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In the paper, page 6, Example 5.1, let $I=(x_2 x_3, x_1 x_3, x_1 x_2) \subseteq K[x_1, x_2, x_3]$. It is said that the Hilbert series of $K[x_1, x_2, x_3]/I$ is $H(I, t) = \frac{3t^2-2t^3}{(1-t)^3}$.

But using Maple, I computed $HilberSeries( [x_2 x_3, x_1 x_3, x_1 x_2], \{x_1, x_2, x_3\}, t ) = \frac{1+2t}{1-t}$. The results are different. Have I made some mistake? Thank you very much.

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    $\begingroup$ That's a completely different ring; the quotient by $I$ is not the subalgebra generated by the generators of $I$. $\endgroup$ – Qiaochu Yuan Nov 26 '17 at 19:52
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$I$ contains all monomials $x_1^a x_2^bx_3^c$ where at least two of $a$, $b$, $c$ are $>0$. So the only monomials it doesn't contain are the $x_1^a$, $x_2^b$, $x_3^c$ and these form a basis for $R/I$. There are $3$ of these basis elements in each dimension $\ge1$ so the Hilbert series is $$1+\sum_{n=1}^\infty 3t^n=1+\frac{3t}{1-t}.$$

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The Hilbert series of the algebra $K[x_1, x_2, x_3]/I$ is $$H(K[x_1, x_2, x_3]/I, t) = 1 + \frac{3t}{1 - t} = \frac{1 + 2t}{1 - t},$$ as explained by Lord Shark the Unknown and computed by Maple.

However, $H(I, t)$ is not the Hilbert series of the algebra $K[x_1, x_2, x_3]/I$, but the Hilbert series of the ideal $I$ (as a graded subspace of the graded vector space $K[x_1, x_2, x_3]$). Note that the (images mod $I$ of the) monomials $x_1^a, x_2^b, x_3^c$ form a graded basis for $K[x_1, x_2, x_3]/I$, and the remainder of the monomials form a graded basis for $I$. Hence, $$\begin{aligned} H(I, t) &= H(K[x_1, x_2, x_3], t) - H(K[x_1, x_2, x_3]/I, t) \\ &= \frac{1}{(1- t)^3} - \frac{1 + 2t}{1 - t}\\ &= \frac{3t^2 - 2t^3}{(1 - t)^3},\end{aligned}$$ as stated in the paper to which you linked.

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