hilbert's nullstellensatz
ofer...@yahoo.com
let x denote [x1,...,xn];
how does one go from the classical form:
let V be the variety of zeros of f1[x],....,fr[x] in R, and I be the
ideal <f1,...,fr>. If g[x] in R is identically zero on V, then some
power of g is in I.
to :
maximal ideals in R are in bijective correspondence with C^n, through
the kernel of substitution map, such that kernel Ma of map Sa : R -> C,
corresponds to the element a in C^n.
eugene
Hi,
Let's show that all the maximal ideals I in C[x_1,x_2,...,x_n] are of
the form
(x_1 - a_1, x_2 - a_2,..., x_n - a_n):
As follows from the Hilbert Nullstellensatz : V(I) is not empty and,
thus, contains at least 1 point, say, (a_1,a_2,...,a_n). Note, I(
{a_1,a_2,...,a_n} ) = (x_1 - a_1, x_2 - a_2,..., x_n - a_n), and since
(a_1,a_2,...,a_n) in V(I) we have that
(x_1 - a_1, x_2 - a_2,..., x_n - a_n) contains(=) I(
{a_1,a_2,...,a_n} ) contains(=) I(V(I)) =
= rad(I) = I (we have I in(=) rad (I) and since I is maximal I =
rad(I).
Here rad(I) is the radical ideal of I.
Jannick Asmus
> ofer...@yahoo.com wrote:
>> let R=C[x1,...,xn]
>> let x denote [x1,...,xn];
>>
>> how does one go from the classical form:
>>
>> let V be the variety of zeros of f1[x],....,fr[x] in R, and I be the
>> ideal <f1,...,fr>. If g[x] in R is identically zero on V, then some
>> power of g is in I.
>>
>> to :
>>
>> maximal ideals in R are in bijective correspondence with C^n, through
>> the kernel of substitution map, such that kernel Ma of map Sa : R -> C,
>> corresponds to the element a in C^n.
>
> Hi,
>
> Let's show that all the maximal ideals I in C[x_1,x_2,...,x_n] are of
> the form
> (x_1 - a_1, x_2 - a_2,..., x_n - a_n):
>
> As follows from the Hilbert Nullstellensatz : V(I) is not empty and,
How do you draw this conclusion by the version of Hilbert's
Nullstellensatz as stated above?
eugene
Well, the version of Hilbert Nullstellensatz stated above say I(V(I)) =
rad(I), so if V(I) is empty set, then I(V(I)) = (constant) =
C[x_1,x_2,...,x_n], so rad(I) = C[x_1,x_2,...,x_n] and
I = C[x_1,x_2,...,x_n] - contradiction.
Timothy Murphy
I'm not expert in this area,
but I'm wondering if you need the NullStellenSatz to prove this result.
If M is a maximal ideal in R then M/R is a field.
This is a finitely-generated extension of C, and so is C itself.
Hence one has a homomorphism f: R -> C with kernel M.
For each coordinate x_i, let f(x_i) = a_i.
Then x_i - a_i is in M for i = 1,...,n,
and it follows easily enough that these linear polynomials generate M.
--
Timothy Murphy
e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
tel: +353-86-2336090, +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Jannick Asmus
Right. Thanks. I was a bit blind.
J.
Jannick Asmus
> ofer...@yahoo.com wrote:
>
>> let R=C[x1,...,xn]
>> let x denote [x1,...,xn];
>>
>> how does one go from the classical form:
>>
>> let V be the variety of zeros of f1[x],....,fr[x] in R, and I be the
>> ideal <f1,...,fr>. If g[x] in R is identically zero on V, then some
>> power of g is in I.
>>
>> to :
>>
>> maximal ideals in R are in bijective correspondence with C^n, through
>> the kernel of substitution map, such that kernel Ma of map Sa : R -> C,
>> corresponds to the element a in C^n.
>
> I'm not expert in this area,
> but I'm wondering if you need the NullStellenSatz to prove this result.
>
> If M is a maximal ideal in R then M/R is a field.
> This is a finitely-generated extension of C, and so is C itself.
> Hence one has a homomorphism f: R -> C with kernel M.
> For each coordinate x_i, let f(x_i) = a_i.
> Then x_i - a_i is in M for i = 1,...,n,
> and it follows easily enough that these linear polynomials generate M.
You are using another version of Hilbert's Nullstellensatz (the
algebraic one). It translates to the version given above by the OP.
J.
Timothy Murphy
>> If M is a maximal ideal in R then M/R is a field.
>> This is a finitely-generated extension of C, and so is C itself.
>> Hence one has a homomorphism f: R -> C with kernel M.
>> For each coordinate x_i, let f(x_i) = a_i.
