Projective varieties

Let be an algebraically closed field. We want to parametrize the 1-dimensional subspaces of . In coordinates, we take away the origin and identify two points and if there exists such that . The resulting space is called the projective space, denoted by or just . The coordinates are denoted as , called the homogeneous coordinates.

Again we want to consider polynomials as functions. Contrary to the affine case, polynomials aren’t really functions on a projective space, because of the scalar factor. However, if we consider only the homogeneous polynomials (i.e. those with each monomial having the same degree)
we can define the zero set.

Definition 1 A projective variety is a subset of , that is the common zeros of a set of homogeneous polynomials.

Remark

1. The set of polynomials can be taken to be finite. (Hilbert Basis Theorem)

Some terminologies

1. the Zariski topology

Same as the affine case – consider the topology on with the closed sets being projective varieties. It is easy to verify that this is a topology:

• ,
• Union of two closed sets:
• Intersection of closed sets: just put all the equations together.

2. the ideal of X

This is the ideal generated by all the homogeneous polynomials in that vanish on . (Remember that in general, a polynomial as a function in projective space does not make sense)

It is immediate that is a graded ideal/homogeneous ideal, i.e. for each , each of its homogeneous components also lies in .

3. the projective coordinate ring

Consider as the coordinates of the projective space , and restrict it on . If we allow free multiplication and addition, this is what we will get.

The Nullstellensatz

Similar to the affine case, we want to develop a correspondence such that:

• for a subset , we have the homogeneous ideal generated by all the homogeneous polynomials vanishing on .
• for a homogeneous ideal , take a generating set of homogeneous polynomials, and call their common zeros .

Similarly, we have

• If , then
• If are homogeneous ideals of , then
• If is a homogeneous ideal of ,
• If , then (closure in Zariski topology)

Again, we want to find a Nullstellensatz such that the third condition is an equality. However, this time it CANNOT be as nice as the affine case, for such an equality would imply a one-one correspondence between projective varieties of and radical homogeneous ideals of , which is false. (e.g. common zero of is empty, but this ideal is not the entire ring)

This counterexample is actually the worse we can get, as we shall see.

Theorem 1 (Hilbert’s Nullstellensatz, projective case) Let be a homogeneous ideal, where is algebraically closed field. If , then .

Proof

It suffices to show that .

Let be a homogeneous polynomial such that for some , vanish on . If we are done. Suppose not, then we see that cannot be a constant since . So is homogeneous of degree at least 1, which means that it would vanish at when regarded as a function on . Regard as an usual ideal in and apply the usual Nullstellensatz, we are done.

What if ? Regarding as an ideal of and as a subset of , we see that or .

So any homogeneous ideal that contains a power of would have empty common zero set in projective space. As in the affine case, it seems more natural to correspond and . So is called the irrelevant maximal ideal, and if we ignore all the radical ideals that contain , we have

Corollary 1 Let be algebraically closed. The association is a 1-1 correspondence between projective subvarieties of and radical homogeneous ideals of that do not contain . Moreover, is irreducible iff is prime.

Affine cover of projective varieties

The projective space has a cover by affine space: for the piece , every point   corresponds to in . In the other direction, a point corresponds to where 1 is inserted to the -th position.

We can also carry this cover to a projective subvariety. Let be an irreducible projective subvariety. Then is closed in , which is like an affine space. If is an affine variety, how would its defining equations be related to that of ?

Let , where each are homogeneous polynomials. is then defined as

where 1 is inserted to the -th position. This is exactly the process of dehomogenenization. It is then clear that is exactly the dehomogenization of .

Notice that if is irreducible, then the closure of in is exactly . (More generally, an open subset of an irreducible set is dense. If not, is a nontrivial decomposition) So we can also work on the “inverse problem”: Given an irreducible affine variety , embed it into (say, by identifying the affine space with ), then how would the defining equations of be related to that of ?

We can guess the answer – it should be the inverse process of dehomogenization. Given a polynomial in , we can of course insert the powers of in each term such that the polynomial becomes homogeneous. This is called homogenization.

