Cheap Nullstellensatz

I was recently asked to prove a non-exhaustive set of cases for the weak Nullstellensatz, and the proof was so deceptively simple I decided to share it. The main theorem goes along the lines of this.

Nullstellensatz: Let K be an algebraically closed field and \mathfrak{m} an ideal of K[X_1, \dots, X_n]. Then \mathfrak{m} is maximal if and only if \mathfrak{m} = \langle X_1 - \alpha_1, \cdots, X_n - \alpha_n \rangle.

It is not hard to see that \mathfrak{m} = \langle X_1 - \alpha_1, \cdots, X_n - \alpha_n \rangle is a maximal ideal. We can compute an isomorphism \varphi : K[X_1, \dots, X_n] / \mathfrak{m} \to K by looking at the surjective ring map X_i \mapsto \alpha_i. It remains to show that all maximal ideals are of this form. We will prove a subset of cases for K as follows.

Cheap Nullstellensatz: Let K be an algebraically closed field of uncountable cardinality and \mathfrak{m} an ideal of K[X_1, \dots, X_n]. If \mathfrak{m} is maximal then \mathfrak{m} = \langle X_1 - \alpha_1, \cdots, X_n - \alpha_n \rangle.

Proof: We do it in a few parts.

  1. Every field extension F / K has an embedding K(x) \hookrightarrow F, where K(x) is the set of rational polynomials of K.

    If F is a non-trivial extension of K (i.e. F \neq K), then F must have a transcendental element \alpha, and P(\alpha) \neq 0 for all non-zero P \in K[X]. Thus, we can define the map f \mapsto f(\alpha) and we claim this is an embedding of K(x) into F. It is clearly well-defined because if f = \frac{p}{q}, then \frac{p(\alpha)}{q(\alpha)} does not divide by zero. It is also easy to check that this is a ring map and because f(\alpha) = 0 \iff f = 0, we can see that it is injective because its kernel is \lbrace 0 \rbrace.

  2. K(x) is a K-vector space and has uncountable dimension. It is not too hard to see that K(x) is a vector space, and to show that its dimension is uncountable, we show that the set of functions of the form \dfrac{1}{x - \beta}, where \beta \in K, is a linearly independent set. Thus, we wish to find coefficients a_i such that

        \[ \sum_{i = 1}^n \frac{a_i}{x - \beta_i} = 0. \]

    We clear denominators and call the resulting function \Phi, or

        \[ \Phi = \sum_{i = 1}^n \left( a_i \prod_{j \neq i} (x - \beta_j) \right) = 0. \]

    Now consider the ring homomorphisms \varphi_i sending x to \beta_i and for all i, compute

        \[ \varphi_i(\Phi) = a_i \prod_{j \neq i} (\beta_i - \beta_j) \right) = \varphi_i(0) = 0. \]

    The product is a product of non-zero elements and thus invertible, so we can calculate a_i = 0 for all i, and thus the set is linearly independent of uncountable cardinality, so K(x) has uncountable dimension over K.

  3. Let R be a finitely generated K-algebra. Then R has countable dimension over K as a vector space. To see this, let x_1, \dots, x_n be the generators of R. We can write r \in R as

        \[ r = P_r(x_1, \dots, x_n) \]

    for some P_r \in K[X_1, \dots, X_n] so R is some sub-algebra of the free algebra K \langle X_1, \dots, X_n \rangle. The free algebra has countable basis \lbrace X_1^{e_1} \cdots X_n^{e_n} \rbrace so R must also have countable basis.

We now have all the building blocks to finish the proof. We notice that K[X_1, \dots, X_n] is a finitely generated K-algebra, and its quotient field K' = K[X_1, \dots, X_n] / \mathfrak{m} must also be a finitely generated field extension. If K' is a transcendental extension, then it has uncountable dimension over K because it contains K(x) as a subfield, but this dimension is also countable because K' is finitely generated. Thus K' cannot be transcendental so it must be an algebraic extension, but K is algebraically closed so K' = K.

This entry was posted in Math. Bookmark the permalink.

Leave a Reply

Your email address will not be published. Required fields are marked *