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Let $A=\mathbb C[x_0,\dots,x_{m-1}]$ be the polynomial ring on $m$ variables.

Define $X(u)=\sum_{i=0}^{m-1} x_i u^{i+1}$ and denote by $(X(u)^r)_n$ the coefficient of $u^n$ in the expansion of the $r$-th power of $X(u)$, i.e $X(u)^r$.

Set $I=\langle (X(u)^{r})_s\mid 1\le r \le m+1, \ s\ge m+1 \rangle$.

I am trying to find a basis for $A/I$ and I am guided by some questions:

1) Is that possible to calculate the dimension of $A/I$? In fact, I am happy if I find a way to prove that it is at most $2^m$.

2) What is a Gröbner basis for $I$?

3) What is a linear basis for $A/I$?

Any help are welcome!


Added: If we let $m=2$, then $$X(u) = x_0u +x_1u^2$$ $$X(u)^2 = x_0^2u^2 + 2x_0x_1u^3 + x_1^2 u^4$$ $$X(u)^3 = x_0^3 u^3+ 3x_0^2x_1u^4 + 3 x_0x_1^2u^5+ x_1^3u^6$$ so then $I$ would be generated by $$\{x_0x_1,\ x_1^2, \ x_0^3,\ x_0^2x_1, \ x_0x_1^2,\ x_1^3 \}.$$ Therefore we conclude that $I$ is the ideal generated by $\{ x_0x_1,\ x_1^2, \ x_0^3 \}$, so that $A/I$ is a $\mathbb C$-space with basis given by the image (with respect to the natural projection) of $\{ 1, \ x_0, \ x_0^2, \ x_1 \}$.

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    Let me make sure I'm understanding your definitions right. If we let $m=2$, then $$\begin{array}{lcl} X(u) &=& x_0+u x_1 \\ X(u)^2 &=& x_0^2+2 x_0x_1 u+x_0^2 u^2 \\ X(u)^3 &=& x_0^3+3 x_0^2 x_1 u +3 x_0 x_1^2u^2+ x_1^3u^3 \end{array}$$ so then $I$ would be generated by $0$ and $x_1^3$, right?2012-12-27
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    @Alexander Gruber: I am sorry, I forgot to lift the power of $u$ in the definition of $X(u)$. As you have spent your time doing calculations, I added in the question the example of the case $m=2$. Thanks.2012-12-27
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    Alright, cool, after lifting the $u$ that is what I get for $m=2$ also.2012-12-27
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    In commutative algebra *dimension* of a ring stands for the Krull dimension. Since you want in (1) the dimension of $A/I$ as a $\mathbb C$-vectorspace, maybe it would be good to say it explicitely.2012-12-28

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