Proposition 1 Given an irreducible affine variety, and embed it into via . Then is the ideal generated by the homogenization of each term of .

Proof

Denote the ideal generated by the homogenization of each term of as . We want to show that it is exactly . Note that is generated by homogeneous polynomials, so it is a homogeneous ideal.

Clearly, , so , which implies contains .

For the other direction, consider . If we dehomogenize this polynomial by substituting by 1, the new polynomial . Then when we homogenize back, , implying that , showing the equality.

So far we talked about projective varieties and affine varieties. Just now we have seen that affine varieties can be embedded into the projective space and is a locally closed subset of . (meaning that it is an intersection of closed subset and open subset) This is the class of varieties we would focus on.

Definition 2 A quasi-projective variety in is an intersection of a closed subset and an open subset in Zariski topology.

From now on a variety would mean a quasi-projective variety.

Regular functions of quasi-projective varieties

In the affine variety case, we first looked at the quotient of polynomials (with the denominator not vanishing anywhere on the variety), and proved that it’s the same as polynomial function.

In the projective case, our analogue of polynomial is homogeneous polynomials. However, when we want to do quotient, we would want the numerator and the denominator to have the same degree – otherwise the function is still not well defined. So our first attempt is, for a quasi-projective variety , a regular function is a quotient of homogeneous polynomials of the same degree with denominator nonvanishing on . This, however, has some deficiencies.

Motivating case

Consider minus the point . The function defines a regular map from minus , for the denominator vanishes when , i.e. ([0,1,1]) or ([0,1,-1]).

has another expression on , that is . It is unnatural to distinguish these two as functions, while their denominators do vanish at different places. In particular, the expression “extends” the definition to the point , so should make sense as a function on , even though the value at is not initially defined.

This suggests the following definition,

Definition 3 Let be a variety. A regular function from to is a map such that for each , there exists a neighborhood around and homogeneous polynomials of the same degree such that is nonvanishing on and on .

We can show that this is consistent with our definition for the affine case. In fact,

Proposition 2

• For ( is a polynomial, and an affine variety), a regular map is of the form
• For ( is a homogeneous polynomial, and is a projective variety), a regular map is of the form , where the numerator has the same degree as the denominator.

Proof

We will prove the first statement only as these statements are analogous.

First notice that the sets (as varies) actually form a basis of the Zariski topology on $X$. (For this reason they are called the distinguished open sets)

A better local representation of regular function

Consider an arbitrary regular function . For each , there exists a neighborhood such that on . Since form a basis, we can assume that by shrinking if needed.

does not vanish at all on , meaning that . Taking on both sides, , i.e. For some , on . Therefore the regular function has another local expression

Notice furthermore that . So let us replace by . That means that for any , there exists a neighorhood such that on .

Putting the presentations together

Notice that is compact, for if where are polynomials and is an index set, then , where is the ideal generated by all . Applying on both sides, , i.e. there exists , . Say , where , as functions on . Then , showing compactness.

So suppose that is covered by , where are chosen as before. As before, , so there exists polynomials such that

This tells us how to patch the local functions together. They are supposed to be same fraction except that they are defined on different domains, so these functions should all be the same as

which finishes the proof.

Remarks

1. For the projective case, is also a basis.
2. The regular functions for a variety clearly form a ring, denoted as .

Corollary 2

• For an affine variety, a regular function is a polynomial map.
• For an irreducible projective variety, a regular function is constant.

Proof

Take .

Proposition 3 Let be an irreducible variety and be two regular maps. If they agree on an open set, then they are the same.

Proof

The set is closed because is regular, and it contains an open set, which must be dense in . So .

Maps between quasi-projective varieties

The definition of regular functions suggests a local definition for regular maps as well, so let us define

Definition 4 Let be varieties. For a map ,

• If is quasi-affine (intersection open and closed set in ), is regular if each component function is regular.
• If is quasi-projective, consider the affine cover of . is regular if the restrictions of to is regular.

