## 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.