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The following statements are all named the Hilbert's Nullstellensatz, but they appear at first to be completely unrelated to each other. What is the relationship between them exactly?

  1. (Theorem 1.3A on page 4 of Hartshorne's Algebraic Geometry) Let $k$ be an algebraically closed field, $a$ be an ideal in $A = k[x_1, ... x_n]$, and $f \in A$ be a polynomial which vanishes on all points of $Z(a)$, then $f^r \in a$ for some integer $r > 0$

  2. ("weak Nullstellensatz" listed on Wikipedia) An ideal $I \subseteq k[x_1, ... x_n]$ contain 1 iff the polynomials in $I$ do not contain any common zeroes in $k^n$

  3. (Theorem 3.2.4 on page 107 of Vakil's notes) The only maximal ideals in the ring $k[x_1, ... x_n]$ are of the form $(x_1 - a_1, ... x_n - a_n)$, for $(a_1, ... a_n) \in k^n$

  4. (Theorem 3.2.5 on page 107 of Vakil's notes) If $k$ is any field, every maximal ideal of $k[x_1, ... x_n]$ has residue field a finite extension of $k$. Any field extension of $k$ that is finitely generated as a ring is also finitely generated as a module

  5. (Theorem 3.7.1 on page 128 of Vakil's notes) Let $A$ be a commutative ring with identity, then $V(.)$ and $I(.)$ give an inclusion reversing bijection between closed subsets of $Spec(A)$ and radical ideals of $A$

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1 Answer 1

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Number 4. is sometimes called Zariski's lemma. The relation to Nullstellensatz is that if $k$ is algebraically closed, then the residue field is a finite algebraic extension of $k$, which must be $k$ itself. From this follows 3.; it is not hard to see that $k[x_1, \dots, x_n]/ \mathfrak{m} = k$ implies that $\mathfrak{m}$ is of the form $(x_1 -a_1, \dots, x_n - a_n)$. Therefore if $k$ is closed, then $k[x_1, \dots, x_n]/\mathfrak{m}= k$ for any maximal ideal (by 4), and hence all maximal ideals are of this form.

Item 2. is an immediate corrollary. You know that $V(\cdot)$ and $I(\cdot)$ are order reversing, and every ideal is contained in a maximal ideal. Clearly $V(\mathfrak{m})$ when $\mathfrak{m} = (x_1 -a_1, \dots, x_n - a_n)$ is maximal contains a point, and so $V(\mathfrak{A})$ for any other ideal contains at least a point. Thus if $V(I)$ is empty, then $I$ contains $(1)$. The reverse is obvious. Items 2,3,4 are variously/collectively called the weak Nullstellensatz by different authors.

The first one is proved from 2, 3, 4 by algebraic trickery here and is typically called the strong Nullstellensatz.

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    $\begingroup$ I think that in the form 5 is stated, it is independent from the Nullstellensatz and only requires the Scheinnullstellensatz ($\mathrm{rad}(I)=\bigcap_{\mathfrak p\supset I}\mathfrak p$). You probably refer to the bijection between algebraic sets in $\mathbb{A}^n(k)$ (i.e. zero loci of ideals) and radical ideals in $k[X_1,\dots,X_n]$. $\endgroup$
    – asdq
    Jun 9, 2018 at 8:36
  • $\begingroup$ That's funny. Supposedly "Scheinnullstellensatz" is a German derivation but you will never see it in any German textbook (otherwise I would be interested in a reference). $\endgroup$
    – quantum
    Jun 10, 2018 at 5:54
  • $\begingroup$ @asdq you are right, I misread $\endgroup$ Jan 24, 2020 at 7:52

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