There is another definition of regular maps, that makes use of the regular functions and is analogous to differentiable functions.

Proposition 4 Let be varieties. A map is regular if and only if is continuous and for any regular function ( open in ), the composition is also regular.

Proof

We first show that if is quasi-affine, a regular map is continuous.

Let be a closed subset of , and we want to show that is closed. Consider any . We will show that there exists a neighborhood such that is closed in , then we are done by this lemma.

Lemma 1 Let be a topological space with an open cover . A subset is closed iff is closed in for all .

Let has coordinate functions . We can pick a neighborhood such that all these are of the form , where are homogeneous of the same degree. If is the common zeros of polynomials , then is the common zeros of these compose with $latex\displaystyle \frac{P_i}{Q_i}$. This is closed once we clear the denominators.

For the regular function part, take any regular function . We have an open cover of such that each is quasi-affine, and by definition it is clear that is regular if and only if each restriction to is regular. Thus we may assume that is quasi-affine by shrinking if needed. The regularity of can be shown using a similar argument of closedness above.

If is quasi-affine, note that the projection maps are regular functions, so is also regular by the hypothesis.

For quasi-projective case, we are done by the following lemma,

Lemma 2 Let be varieties, and be a continuous map such that for any regular function ( is open in ), is also regular. If is open and is open such that , then the restriction of on also satisfies the fore mentioned property.

Proof of Lemma 2

The key is that the restriction of a regular function on an open subset is still regular, which is clear.

Corollary 3 The composition of regular maps is regular.

Knowing how to define a regular map, we now have the notion of isomorphism of varieties. From now on a variety that is isomorphic to an affine variety will be called affine. Similarly a variety that is isomorphic to a projective variety will be called projective.

Regular maps and ring of regular functions

A regular map induces a -algebra homomorphism .

Can we generalize Proposition 2 in the last post? Examining the proof, we see that we used the coordinates of . For there, we only regard it as regular functions on . Thus the proof generalizes to give

Proposition 5 For two varieties , where is affine, we have a natural isomorphism

Projective coordinate ring and ring of regular functions

For affine varieties we have seen that . However for projective coordinate ring, this is not the case. For example in proposition 2, we see that , while . This shows that affine varieties are quite special.

In general, the ring of regular functions could be wild. It is mentioned in Shafarevich’s book that Rees and Nagata constructed examples of quasiprojective varieties such that is not finitely generated, though I can’t find these examples anywhere.

Distinguished open sets

We used this basis in the proof of proposition 2. Since the ring of regular functions on the basis is nice (as shown in the proposition), these sets are important.

Proposition 6 Every point of a variety has an affine neighborhood.

Proof

If is quasi-projective, it has an affine open cover, so WLOG assume that is quasi-affine. As we have seen that is a basis, we would be done if we can show that is affine.

Let , where is affine. Suppose that is the common zeros of . Then notice that is isomorphic to the set defined by , , which is affine.

Affine varieties

Let be a field. It is clear what points in , polynomial functions on etc mean.

Definition 1 An affine variety is a subset of , that is the common zeros of a set of polynomials.

Remark

1. The set of polynomials can be taken to be finite. This is because is Noetherian (Hilbert basis theorem), so the ideal generated by a set of polynomials is finitely generated, and their common zero set are the same.
2. As a variety, is more often denoted by .

Some terminologies

1. the Zariski topology

This formalism allows us to talk about varieties more easily. Consider the topology on with the closed sets being affine varieties. It is easy to verify that this is a topology:

• ,
• Union of two closed sets:
• Intersection of closed sets: just put all the equations together.

2. the ideal of X

This is the ideal of all the polynomials in that vanish on .

3. the affine coordinate ring

Consider as the coordinates of the affine space , and restrict it on . If we allow free multiplication and addition, this is what we will get.

The Nullstellensatz

It is clear that when we have a subset , we can define the corresponding ideal exactly like above. On the other hand, given an ideal , we can define to be the common zeroes of all .

