# Hilbert Series of $\mathbb{C}^2$?

Consider the following ideal in the polynomial ring $\mathbb{C} [x,y,z]$:

$$I = \langle z^2, yz \rangle$$

One can compute the Hilbert series of the affine varieties defined by this ideal. This could be done by hand or by the computer software such as $\texttt{Macaulay2}$, giving

$$HS(t;\mathbb{C} [x,y,z]/I~) = \frac{1+t-t^2}{(1-t)^2}$$

However as far as I could see that the above ideal obviously defines the same variety as the ideal

$$I' = \langle z \rangle$$ which is nothing but the complex $2$-plane $\mathbb{C}^2$. Then the Hilbert series of this should be

$$HS(t;\mathbb{C} [x,y,z]/I'~) = \frac{1}{(1-t)^2}$$

So my confusion would be: why this happened? How can we make proper transformation ($e.g.$ turning for the help of Groebner basis) so that we can unambiguously tell the Hilbert series of a given variety?

## migrated from mathoverflow.netJun 10 '15 at 7:34

This question came from our site for professional mathematicians.

• This is definitely not the same variety, the first one has an embedded point. – abx Jun 10 '15 at 6:19
• Actually, it's an embedded line (as the $x$-independence makes obvious), along $y=z=0$. Note that the difference in Hilbert series is $t/(1-t)$, the $1/(1-t)$ factor from the fact that it's one-dimensional, and the $t$ because it's "next to" the $y=z=0$ line in the big component. When dealing with monomial ideals, it's instructive to look at the set of monomials not in the ideal; your second one has a quadrant $\{x^i y^j\}$ worth, but the first has also a ray $\{ z x^i \}$. – Allen Knutson Jun 10 '15 at 12:37
• @AllenKnutson Thank you for the comment! Is there any other more intuitive way to see this is an embedded line? – Kevin Ye Jun 10 '15 at 15:31
• Let $R=\mathbb C[x,y,z]$, and instead of the $R$-module $R/I$ consider its associated graded $R/\sqrt{I} \oplus \sqrt{I}/I$ (with the same Hilbert function; here $\sqrt{I} = (z)$). The first is your plane $R/(z)$. The second is $R$-isomorphic to $R/(y,z)$ under the map $1 \mapsto z$, $R/(y,z) \to \sqrt{I}/I$, and $R/(y,z)$ is the coordinate ring of this $x$-axis. Note that the map $1\mapsto z$ changes the grading by $1$; this accounts for the $t^1$ in the numerator. – Allen Knutson Jun 10 '15 at 19:36

these two ideals define the same closed subset of $\mathbf C^3$, but they do not define the same subscheme. (Proof: as abx (corrected by Allen Knutson) says, the first one has an embedded point line.)
The ideal $$\langle z^2, yz \rangle$$ is not a radical ideal. Its radical is $$\langle z \rangle$$. Hilbert series depends on which ideal you choose. If you take the radical ideal of a variety when computing Hilbert series, then there is no ambiguity.