A question about initial ideals.
$begingroup$
Let $R = k[x_1, dots, x_m]$ be a polynomial ring over a field $k$ and $I, J$ be ideals of $R$. Further assume that $J$ is generated by the polynomials $f_1, dots, f_r$. Fix a monomial order $<$ on the monomials of $R$. Is it true that, with respect to the fixed order,
$LT (I + J') subset LT (I + J)$, where $J'$ is generated by the initial terms of $f_1, dots, f_r$ and $LT (I)$ denote the usual initial ideal of $I$ with respect to $<$ for any ideal $I$?
commutative-algebra monomial-ideals
edited Jan 13 at 9:42
user26857
39.3k124183
asked Jan 8 at 16:12
M. D.M. D.
283
$endgroup$
add a comment |
$begingroup$
Let $R = k[x_1, dots, x_m]$ be a polynomial ring over a field $k$ and $I, J$ be ideals of $R$. Further assume that $J$ is generated by the polynomials $f_1, dots, f_r$. Fix a monomial order $<$ on the monomials of $R$. Is it true that, with respect to the fixed order,
$LT (I + J') subset LT (I + J)$, where $J'$ is generated by the initial terms of $f_1, dots, f_r$ and $LT (I)$ denote the usual initial ideal of $I$ with respect to $<$ for any ideal $I$?
commutative-algebra monomial-ideals
edited Jan 13 at 9:42
user26857
39.3k124183
asked Jan 8 at 16:12
M. D.M. D.
283
$endgroup$
add a comment |
$begingroup$
Let $R = k[x_1, dots, x_m]$ be a polynomial ring over a field $k$ and $I, J$ be ideals of $R$. Further assume that $J$ is generated by the polynomials $f_1, dots, f_r$. Fix a monomial order $<$ on the monomials of $R$. Is it true that, with respect to the fixed order,
$LT (I + J') subset LT (I + J)$, where $J'$ is generated by the initial terms of $f_1, dots, f_r$ and $LT (I)$ denote the usual initial ideal of $I$ with respect to $<$ for any ideal $I$?
commutative-algebra monomial-ideals
edited Jan 13 at 9:42
user26857
39.3k124183
asked Jan 8 at 16:12
M. D.M. D.
283
$endgroup$
Let $R = k[x_1, dots, x_m]$ be a polynomial ring over a field $k$ and $I, J$ be ideals of $R$. Further assume that $J$ is generated by the polynomials $f_1, dots, f_r$. Fix a monomial order $<$ on the monomials of $R$. Is it true that, with respect to the fixed order,
$LT (I + J') subset LT (I + J)$, where $J'$ is generated by the initial terms of $f_1, dots, f_r$ and $LT (I)$ denote the usual initial ideal of $I$ with respect to $<$ for any ideal $I$?
commutative-algebra monomial-ideals
commutative-algebra monomial-ideals
edited Jan 13 at 9:42
user26857
39.3k124183
asked Jan 8 at 16:12
M. D.M. D.
283
edited Jan 13 at 9:42
user26857
39.3k124183
asked Jan 8 at 16:12
M. D.M. D.
283
edited Jan 13 at 9:42
user26857
39.3k124183
edited Jan 13 at 9:42
user26857
39.3k124183
edited Jan 13 at 9:42
user26857
39.3k124183
39.3k124183
asked Jan 8 at 16:12
M. D.M. D.
283
asked Jan 8 at 16:12
M. D.M. D.
283
asked Jan 8 at 16:12
M. D.M. D.
283
283
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
No, this is not true, even in $k[x]$. Let $I = J = (1 +x)$. Then $J' = (x)$. We have $I + J' = k[x]$, and $I + J = (1+x)$.
The opposite inclusion also does not usually hold. Let $I = (x)$ and $J = (1+x)$. Then $I + J' = (x)$ and $I + J = k[x]$
Can you find an example where $LT(I +J')$ and $LT(I + J)$ are incomparable?
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No, this is not true, even in $k[x]$. Let $I = J = (1 +x)$. Then $J' = (x)$. We have $I + J' = k[x]$, and $I + J = (1+x)$.
The opposite inclusion also does not usually hold. Let $I = (x)$ and $J = (1+x)$. Then $I + J' = (x)$ and $I + J = k[x]$
Can you find an example where $LT(I +J')$ and $LT(I + J)$ are incomparable?
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
$endgroup$
add a comment |
$begingroup$
No, this is not true, even in $k[x]$. Let $I = J = (1 +x)$. Then $J' = (x)$. We have $I + J' = k[x]$, and $I + J = (1+x)$.
The opposite inclusion also does not usually hold. Let $I = (x)$ and $J = (1+x)$. Then $I + J' = (x)$ and $I + J = k[x]$
Can you find an example where $LT(I +J')$ and $LT(I + J)$ are incomparable?
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
$endgroup$
add a comment |
$begingroup$
No, this is not true, even in $k[x]$. Let $I = J = (1 +x)$. Then $J' = (x)$. We have $I + J' = k[x]$, and $I + J = (1+x)$.
The opposite inclusion also does not usually hold. Let $I = (x)$ and $J = (1+x)$. Then $I + J' = (x)$ and $I + J = k[x]$
Can you find an example where $LT(I +J')$ and $LT(I + J)$ are incomparable?
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
$endgroup$
No, this is not true, even in $k[x]$. Let $I = J = (1 +x)$. Then $J' = (x)$. We have $I + J' = k[x]$, and $I + J = (1+x)$.
The opposite inclusion also does not usually hold. Let $I = (x)$ and $J = (1+x)$. Then $I + J' = (x)$ and $I + J = k[x]$
Can you find an example where $LT(I +J')$ and $LT(I + J)$ are incomparable?
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
answered Jan 9 at 4:22
Badam BaplanBadam Baplan
4,586722
4,586722
add a comment |
add a comment |
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