§ Weak and Strong Nullstllensatz
§ Weak Nulstellensatz: On the tin
For every maximal ideal there is a unique
such that . This says that any maximal ideal
is the ideal of some point.
§ Weak Nullstellensatz: implication 1 (Solutions)
every ideal, since it is contained in a maximal ideal, will have zeroes.
Zeroes will always exist in all ideas upto the maximal ideal.
- It simply says that for all ideals in , we have
- Corollary: and are mutual inverses of inclusions between algebraic sets and radical ideals.
§ Weak Nullstellensatz: Implication 2 (Non-solutions)
If an ideal does not have zeroes, then it must be the full ring. Hence, 1 must
be in this ideal. So if and the system has
no solutions, then cannot be included in any maximal ideal, hence .
Thus, , and there exist such that
§ Strong Nullstellensatz: On the Tin
For every ideal , we have that . I am adopting
the radical (heh) notation for the radical, because this matches
my intuition of what the radical is doing: it's taking all roots , not just square roots .
For example, in .
§ Strong Nullstellensatz: Implication 1 (solutions)
Let . If is zero on , then .
Unwrapping this, .
§ Weak Nullstellensatz: Proof
- Let be a maximal ideal.
- Let be the quotient ring .
- See that is a field because it is a ring quotiented by a maximal ideal.
- Consider the map , or by sending elements into the quotient.
- We will show that is an evaluation map, and . So we will get a function that evaluates polynomials at a given point, which will have a single point as a solution.
- Core idea: See that . Hence is a field that contains . But is algebraically closed, hence .
- First see that , or that preserves [ie, ]. note that no complex number can be in . If we had a complex number in , then we would need to have in (since an ideal is closed under multiplication by the full ring), which means , due to which we get is the full ring. This can't be the case because is a proper maximal ideal.
- Hence, we have or .
- Thus the map we have is .
- Define . Now we get that . That is, we have an evaluation map that sends .
- CLAIM: The kernel of an evaluation map is of the form .
- PROOF OF CLAIM:TODO
- The kernel is also . Hence, , and point that corresponds to the maximal ideal is .
§ Strong Nullstellensatz: Proof
We use the Rabinowitsch trick .
- Suppose that wherever simultaneously vanish, then so does . [that is, where ].
- Then the polynomials have no common zeros where is a new variable into the ring.
- Core idea of why they can't have common zeros: Contradiction. assume that , and all the vanish at some point. Then we need which mean , so cannot vanish, so . However, since all the vanish, also vanishes as . This is contradiction.
- Now by weak Nullstellensatz, the ideal cannot be contained in a maximal ideal (for then they would simultaneously vanish). Thus, and .
- This means there are coefficients such that
Since this holds when are arbitrary variables, it continues to hold
on substituting , the coefficient disappears. This gives:
since , we can write . By clearing denominators, we get:
This means that
§ Strong Nullstellensatz: algebraic proof
- We have .
- We want to show that in .
- This is the same as showing that in . ( in the quotient ring).
- If is nilpotent in , then the becomes the trivial ring . [Intuitively, if is nilpotent and a unit, then we will have , that is unit raised to some power is 0, from which we can derive ].
- Localising at is the same as computing .
- But we have that . Weak Nullstellensatz implies that .
- This means that . Thus, has as nilpotent, or in .
§ Relationship between strong and weak
Strong lets us establish what functions vanish on a variety . Weak let us establish
what functions vanish at a point .
§ Strong Nullstellensatz in scheme theory
Holy shit, scheme theory really does convert Nullstellensatz-the-proof into
Nullstellensatz-the-definition! I'd never realised this before, but this.. is crazy.
Not only do we get easier proofs, we also get more power! We can reason about generic
points such as or which don't exist in variety-land. This is really really cool.
- Same statement: .
- is the set of points on which vanihes. Evaluation is quotienting. So it's going to be set of prime ideals such that . So . This means that .
- is the set of functions that vanish over every point in . The functions that vanish at are the elements of . So the functions that vanish over all points is .
- Unwrapping, this means that is the intersection of all ideals in , which is the intersection of all primes that contains , which is the radical of .