrak. We now prove (*). If ''B'' contains an element that is transcendental over ''A'', then it contains a polynomial ring over ''A'' to which ''φ'' extends (without a requirement on ''a'') and so we can assume ''B'' is algebraic over ''A'' (by Zorn's lemma, say). Let x_1, \dots, x_r be the generators of ''B'' as ''A''-algebra. Then each x_i satisfies the relation :a_x_i^n + a_x_i^ + \dots + a_ = 0, \, \, a_ \in A where ''n'' depends on ''i'' and a_ \ne 0. Set a = a_a_ \dots a_. Then B ^/math> is integral over A ^/math>. Now given \phi: A \to K, we first extend it to \widetilde: A ^\to K by setting \widetilde(a^) = \phi(a)^. Next, let \mathfrak = \operatorname\widetilde. By integrality, \mathfrak = \mathfrak \cap A ^/math> for some maximal ideal \mathfrak of B ^/math>. Then \widetilde: A ^\to A ^\mathfrak \to K extends to B ^\to B ^\mathfrak \to K. Restrict the last map to ''B'' to finish the proof. \square


Notes


Sources

* * * {{refend Lemmas in algebra Theorems about algebras