Determinantal ideal theorem: Difference between revisions

Revision as of 08:37, 10 August 2007

History

The result was proved for polynomial rings by Macaulay and for arbitrary Noetherian rings by Eagon.

Statement

Let $M$ be a $p\times q$ matrix with entries over a Noetherian ring $R$ . Denote by $I_{k}(M)$ the ideal generated by the $k\times k$ minors of $M$ . Then, the codimension of any prime ideal minimal over $I_{k}(M)$ is at most $(p-k+1)(q-k+1)$ .

The case $k=p=1$ yields Krull's height theorem and the case $p=q=k=1$ yields Krull's principal ideal theorem.

@@ Line 5: / Line 5: @@
 ==Statement==
-Let <math>M</math> be a <math>p \times q</math> matrix with entries over a [[Noetherian ring]] <math>R</math>. Denote by <math>I_k(M)</math> the ideal generated by the <math>k \times k</math> minors of <math>M</math>. Then, the [[codimension of a prime ideal|codimension of any prime ideal]] minimal over <math>I_p(M)</math> is at most <math>q - p + 1</math>.
+Let <math>M</math> be a <math>p \times q</math> matrix with entries over a [[Noetherian ring]] <math>R</math>. Denote by <math>I_k(M)</math> the ideal generated by the <math>k \times k</math> minors of <math>M</math>. Then, the [[codimension of a prime ideal|codimension of any prime ideal]] minimal over <math>I_k(M)</math> is at most <math>(p - k + 1)(q - k + 1)</math>.
-The case <math>p = 1</math> yields [[Krull's height theorem]] and the case <math>p = q = 1</math> yields [[Krull's principal ideal theorem]].
+The case <math>k = p = 1</math> yields [[Krull's height theorem]] and the case <math>p = q = k = 1</math> yields [[Krull's principal ideal theorem]].