Mixing
Basic application of the Nullstellensatz to locus of hypersurfaces
$begingroup$
Let $k$ be an algebraically closed field. For $f in k[x_1,...,x_n]$, write $V(f) subset k^n$ for the hypersurface defined by $f=0$.
As an application of the NSS, I want to prove the following.
If $f$ is irreducible and $f$ does not divide $g$, then $V(f) notsubset V(g)$.
My idea is to reason by contradiction and to suppose that $V(f) subset V(g)$.
Now I consider the ideal generated by $f$. Since $f$ is irreducible, $(f)$ is prime and thus $(f) neq k[x_1,...,x_n]$.
Since I supposed that $V(f) subset V(g)$, the NSS (which I can apply since $k$ is algebraically closed) tells me that there exists $n in mathbb{N}$ such that $g^n in (f)$. But this is a contradiction to the fact that $f$ does not divide $g$.
I am really unsure about my reasoning and my understanding of the NSS.
Any comments would be appreciated.
abstract-algebra proof-verification commutative-algebra
asked Jan 30 at 21:00
riri92riri92
2018
$endgroup$
add a comment |
$begingroup$
Let $k$ be an algebraically closed field. For $f in k[x_1,...,x_n]$, write $V(f) subset k^n$ for the hypersurface defined by $f=0$.
As an application of the NSS, I want to prove the following.
If $f$ is irreducible and $f$ does not divide $g$, then $V(f) notsubset V(g)$.
My idea is to reason by contradiction and to suppose that $V(f) subset V(g)$.
Now I consider the ideal generated by $f$. Since $f$ is irreducible, $(f)$ is prime and thus $(f) neq k[x_1,...,x_n]$.
Since I supposed that $V(f) subset V(g)$, the NSS (which I can apply since $k$ is algebraically closed) tells me that there exists $n in mathbb{N}$ such that $g^n in (f)$. But this is a contradiction to the fact that $f$ does not divide $g$.
I am really unsure about my reasoning and my understanding of the NSS.
Any comments would be appreciated.
abstract-algebra proof-verification commutative-algebra
asked Jan 30 at 21:00
riri92riri92
2018
$endgroup$
add a comment |
$begingroup$
Let $k$ be an algebraically closed field. For $f in k[x_1,...,x_n]$, write $V(f) subset k^n$ for the hypersurface defined by $f=0$.
As an application of the NSS, I want to prove the following.
If $f$ is irreducible and $f$ does not divide $g$, then $V(f) notsubset V(g)$.
My idea is to reason by contradiction and to suppose that $V(f) subset V(g)$.
Now I consider the ideal generated by $f$. Since $f$ is irreducible, $(f)$ is prime and thus $(f) neq k[x_1,...,x_n]$.
Since I supposed that $V(f) subset V(g)$, the NSS (which I can apply since $k$ is algebraically closed) tells me that there exists $n in mathbb{N}$ such that $g^n in (f)$. But this is a contradiction to the fact that $f$ does not divide $g$.
I am really unsure about my reasoning and my understanding of the NSS.
Any comments would be appreciated.
abstract-algebra proof-verification commutative-algebra
asked Jan 30 at 21:00
riri92riri92
2018
$endgroup$
Let $k$ be an algebraically closed field. For $f in k[x_1,...,x_n]$, write $V(f) subset k^n$ for the hypersurface defined by $f=0$.
As an application of the NSS, I want to prove the following.
If $f$ is irreducible and $f$ does not divide $g$, then $V(f) notsubset V(g)$.
My idea is to reason by contradiction and to suppose that $V(f) subset V(g)$.
Now I consider the ideal generated by $f$. Since $f$ is irreducible, $(f)$ is prime and thus $(f) neq k[x_1,...,x_n]$.
Since I supposed that $V(f) subset V(g)$, the NSS (which I can apply since $k$ is algebraically closed) tells me that there exists $n in mathbb{N}$ such that $g^n in (f)$. But this is a contradiction to the fact that $f$ does not divide $g$.
I am really unsure about my reasoning and my understanding of the NSS.
Any comments would be appreciated.
abstract-algebra proof-verification commutative-algebra
abstract-algebra proof-verification commutative-algebra
asked Jan 30 at 21:00
riri92riri92
2018
asked Jan 30 at 21:00
riri92riri92
2018
edited Jan 30 at 21:19
riri92
asked Jan 30 at 21:00
riri92riri92
2018
asked Jan 30 at 21:00
riri92riri92
2018
asked Jan 30 at 21:00
riri92riri92
2018
2018
add a comment |
add a comment |
1 Answer
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$begingroup$
It seems reasonable to start with a contradiction, i.e., that $V(f)subseteq V(g)$. At this point, however, it may be better to observe that $I(V(f))supseteq I(V(g))$. You are essentially making such an observation, but the ideal $I(V(f))$ should be identified directly.
