Is the Hilbert series of the monomial ideal $\langle x^3, xyz \rangle$ equal to $\frac{2x^3-x^5}{(1-x)^3}$?

Question

Let $R$ be a polynomial ring of 3 variables over a field and take the monomial order lex.
Let $I = \langle x^3, xyz \rangle$

There is just one syzygy $\left[\begin{matrix}yz \\-x^2\end{matrix}\right]$ and when I make the free resolution

$$0 \to R(-5) \xrightarrow[]{ \left[\begin{matrix}yz \\-x^2\end{matrix}\right]} R(-3)^2 \xrightarrow[]{\left[\begin{matrix}x^3 & xyz\end{matrix}\right]} R \to R/I \to 0$$

I get

$$H_{R/I}(x) = \frac{1-2x^3+x^5}{(1-x)^3}$$

so

$$H_{I}(x) = H_R(x) - H_{R/I}(x) = \frac{2x^3-x^5}{(1-x)^3}$$

But I tried to check my answer with this SageMath code

P.<x,y,z> = PolynomialRing(QQ, 3, order='lex')
I = Ideal(P, [x^3, x*y*z])
H = I.hilbert_series()

print (H)
#difference with my answer should be zero
#is there a better way to check?
#H == (2*t^3-t^5)/(1-t)^3 doesn't work
t = var("t")
plot(H - (2*t^3-t^5)/(1-t)^3, (t, -0.5, 0.5))

And it gives

$$H_I (x) = \frac{x^4 + x^3 - x^2 - x - 1}{-x^2 + 2x - 1}$$

Where is my mistake?

EDIT: So, I am pretty confident that my calculation is correct, since I did this as an example to the recursive calculation of Hilbert series of monomial ideals via the formula

$$H_{R/I}(x) = H_{R/I'}(x) - x^{\deg{m_k}} H_{R/J}(x)$$

where $I = (m_1, \dots, m_k)$, $I' = (m_1, \dots, m_{k-1})$ and $J=(\frac{m_1}{\gcd(m_1, m_k)}, \dots, \frac{m_{k-1}}{\gcd(m_{k-1}, m_k)})$

I finally managed to prove that and then calculated ($I' = (x^3)$ and $J=(x^2)$ in this case)

$$H_{R/I'} (x) = \frac{1-x^3}{(1-x)^3}$$ $$H_{R/J} (x) = \frac{1-x^2}{(1-x)^3}$$

So we have

$$H_{R/I'} - x^3 H_{R/J} = \frac{2x^3-x^5}{(1-x)^3}$$

Hence the question is why did my code give the wrong result?

Answer 1

For certain computer algebra systems, specifying certain computations for an ideal $I$ is understood to be asking for such a computation on $R/I$, e.g, asking Sage for the Hilbert series of $I$ is going to be interpreted as asking for $H_{R/I}(x)$. If you look at the $H_{R/I}(x)$ that you computed by hand, you'll notice the numerator is divisible by $(1-x)$ and that dividing out this factor from numerator and denominator, you get exactly the printed series from your code.