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If $R$ is a local Noetherian regular ring and $I$ is an ideal contained in the maximal ideal. Can we compare the number of minimal set of generator of $I$ and $I^2$?
Thanks a lot for helping me.

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What kind of result are you expecting? – Youngsu Apr 3 '13 at 20:57
Let $I$ be generated by $n$ elements. Then clearly $I^2$ requires at most $n^2$ generators, and clearly this bound is realized for the local ring of the origin in affine $n$-space. So what more might you hope to say? – Steven Landsburg Apr 3 '13 at 22:18
Steven, in a commutative ring you can get away with $n+1 \choose 2$ generators for $I^2$. This works for your example as well. – Sándor Kovács Apr 3 '13 at 22:45
Sandor: that's indeed what I meant to say. Thanks. – Steven Landsburg Apr 4 '13 at 4:16
Since the ring is assumed to be regular, $\dim I^2 / I^3 = \binom{n+1}{2}$ where $n = \dim I / I^2$ is the minimal number of generators of $I$. So $I^2$ needs at least $\binom{n+1}{2}$ generators. – user91132 Oct 30 '15 at 8:39

Confirming Sandor's answer, I would say that the bound $n+1 \choose 2$ is the best answer in many cases. The ring may be only Noetherian local. Infact there exists always a surjective map from symmetric power to the ordinary power $Sym^2(I)\to I^2$. Tensoring this into $R/\frak{m}$ we see that num of gens of $I^2$ is at most num gens of $Sym^2(I/{\frak{m}} I)$ the latter is degree $2$ monomials in the polynomial ring with $\mu(I)=n$ variables, so that we have $\mu(I^2)\leq {n+1 \choose 2}$(Obviously ;)). The non-obvious fact is that, the equality happens if $Sym^2(I)\to I^2$ is an isomorphism. The class of ideals with this property contains the class of "ideals of linear type" and it is called "syzygetic ideals". Actually $ker(Sym^2(I)\to I^2)=T_2(Spec(R),Spec(R/I))$. For example if an ideal is "Strongly Cohen-Macaulay" and "generically Complete Intersection" then $T_2=0$. A concrete example is the following: The ideal of the definition of the image of the map $P^1\to P^4$ given by $(s^8:s^5t^3:s^4t^4:s^3t^5:t^8)$.

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