# On the generator of power of ideal

Let $I$ be a graded ideal in a polynomial ring $R$, which is generated minimally by $x_1,...,x_k$. Then the power of $I$, i.e $I^t$ is generated by monomials of the form $x_{1}^{a_1}...x_{n}^{a_{n}}$ where $a_1+...+a_n=t$. Denote this set by $S$.

Can we say anything (others than above)about the minimal generating set of $I^{t}$? Is it $S$ ?

Given a minimal generating set for a graded ideal in a graded commutative ring, from these how much do we know about the minimal generating set for the power of it?

Edit : Here is an example for precising my question :

In the polynomial ring $k[x,y,z]$ let $I=(x^2, xy^3, y^2z^3)$, then $I^2=(x^4, x^2y^6, y^4z^6, x^3y^3, xy^5z^3, x^2y^2z^3)$

Is $\lbrace x^4, x^2y^6, y^4z^6, x^3y^3, xy^5z^3, x^2y^2z^3\rbrace$ a minimal generating set for $I^2$ ?

Update There are some typing mistake that I have not noticed. I have change my question. This time, the generating set of $I$ is minimal. So what can we say about the generating set for $I^2$ above ? Is it minimal? Thank you everyone for helping me answer my question!

• I think you don't mean $I : I^t$, since that is always $R$. (Note $I^t \subseteq I$). Perhaps you mean $I^t : I$? Oct 9, 2012 at 4:30
• In your example, you do not need $x^6y^2$ because you have $x^4$. Can you see a pattern in this, at least for monomial ideals? Oct 9, 2012 at 5:45
• Let $I$ be a graded ideal of a graded commutative ring $A$. Let $R=A/I^t$. Then the minimal generating set for (the image of) $I^t$ in $R$ is empty. Oct 9, 2012 at 13:21
• Richard's example suggests that you cannot deduce the minimal generating set of $I^t$ from the generators of $I$ without knowing more about your ring. Oct 11, 2012 at 8:56
• Still more typos x^3y^5z^3. Keep at it until it is right. Gerhard "If At First You Mistype..." Paseman, 2012.10.11 Oct 12, 2012 at 1:30

In general, the answer to your first question (second paragraph) is NO, it is not $S$, even for monomial ideals in a polynomial rings. Take the ideal $I$ generated by $x_1 = a^4b, x_2=b^4a, x_3=a^3b^3$. Then $x_3^2$ is not a minimal generator for $I^2$ since it is divisible by $x_1x_2$.

One particular case when the answer is YES is when the $x_1,\cdots, x_n$ form a regular sequence.

For specific examples, it may be worth learning some program such as Macaulay 2. In your specific example $S$ is the minimal generating set for $I^2$, provided that you fix the fifth entry as Gerhard pointed out.

For a monomials ideal one can also visualize the minimal generators as the points on the convex hull of the set of degrees of the ideal.

As my first example shows, one can not make very good statement about a specific power of $I$. However, asymptotically we can say quite a bit:

The minimal number of generators of $I^n$ for $n>>0$ is a polynomial in $n$. The degree of this polynomial is called the analytic spread of $I$ (geometrically it is one more than the dimension of the exceptional fibre of the blow-up).

• Dear Hailong Dao, firstly I want to say thank for your answer, it helps me alot. Secondly, I do not care about the fact that the question may let people think I am stupid, I just want to know the answer and how to think about it, it is more worthy for me. About your answer, can I ask(may be another stupid one) a question : If $x_1,..x_n$ minimally generate a graded ideal $I$, and they form a regular sequence, then S is the minimal generating set ?
– Knot
Oct 13, 2012 at 15:24
• The answer is yes. In Bruns-Herzog, I think somewhere in the first couple sections of chapter 1, there's a theorem (or perhaps an exercise? a student has my book at the moment) that says that if an ideal $I$ is generated by a regular sequence of length $n$, then the graded ring associated to $I$ is graded-isomorphic to the polynomial ring over $R/I$ in $n$ variables. The $t$th graded piece is then essentially the free module over $R/I$ with generators the image of the set $S$. The result follows. Oct 14, 2012 at 16:56
• @Neil Epstein : What will happen if I is a monomial ideal ?
– Knot
Oct 15, 2012 at 13:43