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

\begin{equation} I = \langle z^2, yz \rangle \end{equation}

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

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

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

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

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

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.net Jun 10 '15 at 7:34

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  • 3
    $\begingroup$ This is definitely not the same variety, the first one has an embedded point. $\endgroup$ – abx Jun 10 '15 at 6:19
  • 1
    $\begingroup$ 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 $\{ z x^i \}$. $\endgroup$ – Allen Knutson Jun 10 '15 at 12:37
  • $\begingroup$ @AllenKnutson Thank you for the comment! Is there any other more intuitive way to see this is an embedded line? $\endgroup$ – Kevin Ye Jun 10 '15 at 15:31
  • 1
    $\begingroup$ 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. $\endgroup$ – Allen Knutson Jun 10 '15 at 19:36

To expand on abx's comment:

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.)

Since the Hilbert series is an invariant of the ring, it is sensitive to the scheme structure, and not just the closed subset.

  • $\begingroup$ Thank you for the explanation! Could you elaborate the scheme structure in the answer so that I can accept it? $\endgroup$ – Kevin Ye Jun 10 '15 at 23:31

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.


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