Some natural consequences:

• If , then
• If are ideals of , then
• If is an ideal of ,
• If , then (closure in Zariski topology)

Only the proof of is nontrivial. It is clear that . For the other side, suppose that is Zariski closed and . Then , thus . This shows that is the smallest Zariski-closed set containing .

If is an equality, then we can establish a one-one correspondence between the affine varieties in and the ideals of . This is desirable because as in the case manifolds, we want to define “varieties” without an embedding into an affine space. As we shall see, varieties are isomorphic iff their affine coordinate rings are isomorphic as -algebras. Thus once we have this correspondence, we can encode the information of points into the affine coordinate ring as the first step of the definition of “varieties”.

This correspondence exists when is algebraically closed.

Theorem 1 (Hilbert’s Nullstellensatz) Let , where is algebraically closed field. Then .

In other words, if vanishes on the common zeros of an ideal , then for some , .

Regular functions

Basically we want to deal with polynomial maps. However, when we restrict on affine varieties, there are some more possibilities. For example, if is an affine variety, and is a polynomial that never vanish on , then makes sense for any polynomial .

However, the following proposition tells us that allowing such division does not enlarge the class of maps concerned, when is algebraically closed.

Proposition 1 Suppose that is a polynomial that never vanishes on an affine variety , where is algebraically closed. Then is a unit in .

Proof

Let . As never vanishes on , have no common zeros. By the Nullstellensatz, . This means that there exists polynomials such that

Modulo gives the desired result.

Thus maps of the form on can also be represented by , i.e. we are still working with polynomial maps.

Definition 2 (Regular functions) Let be an affine variety. A regular function is a polynomial map from to .

It is clear that the regular functions form a ring, which is exactly , the affine coordinate ring.

Maps between affine varieties

Naturally, just as in the case for differentiable functions,

Definition 2 For two affine varieties , , a map is called regular if are regular functions on .

This immediately gives us a notion of isomorphism: Two affine varieties and are isomorphic if there exists regular maps and such that and .

Regular maps and ring of regular functions

Notice that composition of regular maps is still regular. In particular a regular map induces a -algebra homomorphism .

On the other hand, a -algebra homomorphism has to come from a regular map. For are the coordinates of , and tells you what they are in terms of . Thus consider defined by

Then one can see that . In fact we have

Proposition 2 For two affine varieties and , we have a natural isomorphism

Corollary 1 Two affine varieties and are isomorphic if and only if is isomorphic to as -algebra.

Examples

1. is an affine variety.

2. is an affine variety.

3. (Let be algebrically closed) The map defined by is regular. It is regular, bijective, yet not an isomorphism, since a polynomial in and can’t possibly give .

What algebras are coordinate rings?

This is answered by the following corollary of the Nullstellensatz.

Corollary 2 Let be algebraically closed. The association is a 1-1 correspondence between affine subvarieties of and radical ideals of . Moreover, is irreducible iff is prime.

Recall that a topological space is irreducible if it cannot be written as a union of two proper closed subsets.

Proof

The 1-1 correspondence was established already.

Suppose that is reducible, i.e. , being closed and proper. Then . If is prime, then or , meaning or , contradiction. Therefore if is reducible, is not prime.

If is not prime, let but .  (i.e. vanishes entirely on , but neither or do). Then , and both are proper, closed subsets. So if is not prime, is reducible.

Corollary 3 Let be algebraically closed. Then an -algebra is an affine coordinate ring if and only if it is finitely generated over and admits no nilpotents.

Quotient by group action

(Assumption: is an algebraically closed field)

Let be an affine variety, and be a finite group of automorphisms of . can also be realized as a group of -automorphisms of . We want to give the -orbits a variety structure.

What is the natural definition of this? Well, mimicking quotient topology, we would want a natural regular map if is the variety representing the -orbits. As we already know that the affine coordinate ring is an invariant, we can first study the functions on , and try to see if there is any nice -algebra homomorphism .

The natural class of functions on should be those functions on , who have consistent value on each -orbit. This is exactly , the -invariants of .

