# Dimension of a certain quotient ring of $\mathbb{C}[x_0,\ldots,x_{m-1}]$.

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? –  Alexander Gruber Dec 27 '12 at 5:32
@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. –  Binai Dec 27 '12 at 12:57
Alright, cool, after lifting the $u$ that is what I get for $m=2$ also. –  Alexander Gruber Dec 27 '12 at 13:09
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. –  user26857 Dec 28 '12 at 17:25

So, if you happen to have access to MAGMA, you can run this code.

m := 5;
P<[x]> := PolynomialRing(RationalField(), m+1);

p := P!0;
for i in [1..m] do
p := p + x[i]*x[m+1]^i;
end for;

Igen := [];
for r in [1..m+1] do
C := Coefficients(p^r,x[m+1]);
for s in [m+2..#C] do
Append(~Igen,C[s]);
end for;
end for;

I := ideal<P|Append(Igen,x[m+1])>;

GroebnerBasis(EasyIdeal(I));

Q := P/I;

Dimension(Q);


Using this I've found that $\text{dim}(A/I)=2^m$ exactly for $m\leq 9$ and computed the corresponding Groebner bases and quotient rings. I am not sure how this could be proven for all $m$. Aside from actually computing it, I don't know if there is any way to get the basis of $I$ or $A/I$ in general, as the output seems quite complicated. MAGMA uses the $\rm F_4$ algorithm to find Groebner bases, which seems to be the fastest method available at this point in time. If you try to implement a Groebner basis algorithm yourself, however, I would suggest Mutant XL since it is intuitively very easy and still pretty fast.

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Hope this helps! Feel free to leave the question open if you need to, especially if you don't have access to MAGMA. –  Alexander Gruber Dec 27 '12 at 14:12
Thanks for the suggestion by using MAGMA. I wrote a procedure to calculate the Grobner basis for $I$ and the linear basis for $A/I$ and I am obtaining the answer until $m=9$ too. I will leave open the question to see if anyone can suggest anything else. –  Binai Dec 27 '12 at 14:23