Existence of a polynomial not vanishing on subvarietiesIsomorphism including a tensor productProof of Proposition 2.1.1 in Bruns and HerzogEquivalence of weak forms of Hilbert's Nullstellensatz$k$-algebra homomorphism of the polynomial ring $k[x_1,dots,x_n]$Integer polynomials having no common roots in $mathbb C^n$Singular homogeneous polynomialIs the ideal of a variety the annihilator of a subspace of the symmetric algebra?proof: if $K(x_1,ldots,x_k)/K$ is a field extension, and the $x_j$ are algebraically independent over K, M cannot be a finitely generated K-algebraThe existence of a linear map onto an affine algebraic setInequality involving Ideals generated by forms.
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Existence of a polynomial not vanishing on subvarieties
Isomorphism including a tensor productProof of Proposition 2.1.1 in Bruns and HerzogEquivalence of weak forms of Hilbert's Nullstellensatz$k$-algebra homomorphism of the polynomial ring $k[x_1,dots,x_n]$Integer polynomials having no common roots in $mathbb C^n$Singular homogeneous polynomialIs the ideal of a variety the annihilator of a subspace of the symmetric algebra?proof: if $K(x_1,ldots,x_k)/K$ is a field extension, and the $x_j$ are algebraically independent over K, M cannot be a finitely generated K-algebraThe existence of a linear map onto an affine algebraic setInequality involving Ideals generated by forms.
$begingroup$
Let $K$ be an algebraically closed field and $X_1,dots,X_m$ be subvarieties of a variety $Xsubset mathbb A^n$. Let $f_1,dots,f_kin K[X]$ such that for every $X_i$ there exists some $f_j$ not vanishing on $X_i$. Prove there exists $g_1,dots,g_k$ such that $sum_a=1^k f_ag_a$ doesn't vanish on $X_i$ for all $i$.
abstract-algebra commutative-algebra
edited yesterday
user26857
39.3k124183
asked Mar 3 at 21:34
ryanriessryanriess
654
$endgroup$
add a comment |
$begingroup$
Let $K$ be an algebraically closed field and $X_1,dots,X_m$ be subvarieties of a variety $Xsubset mathbb A^n$. Let $f_1,dots,f_kin K[X]$ such that for every $X_i$ there exists some $f_j$ not vanishing on $X_i$. Prove there exists $g_1,dots,g_k$ such that $sum_a=1^k f_ag_a$ doesn't vanish on $X_i$ for all $i$.
abstract-algebra commutative-algebra
edited yesterday
user26857
39.3k124183
asked Mar 3 at 21:34
ryanriessryanriess
654
$endgroup$
$begingroup$
$k$ is an index, so $K$ is the field.
$endgroup$
– user26857
yesterday
add a comment |
$begingroup$
Let $K$ be an algebraically closed field and $X_1,dots,X_m$ be subvarieties of a variety $Xsubset mathbb A^n$. Let $f_1,dots,f_kin K[X]$ such that for every $X_i$ there exists some $f_j$ not vanishing on $X_i$. Prove there exists $g_1,dots,g_k$ such that $sum_a=1^k f_ag_a$ doesn't vanish on $X_i$ for all $i$.
abstract-algebra commutative-algebra
edited yesterday
user26857
39.3k124183
asked Mar 3 at 21:34
ryanriessryanriess
654
$endgroup$
Let $K$ be an algebraically closed field and $X_1,dots,X_m$ be subvarieties of a variety $Xsubset mathbb A^n$. Let $f_1,dots,f_kin K[X]$ such that for every $X_i$ there exists some $f_j$ not vanishing on $X_i$. Prove there exists $g_1,dots,g_k$ such that $sum_a=1^k f_ag_a$ doesn't vanish on $X_i$ for all $i$.
abstract-algebra commutative-algebra
abstract-algebra commutative-algebra
edited yesterday
user26857
39.3k124183
asked Mar 3 at 21:34
ryanriessryanriess
654
edited yesterday
user26857
39.3k124183
asked Mar 3 at 21:34
ryanriessryanriess
654
edited yesterday
user26857
39.3k124183
edited yesterday
user26857
39.3k124183
edited yesterday
user26857
39.3k124183
39.3k124183
asked Mar 3 at 21:34
ryanriessryanriess
654
asked Mar 3 at 21:34
ryanriessryanriess
654
asked Mar 3 at 21:34
ryanriessryanriess
654
654
$begingroup$
$k$ is an index, so $K$ is the field.
$endgroup$
– user26857
yesterday
add a comment |
$begingroup$
$k$ is an index, so $K$ is the field.
$endgroup$
– user26857
yesterday
$begingroup$
$k$ is an index, so $K$ is the field.
$endgroup$
– user26857
yesterday
$begingroup$
$k$ is an index, so $K$ is the field.
$endgroup$
– user26857
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I suppose your varieties are irreducible. Assume the contrary. Then consider the ideal $I=(f_1,...,f_k)$. Then this implies any element of the ideal vanishes on some $X_i$ and hence is in $mathscr I (X_i)$ for some $i$. Thus you get $Isubset bigcup_i=1 ^n mathscr I(X_i)$. Prime avoidance tells you $I$ is in some $mathscr I(X_i)$ which means each $f_j$ vanishes on $X_i$. Contradiction.
edited yesterday
user26857
39.3k124183
answered yesterday
Soumik GhoshSoumik Ghosh
473110
$endgroup$
$begingroup$
"ur", "u" are more suitable on Facebook. Let's respect this site, and mathematics as well. Thank you!
$endgroup$
– user26857
yesterday
$begingroup$
surprising how a colloquialism implied disrespect lol
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
"lol" is also a Facebook way of talking. This is definitely not colloquialism, it's something close to rudeness on a math site. If can't change this attitude I suggest you to go back to Facebook.
