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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.

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    $\begingroup$ 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. $\endgroup$ – user26857 Dec 5 '18 at 14:45

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