>> Then x_i - a_i is in M for i = 1,...,n,
>> and it follows easily enough that these linear polynomials generate M.
>
> You are using another version of Hilbert's Nullstellensatz (the
> algebraic one). It translates to the version given above by the OP.
I don't think so.
The argument I used is much simpler than the Nullstellensatz.
If you think they are equivalent can you explain
how to deduce the Nullstellensatz from the result I gave?
Jannick Asmus
> Jannick Asmus wrote:
>
>>> If M is a maximal ideal in R then M/R is a field.
>>> This is a finitely-generated extension of C, and so is C itself.
>>> Hence one has a homomorphism f: R -> C with kernel M.
>>> For each coordinate x_i, let f(x_i) = a_i.
>>> Then x_i - a_i is in M for i = 1,...,n,
>>> and it follows easily enough that these linear polynomials generate M.
>> You are using another version of Hilbert's Nullstellensatz (the
>> algebraic one). It translates to the version given above by the OP.
>
> I don't think so.
> The argument I used is much simpler than the Nullstellensatz.
> If you think they are equivalent can you explain
> how to deduce the Nullstellensatz from the result I gave?
I didn't say they are equivalent (but I think they are), but here is a
sketch of proof of implication you wanted me to give:
Let I be an ideal of polynomial ring A=k[X1,...,Xn] of the algebraically
closed field k. Let V(I) the variety of zeros of all polynomials in I
and Z(V(I)) the ideal of all polynomials in A vanishing on V(I). We
claim that rad(I)=Z(V(I)) where rad(I) denotes the radical of I.
It is easy to see that rad(I) is contained in Z(V(I)). For the converse
inclusion we take f not in rad(I) and show that f is not in Z(V(I)):
Since rad(I) is the intersection of all prime ideals in R containing I,
there is a prime P not containing f. Then the image ff of f in B=A/P is
not zero. Let M be a maximal ideal in the finitely generated k-algebra
C=B[X]/(ff.X-1). By your version of Hilbert's Nullstellensatz, the
homomorphism k -> C/M is an isomorphism, since the field C/M is a
finitely generated k-algebra. Let x1,...,xn in k be the corresponding
images of X1,...,Xn in k via C/M. Then, by construction, x=(x1,...xn) in
k^n is in V(I) with f(x)=/=0.
Note that we are dealing with so called weak and strong versions of
Hilbert's Nullstellensatz here.
Timothy Murphy
>>>> If M is a maximal ideal in R then M/R is a field.
>>>> This is a finitely-generated extension of C, and so is C itself.
>>>> Hence one has a homomorphism f: R -> C with kernel M.
>>>> For each coordinate x_i, let f(x_i) = a_i.
>>>> Then x_i - a_i is in M for i = 1,...,n,
>>>> and it follows easily enough that these linear polynomials generate M.
>>> You are using another version of Hilbert's Nullstellensatz (the
>>> algebraic one). It translates to the version given above by the OP.
>>
>> I don't think so.
>> The argument I used is much simpler than the Nullstellensatz.
>> If you think they are equivalent can you explain
>> how to deduce the Nullstellensatz from the result I gave?
>
> I didn't say they are equivalent (but I think they are), but here is a
> sketch of proof of implication you wanted me to give:
You said my argument used another version of the Nullstellensatz,
which suggests to me that the two versions are equivalent.
Incidentally, you said my version was the "algebraic" one.
How do you describe the other version?
> Let I be an ideal of polynomial ring A=k[X1,...,Xn] of the algebraically
> closed field k. Let V(I) the variety of zeros of all polynomials in I
> and Z(V(I)) the ideal of all polynomials in A vanishing on V(I). We
> claim that rad(I)=Z(V(I)) where rad(I) denotes the radical of I.
>
> It is easy to see that rad(I) is contained in Z(V(I)). For the converse
> inclusion we take f not in rad(I) and show that f is not in Z(V(I)):
> Since rad(I) is the intersection of all prime ideals in R containing I,
> there is a prime P not containing f. Then the image ff of f in B=A/P is
> not zero. Let M be a maximal ideal in the finitely generated k-algebra
> C=B[X]/(ff.X-1). By your version of Hilbert's Nullstellensatz, the
> homomorphism k -> C/M is an isomorphism, since the field C/M is a
> finitely generated k-algebra. Let x1,...,xn in k be the corresponding
> images of X1,...,Xn in k via C/M. Then, by construction, x=(x1,...xn) in
> k^n is in V(I) with f(x)=/=0.
>
> Note that we are dealing with so called weak and strong versions of
> Hilbert's Nullstellensatz here.
I've only heard of one version of the theorem.
The fact is, it is much easier to prove that a maximal ideal
is the set of polynomials vanishing at a point
than it is to prove the Nullstellensatz (as I understand it).