$endgroup$
– user26857
21 hours ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suppose your varieties are irreducible. Assume the contrary. Then consider the ideal $I=(f_1,...,f_k)$. Then this implies any element of the ideal vanishes on some $X_i$ and hence is in $mathscr I (X_i)$ for some $i$. Thus you get $Isubset bigcup_i=1 ^n mathscr I(X_i)$. Prime avoidance tells you $I$ is in some $mathscr I(X_i)$ which means each $f_j$ vanishes on $X_i$. Contradiction.
edited yesterday
user26857
39.3k124183
answered yesterday
Soumik GhoshSoumik Ghosh
473110
$endgroup$
$begingroup$
"ur", "u" are more suitable on Facebook. Let's respect this site, and mathematics as well. Thank you!
$endgroup$
– user26857
yesterday
$begingroup$
surprising how a colloquialism implied disrespect lol
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
"lol" is also a Facebook way of talking. This is definitely not colloquialism, it's something close to rudeness on a math site. If can't change this attitude I suggest you to go back to Facebook.
$endgroup$
– user26857
21 hours ago
add a comment |
$begingroup$
I suppose your varieties are irreducible. Assume the contrary. Then consider the ideal $I=(f_1,...,f_k)$. Then this implies any element of the ideal vanishes on some $X_i$ and hence is in $mathscr I (X_i)$ for some $i$. Thus you get $Isubset bigcup_i=1 ^n mathscr I(X_i)$. Prime avoidance tells you $I$ is in some $mathscr I(X_i)$ which means each $f_j$ vanishes on $X_i$. Contradiction.
edited yesterday
user26857
39.3k124183
answered yesterday
Soumik GhoshSoumik Ghosh
473110
$endgroup$
$begingroup$
"ur", "u" are more suitable on Facebook. Let's respect this site, and mathematics as well. Thank you!
$endgroup$
– user26857
yesterday
$begingroup$
surprising how a colloquialism implied disrespect lol
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
"lol" is also a Facebook way of talking. This is definitely not colloquialism, it's something close to rudeness on a math site. If can't change this attitude I suggest you to go back to Facebook.
$endgroup$
– user26857
21 hours ago
add a comment |
$begingroup$
I suppose your varieties are irreducible. Assume the contrary. Then consider the ideal $I=(f_1,...,f_k)$. Then this implies any element of the ideal vanishes on some $X_i$ and hence is in $mathscr I (X_i)$ for some $i$. Thus you get $Isubset bigcup_i=1 ^n mathscr I(X_i)$. Prime avoidance tells you $I$ is in some $mathscr I(X_i)$ which means each $f_j$ vanishes on $X_i$. Contradiction.
edited yesterday
user26857
39.3k124183
answered yesterday
Soumik GhoshSoumik Ghosh
473110
$endgroup$
I suppose your varieties are irreducible. Assume the contrary. Then consider the ideal $I=(f_1,...,f_k)$. Then this implies any element of the ideal vanishes on some $X_i$ and hence is in $mathscr I (X_i)$ for some $i$. Thus you get $Isubset bigcup_i=1 ^n mathscr I(X_i)$. Prime avoidance tells you $I$ is in some $mathscr I(X_i)$ which means each $f_j$ vanishes on $X_i$. Contradiction.
edited yesterday
user26857
39.3k124183
answered yesterday
Soumik GhoshSoumik Ghosh
473110
edited yesterday
user26857
39.3k124183
edited yesterday
user26857
39.3k124183
edited yesterday
user26857
39.3k124183
39.3k124183
answered yesterday
Soumik GhoshSoumik Ghosh
473110
answered yesterday
Soumik GhoshSoumik Ghosh
473110
answered yesterday
Soumik GhoshSoumik Ghosh
473110
473110
$begingroup$
"ur", "u" are more suitable on Facebook. Let's respect this site, and mathematics as well. Thank you!
$endgroup$
– user26857
yesterday
$begingroup$
surprising how a colloquialism implied disrespect lol
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
"lol" is also a Facebook way of talking. This is definitely not colloquialism, it's something close to rudeness on a math site. If can't change this attitude I suggest you to go back to Facebook.
$endgroup$
– user26857
21 hours ago
add a comment |
$begingroup$
"ur", "u" are more suitable on Facebook. Let's respect this site, and mathematics as well. Thank you!
$endgroup$
– user26857
yesterday
$begingroup$
surprising how a colloquialism implied disrespect lol
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
"lol" is also a Facebook way of talking. This is definitely not colloquialism, it's something close to rudeness on a math site. If can't change this attitude I suggest you to go back to Facebook.
$endgroup$
– user26857
21 hours ago
$begingroup$
"ur", "u" are more suitable on Facebook. Let's respect this site, and mathematics as well. Thank you!
$endgroup$
– user26857
yesterday
$begingroup$
"ur", "u" are more suitable on Facebook. Let's respect this site, and mathematics as well. Thank you!
$endgroup$
– user26857
yesterday
$begingroup$
surprising how a colloquialism implied disrespect lol
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
surprising how a colloquialism implied disrespect lol
$endgroup$
– Soumik Ghosh
yesterday
$begingroup$
"lol" is also a Facebook way of talking. This is definitely not colloquialism, it's something close to rudeness on a math site. If can't change this attitude I suggest you to go back to Facebook.
$endgroup$
– user26857
21 hours ago
$begingroup$
"lol" is also a Facebook way of talking. This is definitely not colloquialism, it's something close to rudeness on a math site. If can't change this attitude I suggest you to go back to Facebook.
$endgroup$
– user26857
21 hours ago
add a comment |
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$begingroup$
$k$ is an index, so $K$ is the field.
$endgroup$
– user26857
yesterday