My professor based on pg. 320 - 321 of Eisenbud, wrote the following:
Let $I = (m_1, \dots, m_l)$ be a minimal set of monomial generators, $I' = (m_1, \dots, m_{l-1}) \subsetneq I,$ and $d = \operatorname{deg}m_l.$
$$S\mu \xrightarrow{\varphi} S/I' \rightarrow S/I \rightarrow 0$$ $$\mu \mapsto m_l + I^{'}$$
Then we have:
$$\operatorname{ker}{\varphi} = (I': m_l) = J = (m_1/\operatorname{gcd}(m_1, m_l), \dots , m_{l-1}/\operatorname{gcd}(m_{l - 1}, m_l))$$
Therefore, $\operatorname{im}(\varphi) = S/J\mu$ and we have the s.e.s.:
$$0 \rightarrow (S/J)\mu \xrightarrow{\tilde{\varphi}} S/I' \rightarrow S/I \rightarrow 0 .$$
Regard $S\mu$ as a graded $S-$module, $\operatorname{deg}\mu = d.$ Then the s.e.s. is graded. So for each $i \in \mathbb Z,$ we get a s.e.s of vector spaces
$$0 \rightarrow (S/J)_{i - d} \rightarrow (S/I')_{i} \rightarrow (S/I)_i \rightarrow 0 $$ which implies that $H_{S/I}(i) + H_{S/J}(i -d) = H_{S/I'}(i)$ and this gives us an algorithm to find $H_{S/I}(i).$
Now, here is her solution to the question I mentioned above:
1- Put $m_l = x_2x_4, I' = (x_1x_3, x_3x_4)$ then
$J = (I': x_2x_4) = (x_1x_3/\operatorname{gcd}(x_1x_3, x_2x_4), x_1x_4/\operatorname{gcd}(x_1x_4, x_2x_4)) = (x_1x_3, x_2x_1)= (x_1)$
And $S/J = k[x_2, x_3, x_4],$ so $H_{S/J}(d) = \binom{3 + d -1}{d} = \frac{(d + 1)(d + 2)}{2}$ where $d = \operatorname{deg} m_l $ and $H_{S/J}(i - 2) = \frac{(i-1)i}{2}, i \geq 2$
2- Put $m_{l}^{'}= x_1x_4$, $I^{''} = (x_1x_3)$ then $\operatorname{deg}m_{l}^{'} = 2,$ and $J' = (I'': x_1x_4) = (x_1x_3/\operatorname{gcd}(x_1x_3, x_1x_4)) = (x_1x_3/x_1) = (x_3)$
And $S/J' = k[x_1, x_2, x_4],$ so $H_{S/J'}(d) = \binom{3 + d -1}{d} = \frac{(1 + d)(2 + d)}{2}$ where $d = \operatorname{deg} m_l^{'} $ and $H_{S/J'}(i - 2) = \frac{(i - 1)i}{2}, i \geq 2$
So, $$H_{S/I^{'}}(i) + H_{S/J^{'}}(i - 2) = H_{S/I^{''}}(i).$$
Now, since $H_{S/J^{'}}(i - 2) = i - 2, i \geq 2,$ it remains to find $H_{S/I^{''}}(i)$.
Since we have $S/I^{''} = \frac{k[x_1, x_3]}{(x_1 x_3)}[x_2, x_4],$ a polynomial ring over the indeterminates $ x_2, x_4,$ monomials: $x_1^a x_{2}^b x_4^c, x_3^ax_2^b x_4^c,$ then $$S/I^{''} = k[x_1, x_2, x_4] + k[x_3, x_2, x_4].$$
But then my professor wrote $H_{S/I^{''}}(d) = 2 \frac{(d + 1)(d + 2)}{2} - (d + 1) = (d + 1)^2$.
She said that multiplication by $2$ occurs because we have $$S/I^{''} = k[x_1, x_2, x_4] + k[x_3, x_2, x_4]$$
is the addition of two polynomial rings (is that reasoning correct ?) and the subtraction of $d+1$ because we have counted $x_2x_4$ twice, but I do not understand why the number of $x_2x_4$ is $d+1,$ could someone explain this to me please?
Also, I know that $H_{S/I}(i)$ should be called Hilbert function of dimension $i$, but what should be called the Hilbert polynomial and it is polynomial in which indeterminates? Could anyone clarify this to me, please?
The process of counting above is a basic use of the inclusion-exclusion principle:
Step 1: The number of monomials $x_1^a x_{2}^b x_4^c$ that has degree $d$ is $d+2 \choose 2$ which is $\frac{(d+1)(d+2)}{2}$. These monomials include monomials of the form $x_2^bx_4^c$ of degree $d$ (when $a=0$).
Step 2: Similarly, he number of monomials $x_3^a x_{2}^b x_4^c$ that has degree $d$ is also $d+2 \choose 2$ which is $\frac{(d+1)(d+2)}{2}$. These monomials also include monomials of the form $x_2^bx_4^c$ of degree $d$ (when $a=0$).
So if we want to count the number of monomials $x_1^a x_{2}^b x_4^c, x_3^ax_2^b x_4^c$ of degree $d$, we have to take the total number of step 1 and step 2 and subtract the number of monomials of the form $x_2^bx_4^c$ of degree $d$ (because when we take the total, we count the number of monomials include monomials of the form $x_2^bx_4^c$ twice).