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Hilbert's Nullstellensatz in Infinite-Dimensional SpaceAuthor(s): Serge LangSource: Proceedings of the American Mathematical Society, Vol. 3, No. 3, (Jun., 1952), pp. 407-410Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/2031893Accessed: 13/06/2008 14:50

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HILBERT'S NULLSTELLENSATZ ININFINITE-DIMENSIONAL SPACESERGE LANG

1. The theorem. Let k be an algebraically closed field. Let A= {a} be an indexing set. Let o=k[xa] be the polynomial ring gen-erated over k by an indexed set of variables xa, aCA. Let a be anideal of o. A zero of a is a set (ta) of elements Sa n some extensionfield of k such that f( )=0 for all fCa. (Of course, a polynomial finvolves only a finite number of the variables xa.) A zero of a willbe called algebraic if all Sa ie in k. The set of all algebraic zeros of anideal a will be called the variety defined by a.

The Hilbert Nullstellensatz is in general not valid if A is an infiniteset. We shall prove however the following theorem:THEOREM. Thefollowing threestatementsare equivalent:S1. If a is an ideal of o and fCo vanishes on the variety defined bya, thenfPCa for some integer p.S2. If a is an ideal of o and a$o, then a has an algebraic zero.S3. A ring extension k[a] by elements (a in some extension field ofk is a field if and only if all ( lie in k.Furthermorethe three statements hold if and only if one of the follow-ing two conditions is satisfied:(i) A is a finite set.(ii) Let A have cardinality a. Let the transcendence degree of k overthe prime field have cardinality b. Then a

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408 SERGE LANG [JuneWhen A is a finite set, the above is the usual Hilbert theorem andis well known. A proof was given recently by Zariski [1] to whom

Statement 3 is due.2. The proof. We first prove the equivalence of the three state-ments.S1-S2. Let a be an ideal without algebraic zero. Then the poly-nomnial1 vanishes (vacuously) on the variety of a. HencelEa and a=o.

S2.*S1. We reproduce here a well known argument due toRabinowitsch. The statement is well known if A is a finite set, so weshall assume that A is an infinite set. Let a be an ideal of o and let fvanish on the variety of a. Let t be a new variable. In the ring o[t]we consider the ideal (a, 1 -tf). The ring o [t] is isomorphic to o, and(a, 1 -tf) has no algebraic zero. Hence by S2 there is an expression ofthe type

A if, + * + Arfr + A(I - tf) = 1

where A^, A(Eo[t] and fi, * * ,f,Ea. We may now let t=I/f andclear denominators to give fPEa for some integer p.S2--S3. Let k[ a] be a field. Let p be the kernel of the mapk[x. ]~k[t] so p is a maximal ideal. p has an algebraic zeroand is the kernel of the map k [xa]k [71k]. Hence(x. -I.) Elp (alla),

and therefore a lies in k.S3+-S2. Let a be an ideal $ o. Let p be a maximal ideal containinga. Then o/p is a field k[ta] and all {, lie in k. It follows that (a) is analgebraic zero of a.We shall now prove that the three statements hold precisely underthe above-mentioned conditions. We assume familiarity with cardinalarithmetic, and the following fact: If a field F has transcendencedegree b over the prime field, and b is not finite, then the set consist-ing of all elements in the algebraic closure of F has cardinality b also.If A is a finite set, the theorem is well known. We suppose that A

is not finite and that a

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1952] HILBERT'S NULLSTELLENSATZ 409PROOF. We consider the set of all subrings k [xa1, , x.,] gen-erated by a finite number of indeterminates. This set has cardinality

a, and we may index it by A again. Let these rings be denoted byOaand let aa=QaQOa. Then a=Uaaa. Each aa has a finite basis andtherefore the union of these basis elements is a basis for a havingcardinality at most a, as was to be shown.Let a be a proper ideal. Let p be a prime ideal containing a. (Amaximal such ideal will always exist by Zorn's Lemma.) It has a basisof at most a elements. Let ko be the algebraic closure of the fieldobtained by adjoining to the prime field all coefficients of elementsin such a basis of p. Then ko has cardinality at most a, and by hy-pothesis k/ko has transcendence degree of cardinality at least a.Let po=ko[xa]Cp. It has the same basis as p. Any algebraic zeroof po will therefore be an algebraic zero of p. We shall now find analgebraic zero of po. The residue class ring ko[Xa]/po is isomorphicto a ring extension ko[b,] under the natural map, and (b,) s a zeroof po. It is easy to see that there exists an isomorphism of ko[Ma]into k which is the identity on ko. If (t') is the image of (Sa)underthis isomorphism, then (t') is the desired algebraic zero. (The above-mentioned isomorphism may be constructed as follows: Consider thefield kO(ta). Select a transcendence base. By our cardinality assump-tion we may map the rational field generated over koby this base iso-morphically into k. It is now well known that this map may be ex-tended to the algebraic closure, and its restriction to the ring ko[M]is the desired isomorphism.)In order to complete the proof we need only show that our condi-tions (i) and (ii) are necessary. In other words, if neither conditionis satisfied, then there exists a proper ideal without algebraic zero.We show how to construct such an ideal.Suppose that A is an infinite set and that b

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410 T. SZELE [Juneof our map is a maximal ideal and we see that Statements S2 andS3 are violated. This concludes the proof of our theorem.

REFERENCE1. 0. Zariski, A new proof of Hilbert's Nullstellensatz, Bull. Amer. Math. Soc. vol.53 (1947) p. 362.PRINCETON UNIVERSITY

ON ORDERED SKEW FIELDST. SZELE

In this paper we shall give a necessary and sufficient conditionthat a skew field can be ordered; moreover, that the ordering of anordered skew field K can be extended to an ordering of L, L being agiven extension of K. The first of these two results generalizes to skewfields a theorem of E. Artin and 0. Schreier [1],I according to whicha commutative field can be ordered if and only if it is formally real.The second result generalizes in the same sense a recent theorem ofJ. P. Serre [2].Our considerations are based on the following definition.DEFINITION. A skew field is said to be ordered if in its multiplica-tive group a subgroup of index 2 is marked out which is also closedunder addition.Hence a skew field can be ordered if and only if its multiplicativegroup has a subgroup of index 2 which is also closed under addition.We shall now prove the following theorem.THEOREM 1. A skewfield K can be orderedif and only if -1 cannotbe represented as a sum of elements of theform

2 2 2(1) ~~~~~~aja2.. ak (as K, i = 1, 2, * ,k).REMARK. This property can be considered as a generalization ofthe notion "formally real" to the case of skew fields.The necessity of the condition in Theorem 1 is obvious. In order

to prove its sufficiency we consider a skew field K in which -1 can-not be represented as a sum of elements (1). We shall show that theReceived by the editors September 20, 1951.1 Numbers in brackets refer to the bibliography at the end of the paper.