Now, the Nullstellensatz (and algebraic closure) comes in to show that $I(V(f))=langle frangle$. This requires the Nullstellensatz and algebraic closure since otherwise $V(f)$ might be empty, so $I(V(f))$ would be larger than $langle frangle$.
Now, if $I(V(f))supseteq I(V(g))$, then $gin I(V(f))=langle frangle$. This, however, contradicts the fact that $f$ does not divide $g$.
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
$endgroup$
$begingroup$
Thank you for your reply.
$endgroup$
– riri92
Jan 30 at 22:04
$begingroup$
Dear Michael, I wondered if I could ask for some more help. I now want to deduce that if $g =$ const $cdot Pi f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. If I understand correctly, this follows from the counterpositive statement of what we showed above right?
$endgroup$
– riri92
Jan 30 at 23:16
1
$begingroup$
@riri92 A follow-up question should be posed as a new question so that others also have a chance to answer. To note: this new question is weaker than the original question as it doesn't require the Nullstellensatz or an algebraically closed field.
$endgroup$
– Michael Burr
Jan 31 at 1:57
$begingroup$
makes sense, sorry !
$endgroup$
– riri92
Jan 31 at 2:28
add a comment |
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1 Answer
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$begingroup$
It seems reasonable to start with a contradiction, i.e., that $V(f)subseteq V(g)$. At this point, however, it may be better to observe that $I(V(f))supseteq I(V(g))$. You are essentially making such an observation, but the ideal $I(V(f))$ should be identified directly.
Now, the Nullstellensatz (and algebraic closure) comes in to show that $I(V(f))=langle frangle$. This requires the Nullstellensatz and algebraic closure since otherwise $V(f)$ might be empty, so $I(V(f))$ would be larger than $langle frangle$.
Now, if $I(V(f))supseteq I(V(g))$, then $gin I(V(f))=langle frangle$. This, however, contradicts the fact that $f$ does not divide $g$.
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
$endgroup$
$begingroup$
Thank you for your reply.
$endgroup$
– riri92
Jan 30 at 22:04
$begingroup$
Dear Michael, I wondered if I could ask for some more help. I now want to deduce that if $g =$ const $cdot Pi f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. If I understand correctly, this follows from the counterpositive statement of what we showed above right?
$endgroup$
– riri92
Jan 30 at 23:16
1
$begingroup$
@riri92 A follow-up question should be posed as a new question so that others also have a chance to answer. To note: this new question is weaker than the original question as it doesn't require the Nullstellensatz or an algebraically closed field.
$endgroup$
– Michael Burr
Jan 31 at 1:57
$begingroup$
makes sense, sorry !
$endgroup$
– riri92
Jan 31 at 2:28
add a comment |
$begingroup$
It seems reasonable to start with a contradiction, i.e., that $V(f)subseteq V(g)$. At this point, however, it may be better to observe that $I(V(f))supseteq I(V(g))$. You are essentially making such an observation, but the ideal $I(V(f))$ should be identified directly.
Now, the Nullstellensatz (and algebraic closure) comes in to show that $I(V(f))=langle frangle$. This requires the Nullstellensatz and algebraic closure since otherwise $V(f)$ might be empty, so $I(V(f))$ would be larger than $langle frangle$.
Now, if $I(V(f))supseteq I(V(g))$, then $gin I(V(f))=langle frangle$. This, however, contradicts the fact that $f$ does not divide $g$.
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
$endgroup$
$begingroup$
Thank you for your reply.
$endgroup$
– riri92
Jan 30 at 22:04
$begingroup$
Dear Michael, I wondered if I could ask for some more help. I now want to deduce that if $g =$ const $cdot Pi f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. If I understand correctly, this follows from the counterpositive statement of what we showed above right?
$endgroup$
– riri92
Jan 30 at 23:16
1
$begingroup$
@riri92 A follow-up question should be posed as a new question so that others also have a chance to answer. To note: this new question is weaker than the original question as it doesn't require the Nullstellensatz or an algebraically closed field.
$endgroup$
– Michael Burr
Jan 31 at 1:57
$begingroup$
makes sense, sorry !
