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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 $x_1^{k},...,x_n^{k}$?

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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 $\rm\: I_j = (x_j).$

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Dear Bill Dubuque, does it preserve the minimal property of the generating set? – Arsenaler Aug 21 '12 at 5:19
@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 $\rm\:xy\in(x^2,y^2)$? – Gone Aug 21 '12 at 14:52
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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 $x_i$.

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