## Archive for the ‘mathematics’ Category

### 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.

### Schonhardt polyhedron and IMS-2013/4 program

28 May, 2012

One cannot always triangulate a non-convex polyhedron using only its vertices, sometimes one need to add more of them. A simple example of this phenomenon isĀ Schonhardt polyhedron. Here is a picture illustrating how one builds it that I drew for a forthcoming paper, using Tikz LaTeX package, which is awesome, but totally overwhelming.

It fact, it’s easy to see that the 6 vertices and 12 edges it has are not enough. Indeed, each pair of non-intersecting edges determines a simplex, but it’s easy to observe that any such selection will include one the forbidden pairs of vertices AC, A’B, or B’C’. (The LaTeX source of the picture is here).

The paper I mention is related to a topic of the program on inverse moment problems at IMS (NUS/Singapore) in late 2013-early 2014 which I co-organize.