In particular your proof of the "equivalence" above
(which uses a different definition of rad(I))
seems to me much more difficult than the proof of the maximal ideal result.
However, I guess this is only a matter of opinion.
Hagen
> but I'm wondering if you need the NullStellenSatz to
> prove this result.
>
> If M is a maximal ideal in R then M/R is a field.
> This is a finitely-generated extension of C, and so
> is C itself.
> Hence one has a homomorphism f: R -> C with kernel M.
> For each coordinate x_i, let f(x_i) = a_i.
> Then x_i - a_i is in M for i = 1,...,n,
> and it follows easily enough that these linear
> polynomials generate M.
>
>
> --
> Timothy Murphy
> e-mail (<80k only): tim /at/ birdsnest.maths.tcd.ie
> tel: +353-86-2336090, +353-1-2842366
> s-mail: School of Mathematics, Trinity College,
> Dublin 2, Ireland
The essential point in your statement is that if in
the field extension L|K the field L is a finitely
generated K-algebra, then L|K is algebraic.
This fact is sometimes called the >weak Hilbert Null-
stellensatz< - see for example Zariski-Samuel, Commutative
Algebra. It is equivalent to Hilbert's Nullstellensatz.
H
Timothy Murphy
>> If M is a maximal ideal in R then M/R is a field.
>> This is a finitely-generated extension of C, and so
>> is C itself.
>> Hence one has a homomorphism f: R -> C with kernel M.
>> For each coordinate x_i, let f(x_i) = a_i.
>> Then x_i - a_i is in M for i = 1,...,n,
>> and it follows easily enough that these linear
>> polynomials generate M.
> The essential point in your statement is that if in
> the field extension L|K the field L is a finitely
> generated K-algebra, then L|K is algebraic.
> This fact is sometimes called the >weak Hilbert Null-
> stellensatz< - see for example Zariski-Samuel, Commutative
> Algebra. It is equivalent to Hilbert's Nullstellensatz.
Sorry to nit-pick, but what exactly do you mean
by saying that two propositions are "equivalent"?
Logically, I suppose, any two correct propositions are equivalent.
In this case, one proposition - the one I used -
seems to me more or less trivial,
while the other one - the nullstellensatz -
is rather difficult.
Doubtless there is a chain of reasoning
which goes through the first proposition
and winds up with the second,
but that hardly shows the propositions are equivalent.
The argument you gave assumed, among other results,
that the radical of an ideal I is the intersection
or the prime ideals containing I,
which is not at all obvious.
Jannick Asmus
> Jannick Asmus wrote:
>
>>>>> If M is a maximal ideal in R then M/R is a field.
>>>>> This is a finitely-generated extension of C, and so is C itself.
>>>>> Hence one has a homomorphism f: R -> C with kernel M.
>>>>> For each coordinate x_i, let f(x_i) = a_i.
>>>>> Then x_i - a_i is in M for i = 1,...,n,
>>>>> and it follows easily enough that these linear polynomials generate M.
>>>> You are using another version of Hilbert's Nullstellensatz (the
>>>> algebraic one). It translates to the version given above by the OP.
>>> I don't think so.
>>> The argument I used is much simpler than the Nullstellensatz.
>>> If you think they are equivalent can you explain
>>> how to deduce the Nullstellensatz from the result I gave?
>> I didn't say they are equivalent (but I think they are), but here is a
>> sketch of proof of implication you wanted me to give:
>
> You said my argument used another version of the Nullstellensatz,
> which suggests to me that the two versions are equivalent.
> Incidentally, you said my version was the "algebraic" one.
> How do you describe the other version?
Topological version - since it can be reworded in terms of the Zariski
topology of the prime spectrum Spec(R). This is a doorway to algebraic
geometry.
>> Let I be an ideal of polynomial ring A=k[X1,...,Xn] of the algebraically
>> closed field k. Let V(I) the variety of zeros of all polynomials in I
>> and Z(V(I)) the ideal of all polynomials in A vanishing on V(I). We
>> claim that rad(I)=Z(V(I)) where rad(I) denotes the radical of I.
>>
>> It is easy to see that rad(I) is contained in Z(V(I)). For the converse
>> inclusion we take f not in rad(I) and show that f is not in Z(V(I)):
>> Since rad(I) is the intersection of all prime ideals in R containing I,
>> there is a prime P not containing f. Then the image ff of f in B=A/P is
>> not zero. Let M be a maximal ideal in the finitely generated k-algebra
>> C=B[X]/(ff.X-1). By your version of Hilbert's Nullstellensatz, the
>> homomorphism k -> C/M is an isomorphism, since the field C/M is a
>> finitely generated k-algebra. Let x1,...,xn in k be the corresponding
>> images of X1,...,Xn in k via C/M. Then, by construction, x=(x1,...xn) in
>> k^n is in V(I) with f(x)=/=0.
