1
$\begingroup$

Let $I$ be an ideal in a Noetherian ring $R$ which is generated by $x_1,...,x_n$. From this system, can we find out what is the generating set for an arbitrary power of $I$: $I^k$? Is it $x_1^{k},...,x_n^{k}$?

2 Answers 2

5

All k-degree monomials in $x_1,x_2,\dotsc,x_n$ work. In general you need as many, by looking at the complex polynomial ring over $x_i$.

1

Hint $\ $ The multinomial theorem for ideals is even simpler than that for ring elements

$\rm (I_1 \!+\, \cdots +I_n)^k\, =\, \smash{\sum_{j_1+\cdots + j_n\, =\, k} I_1^{\,j_1} \cdots I_n^{\,j_n}}$ Your problem is the special case $\rm\: I_j = (x_j).$

  • 0
    @msnaber Hint: consider simple examples, e.g. $\rm\:(x,y)^2 = (x^2,xy,y^2).\:$ What happens if $\rm\:x^2 = 0\:$ or $\rm\:xy = 0,\:$ or $\rm\:xy\in(x^2,y^2)$?2012-08-21