Archive for August, 2012

Weak Nullstellensatz for C

20 August, 2012

Weak Nullstellensatz says, that for an algebraically closed field {\mathbb{K}} each maximal ideal {I} in {R=\mathbb{K}[X_1,\dots,X_n]} has form {I=(X_1-a_1,\dots,X_n-a_n)}, for {a:=(a_1,\dots,a_n)\in \mathbb{K}^n}, i.e. {I=I(\{a\})}, the ideal of an one-element subset of {\mathbb{K}^n}. Its proof in this generality needs quite a bit of commutative algebra. However, if we futher assume that {\mathbb{K}} is uncountable (thus covering a very important case {\mathbb{K}=\mathbb{C}}) we can give a much quicker proof.

Theorem 1 Let {\mathbb{K}} be an uncountable algebraically closed field, e.g. {\mathbb{K}=\mathbb{C}}, and {I} a proper maximal ideal in {R=\mathbb{K}[X_1,\dots,X_n]}. Then there exists {a:=(a_1,\dots,a_n)\in \mathbb{K}^n} such that {I=I(\{a\})}.

Proof: The first step is to show that {R/I\cong\mathbb{K}}. To see this, we will show that every {b\in R/I} is algebraic, i.e. a root a nonzero polynomial {f\in\mathbb{K}[z]}. Note that the dimension of {R/I} as a vectorspace over {\mathbb{K}} is at most countable, as {R/I} is generated by the images {\phi(X^J)} of the monomials {X^J} under the ring homomorphism {\phi:R\rightarrow R/I}, and the exponents {J} form a countable set. Thus for {b\in R/I\setminus\mathbb{K}} the set

\displaystyle  \left\{\frac{1}{b-t}\mid t\in \mathbb{K}\right\}

is linearly dependent, i.e. there exist {\mu_1,\dots,\mu_m} such that { \sum_{i=1}^m\frac{\mu_i}{b-t_i}=0. } Thus

\displaystyle  0=\sum_{i=1}^m\frac{\mu_i}{b-t_i}=\frac{P(b)}{\prod_i{(b-t_i)}},

where {P\in\mathbb{K}[z]} and {P(b)=0}. As {\mathbb{K}} is algebraically closed, we have that {P} is linear, i.e. {R/I\cong\mathbb{K}}.

Next, we observe that {\phi} maps {R} to {\mathbb{K}}, and set {a_i:=\phi(X_i)}, for {1\leq i\leq n}. As {\phi(X_i-a_i)=0}, we see that {X_i-a_i\in I}, for {1\leq i\leq n}. By maximality of {I}, we obtain {I=(X_1-a_1,\dots,X_n-a_n)}, as claimed. \Box

Here one can find reduction of the general case (not assuming non-countability of \mathbb{K}) to this one.