Non-zero divisor and Hilbert function
$begingroup$
Consider $T=k[y_1,ldots,y_n]$ the polynomial ring in $n$ variable over $k$. I want to prove that, if $Isubset T$ 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.
proof-verification commutative-algebra exact-sequence hilbert-polynomial
edited Dec 5 '18 at 14:43
user26857
39.4k124183
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
$endgroup$
add a comment |
$begingroup$
Consider $T=k[y_1,ldots,y_n]$ the polynomial ring in $n$ variable over $k$. I want to prove that, if $Isubset T$ 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.
proof-verification commutative-algebra exact-sequence hilbert-polynomial
edited Dec 5 '18 at 14:43
user26857
39.4k124183
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
$endgroup$
1
$begingroup$
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
add a comment |
$begingroup$
Consider $T=k[y_1,ldots,y_n]$ the polynomial ring in $n$ variable over $k$. I want to prove that, if $Isubset T$ 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.
proof-verification commutative-algebra exact-sequence hilbert-polynomial
edited Dec 5 '18 at 14:43
user26857
39.4k124183
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
$endgroup$
Consider $T=k[y_1,ldots,y_n]$ the polynomial ring in $n$ variable over $k$. I want to prove that, if $Isubset T$ 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.
proof-verification commutative-algebra exact-sequence hilbert-polynomial
proof-verification commutative-algebra exact-sequence hilbert-polynomial
edited Dec 5 '18 at 14:43
user26857
39.4k124183
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
edited Dec 5 '18 at 14:43
user26857
39.4k124183
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
edited Dec 5 '18 at 14:43
user26857
39.4k124183
edited Dec 5 '18 at 14:43
user26857
39.4k124183
edited Dec 5 '18 at 14:43
user26857
39.4k124183
39.4k124183
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
asked Dec 5 '18 at 10:21
christmas_lightchristmas_light
777
777
1
$begingroup$
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
add a comment |
1
$begingroup$
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
1
1
$begingroup$
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
$begingroup$
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
add a comment |
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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