$endgroup$
– riri92
Jan 31 at 2:28
add a comment |
$begingroup$
It seems reasonable to start with a contradiction, i.e., that $V(f)subseteq V(g)$. At this point, however, it may be better to observe that $I(V(f))supseteq I(V(g))$. You are essentially making such an observation, but the ideal $I(V(f))$ should be identified directly.
Now, the Nullstellensatz (and algebraic closure) comes in to show that $I(V(f))=langle frangle$. This requires the Nullstellensatz and algebraic closure since otherwise $V(f)$ might be empty, so $I(V(f))$ would be larger than $langle frangle$.
Now, if $I(V(f))supseteq I(V(g))$, then $gin I(V(f))=langle frangle$. This, however, contradicts the fact that $f$ does not divide $g$.
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
$endgroup$
It seems reasonable to start with a contradiction, i.e., that $V(f)subseteq V(g)$. At this point, however, it may be better to observe that $I(V(f))supseteq I(V(g))$. You are essentially making such an observation, but the ideal $I(V(f))$ should be identified directly.
Now, the Nullstellensatz (and algebraic closure) comes in to show that $I(V(f))=langle frangle$. This requires the Nullstellensatz and algebraic closure since otherwise $V(f)$ might be empty, so $I(V(f))$ would be larger than $langle frangle$.
Now, if $I(V(f))supseteq I(V(g))$, then $gin I(V(f))=langle frangle$. This, however, contradicts the fact that $f$ does not divide $g$.
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
answered Jan 30 at 21:31
Michael BurrMichael Burr
27k23262
27k23262
$begingroup$
Thank you for your reply.
$endgroup$
– riri92
Jan 30 at 22:04
$begingroup$
Dear Michael, I wondered if I could ask for some more help. I now want to deduce that if $g =$ const $cdot Pi f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. If I understand correctly, this follows from the counterpositive statement of what we showed above right?
$endgroup$
– riri92
Jan 30 at 23:16
1
$begingroup$
@riri92 A follow-up question should be posed as a new question so that others also have a chance to answer. To note: this new question is weaker than the original question as it doesn't require the Nullstellensatz or an algebraically closed field.
$endgroup$
– Michael Burr
Jan 31 at 1:57
$begingroup$
makes sense, sorry !
$endgroup$
– riri92
Jan 31 at 2:28
add a comment |
$begingroup$
Thank you for your reply.
$endgroup$
– riri92
Jan 30 at 22:04
$begingroup$
Dear Michael, I wondered if I could ask for some more help. I now want to deduce that if $g =$ const $cdot Pi f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. If I understand correctly, this follows from the counterpositive statement of what we showed above right?
$endgroup$
– riri92
Jan 30 at 23:16
1
$begingroup$
@riri92 A follow-up question should be posed as a new question so that others also have a chance to answer. To note: this new question is weaker than the original question as it doesn't require the Nullstellensatz or an algebraically closed field.
$endgroup$
– Michael Burr
Jan 31 at 1:57
$begingroup$
makes sense, sorry !
$endgroup$
– riri92
Jan 31 at 2:28
$begingroup$
Thank you for your reply.
$endgroup$
– riri92
Jan 30 at 22:04
$begingroup$
Thank you for your reply.
$endgroup$
– riri92
Jan 30 at 22:04
$begingroup$
Dear Michael, I wondered if I could ask for some more help. I now want to deduce that if $g =$ const $cdot Pi f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. If I understand correctly, this follows from the counterpositive statement of what we showed above right?
$endgroup$
– riri92
Jan 30 at 23:16
$begingroup$
Dear Michael, I wondered if I could ask for some more help. I now want to deduce that if $g =$ const $cdot Pi f_i^{n_i}$, then the irreducible components of $V(g)$ are of the form $V(f_i)$. If I understand correctly, this follows from the counterpositive statement of what we showed above right?
$endgroup$
– riri92
Jan 30 at 23:16
1
1
$begingroup$
@riri92 A follow-up question should be posed as a new question so that others also have a chance to answer. To note: this new question is weaker than the original question as it doesn't require the Nullstellensatz or an algebraically closed field.
$endgroup$
– Michael Burr
Jan 31 at 1:57
$begingroup$
@riri92 A follow-up question should be posed as a new question so that others also have a chance to answer. To note: this new question is weaker than the original question as it doesn't require the Nullstellensatz or an algebraically closed field.
$endgroup$
– Michael Burr
Jan 31 at 1:57
$begingroup$
makes sense, sorry !
$endgroup$
– riri92
Jan 31 at 2:28
$begingroup$
makes sense, sorry !
$endgroup$
– riri92
Jan 31 at 2:28
add a comment |
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