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.

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