# Monthly Archives: February 2015

## Minimum number of prime ideals

Let $K$ be an algebraically closed field and $R=K[x,y]$. Let $J=(P(x),Q(y))$ $\subseteq R$ be an ideal. Since $R$ is noetherian, a theorem of Noether says that there are finitely many prime ideals $p_1,\dots,p_n\supseteq J$ such that $p_1\cdots p_n\subseteq J$.

Suppose that $P(x)=(x-a)^A(x-b)^B$ and $Q(y)=(y-c)^C$ are uniquely factored into irreducibles. Then each $p_i$ is equal to $(x-a,y-c)$ or $(x-b,y-c)$. Suppose that $r$ of them are $(x-a,y-c)$ and $s$ of them are $(x-b,y-c)$. Then $\displaystyle p_1\cdots p_n=\left((x-a)^i(x-b)^j(y-c)^k:\begin{cases}0\le i\le r\\ 0\le j\le s\\ 0\le k\le r+s\\ i+j+k=r+s\end{cases}\right)$

To have $p_1\cdots p_n\subseteq J$, therefore, we need $\{k\ge C\}\cup(\{i\ge A\}\cap\{j\ge B\})$.

So we need $k\ge C$ or $\min\{i,j\}\ge\max\{A,B\}$, i.e. $k+\min\{i,j\}\ge C+\max\{A,B\}-1$ \begin{aligned}\Leftrightarrow n=r+s&=k+\min\{i,j\}+\max\{i,j\}\\ &\ge C+\max\{A,B\}-1+\max\{i,j\}.\end{aligned}

As $i,j,k$ vary, we may replace $\max\{i,j\}$ with $\max\{r,s\}$. We therefore deduce:

Bound 1. $\min\{r,s\}\ge C+\max\{A,B\}-1$.

Bound 2. $n\ge 2(C+\max\{A,B\}-1)$.

Remark. Bound 1 is obviously stronger; however, since in general we don’t have information on $r$ and $s$, bound 2 is uniformly the best possible.

The calculation becomes much more subtle for general polynomials.