Find $S$ whith Krull $\dim S = 0$ such that $HS(S,\lambda) = 1 +3\lambda +5\lambda^2 +2\lambda^3 +\lambda^4$

Question

Find a $\mathbb K$-standard algebra $S$ whith Krull $\dim S = 0$ such that

$$HS(S,\lambda) = 1 +3\lambda +5\lambda^2 +2\lambda^3 +\lambda^4.$$

I know the Hilbert function of $R$ is defined by

$H(R, n) = \dim_k R_n$ , for $n\in \mathbb N$

and Hilbert series is

$HS(R, \lambda) = \Sigma H(R, n) \lambda^n \in \mathbb Z[[\lambda]]$

But now how can I solve it with this information?

Answer 1

Since $S_0$ should be dimension 1 and $S_1$ should be dimension 3, we build this algebra out of $R = \mathbb{K}[X, Y, Z]$. To get the right dimensions, we just mod out by monomial ideals one degree at a time. Degrees $0$ and $1$ are taken care of, so we only kill monomials of higher degrees. Now that I've briefly described the general strategy, I'll put a spoiler around each degree calculation so that you can work it out yourself if you'd like. I will however leave the final conclusion unhidden.

Degree 2:

We have a basis for $R_2$ consisting of $X^2, XY, Y^2, YZ, Z^2,XZ$. Now, let's replace $R$ by $R/(Z^2)$ so that the new basis for $R_2$ is $X^2, XY, Y^2, YZ, XZ$, which gives the right dimension count.

Degree 3:

Taking into account that $Z^2 = 0$ in $R$, we can write the basis for $R_3$ by killing the usual monomial terms which have a $Z^2$ term. This will give a basis for $R_3$ consisting of $X^3, Y^3, X^2Y, XY^2, X^2Z, Y^2Z, XYZ$. Now, we mod out by $ZX^2, Y^2Z, XYZ, X^3, Y^3$ so that the new basis for $R_3$ is $\{X^2Y, YX^2\}$, so we can move on to the next dimension.

Degree 4:

Taking into account the earlier monomials which have been killed, the current basis for $R_4$ is just $\{X^2Y^2\}$, so we don't need to mod anything more out.

Degree $\geq 5$:

Any monomial $M$ in this higher degree will be a product of $X^2Y^2$ and a monomial of a smaller degree. Write $M = (X^2Y^2) * M'$ By the step in degree 4, we know that if $M$ is nonzero, $M'$ cannot contain a $Z$, so $M'$ either has $X$ or $Y$ raised to a nonzero power. But then, $M$ will contain either $X$ or $Y$ raised to the $\geq 3$ power, which is killed by the modding out the step in degree 4. Hence there are no nonzero monomials of degree $\geq 5$.

In conclusion, let $S = \mathbb{K}[X,Y,Z]/I$ where $$I = \langle Z^2, X^3, Y^3, ZX^2, Y^2Z, XYZ \rangle.$$ Then, $S$ will be zero-dimensional, since its Hilbert polynomial is the zero function, and by construction, its Hilbert function will have the correct values.