# Non-zero divisor and Hilbert function

Consider $$T=k[y_1,\ldots,y_n]$$ the polynomial ring in $$n$$ variable over $$k$$. I want to prove that, if $$I\subset T$$ is an ideal and $$y_1$$ is not a zero-divisor in $$T/I$$, then $$HF(T/I,t)=\sum_{i=0}^t HF(T/(I+(y_1)), i ).$$

My idea: consider the short exact sequence

$$0 \rightarrow T/I(-1) \xrightarrow{\text{y_1}} T/I \rightarrow T/(I+(y_1)) \rightarrow 0.$$

Then we have that $$HF(T/I,i-1)-HF(T/I,i)+HF(T/(I+(y_1),i)=0$$, so

$$HF(T/I,i)=HF(T/I,i-1)+HF(T/(I+(y_1),i)$$

Using this recursively we obtain the thesis. Now my questions are:

1) Is this proof correct? (It's the first time I study the Hilbert function, so I'd like to see if I understand the mechanic.)

2) Can I replace $$y_1$$ with any other element which is not a zerodivisor in $$T/I$$? I'm not sure I can do this, but I didn't find a counterexample.

• 1) Yes, the proof is correct; 2) Yes, you can replace $y_1$ by any homogeneous (degree one) non-zerodivisor on $T/I$. However, if degree of this element is not one then you get a different equation. – user26857 Dec 5 '18 at 14:45