>>
>> Note that we are dealing with so called weak and strong versions of
>> Hilbert's Nullstellensatz here.
>
> I've only heard of one version of the theorem.
>
> The fact is, it is much easier to prove that a maximal ideal
> is the set of polynomials vanishing at a point
> than it is to prove the Nullstellensatz (as I understand it).
> In particular your proof of the "equivalence" above
> (which uses a different definition of rad(I))
What different definition? If you mean that rad(I) is the intersection
of maximal ideals containing I then this is another version of Hilbert's
Nullstellensatz. In commutative algebra rad(I) is defined to be the set
of all elements with some positive power in I.
> seems to me much more difficult than the proof of the maximal ideal result.
>
> However, I guess this is only a matter of opinion.
Well, I am not sure what you mean by saying it is more difficult. The
algebraic version of Hilbert's Nullstellensatz states that every algebra
finitely generated over a field k which is a field is algebraic or,
equivalently, finite over k (*).
I want to stress that there is a big difference between finitely
generated and finite. So I would like to ask you to give a short sketch
a proof of (*). It is by far not trivial.
There are generalizations of this topic: If k is allowed to be a ring
fulfilling (*) for any finitely generated k-algebra, then watch out for
the notion of Jacobson rings. In the topological category there are
Jacobson spaces defined by the condition that the set of closed points
are *very* dense, i.e. any non-void closed subset is the closure of the
set of its closed points.
All this is founded on the different versions of Hilbert's
Nullstellensatz and the generalization of Jacobson rings and spaces, resp.
HTH.
J.
magya...@yahoo.com
i found this version of nullstellensatz, which appears to be not so
evident form the ones quoted before. Any comments or attempts at proof?
(I don't know if it is actually correct)
Assume F is algebraically closed, A=F[x1,...,xn] / I , where I is a
proper ideal. Assume t in A is not nilpotent (t^k != 0 for any k). Then
there exists an algebra homomorphism
H : A -> F
such that H(t) != 0 (non-zero)
Jannick Asmus
This is a basic lemma in commutative algebra.
Well, let me give some assertions all equivalent to each other. They all
boil down to the classical versions of Hilbert's Nullstellensatz if A is
a field (fulfilling criterion 8. below).
Def.: A ring A fulfilling one, hence all assertions of the
following proposition is called *Jacobson* ring.
Prop.: Let A be a commutative ring with unity. Denote the category of
finitely generated A-algebras by CC. Then the following assertions are
equivalent:
1. Every R in CC which is field it is finite over A.
2. For every R in CC and every maximal ideal M of R, the field R/M is
finite over A.
3. For every R in CC and every proper ideal I in R, rad(I) is equal to
the intersection of all maximal ideals of R containing I.
4. For every R in CC every prime ideal P is the intersection of all
maximal ideals containing P.
5. For every R in CC every prime ideal P which is not maximal is the
intersection of all primes containing P strictly.
6. For every R in CC, every ring homomorphism A -> k* where k* is an
algebraically closed field and every x in R which is not nilpotent,
there is an A-algebra homomorphism f: R -> k* with f(x)=/=0.
7. For every R in CC, any closed subset V of Spec(R) is the closure of
sets of all closed points contained in V.
8. Every prime ideal in A is an intersection of maximal ideals of A.
9. Every prime ideal P in A which is not maximal is equal to the
intersection of the prime ideals which strictly contain P.
The hardest part of the proof is to get back to 1. from 9. which needs a
couple of lemmas in commutative algebra.
J.
Timothy Murphy
That is interesting,
but the original question was about the specific ring A = C[x_1,...,x_n]
of polynomials in n variables over the complex numbers C
(or possibly over an algebraically closed field).
The question was: what does a maximal ideal look like?
My issue was not with the truth of what you (or maybe someone else) said,
but simply that I thought you were making the problem sound
much more difficult than it is.
Jannick Asmus
If I understood you right you were asking whether the different versions
*are* equivalent. I listed a set of different versions of Hilbert's
Nullstellensatz all equivalent to each other. This was a twist of the
OP's question initiated by you and Hagen and me answered to, right?
True that version 1. is easy to state, so is 7. in my opinion.
After all you said, Tim, I am asking myself whether you could possibly
mean that a proof of 1. is trivial by saying repeatedly that other
versions make the problem more difficult than it really is. Is that what
you mean? However, I do not think so. If you do, then could you provide
such a proof? The only one I know involves Noether's normalization lemma
and a couple of basics of commutative algebra.