The first question is, can possibly be an affine coordinate ring? It is clear that has no nilpotents, being a subalgebra of .

Proposition 3 Let be the order of . Suppose that , then is a finitely generated -algebra.

We have a natural inclusion map , which tells us that if is an affine variety such that , then we have a map induced by the above inclusion. We then want to see if really represents the orbits.

Proposition 4 are in the same -orbit if and only if . Furthermore, is onto.

Proof of proposition 3

Define the averaging operator

for .This is a map from to .

Let be the generators of , this means that for arbitrary , we have an expression

where . Since are not fixed by , we hope that we can get something after averaging. Yet does NOT behave well with respect to multiplication.

So first let us deal with the simple case, WLOG, . By definition,

So this is like a power sum. Then by Newton’s identity, we would want to consider the polynomial . The coeffcients of the polynomial lie in , and can be expressed as a sum of (coefficients of * ), where , the order of .

Learning from the monomial case, we tackle the case of a general multinomial in the same fashion. WLOG consider , with . Using the same idea, we see that it can be expressed as the sum of (coefficients of * ), where , the order of .

This shows that in general, an arbitrary monomial can be written as the sum of (coefficients of * ), where , the order of . Notice that is additive, and that there are only finitely many elements in to be of the form (coefficients of * ), where , we see that after taking , these are the desired generators of .

Proof of proposition 4

(I can’t find a good proof for surjectivity)

If are in the same -orbit,

recall that how is constructed. We take a set of generators of , embed it into , and use these regular functions to define . This means that the coordinate functions are all -invariant. Therefore for are in the same orbit.

If are not in the same -orbit,

we want to show that for some set of generators of , one of them would have different values at and . This is equivalent to finding a -invariant polynomial that has different values at and .

Construct a polynomial such that and for all . (This can be done using Lagrange interpolation, analogous to the one-variable case) Then we symmetrize it by considering

giving us one of the desired polynomials that separate and .

For surjectivity, one can proceed using commutative algebra, but I still can’t see conceptually why it is onto.

The inclusion is integral, because for any , the polynomial is one integral polynomial for with coefficients in . It is not difficult to show that is a closed map. Since is an inclusion, the Spec map is dense, thus onto.

This post will prove the uniqueness of Jordan form, Smith normal form and Rational canonical form in an equivalence class of similar matrices. This is why they are called canonical.

Theorem 1 Let be an matrix (over an algebraically closed field ). Then there exists a unique matrix among all the similar matrices of , that is in Jordan form. (up to permutation of Jordan blocks)

Proof

We have seen in a previous post that how determines the Jordan blocks. This determines the uniqueness.

Alternatively, notice that if we regard to be the matrix of a linear transformation , then a Jordan form corresopnds to the primary decomposition of finitely generated modules over PID, which is unique up to permutation.

Theorem 2 Let be an matrix (over a PID). Then there exists a unique matrix among all the similar matrices of , that is in Smith normal form. (up to associatedness of elementary divisors)

Proof

Define to be the greatest common divisor of all minors of of order for .

Once we have put the matrix in its Smith normal form = , it is clear that

(if we set if ) Therefore the theorem is proved once we have

Lemma 1 (Invariance of ) Let , where are invertible matrices. Then for any , .

Proof Clear from Binet-Cauchy formula.

Theorem 3 Let be an matrix (over a field ). Then there exists a unique matrix among all the similar matrices of , that is in rational canonical form.

Proof

If we regard to be the matrix of a linear transformation , then rational canonical form corresopnds to the decomposition of finitely generated modules over PID, which is unique.

This gives us some methods to check when two matrices are similar.

Corollary 1 Two matrices and over a field iff they have the same Jordan form/rational canonical form.

Smith normal form also helps determine the similarity of matrices.

Corollary 2 Two matrices and over a field are similar iff and has the same Smith normal form.

Proof

Regard to be the matrices of linear transformations . Remember that the Smith normal form of () represents exactly the decomposition of -module. (-module) Therefore it suffices to show that , are similar iff the -module structure of is the same as its -module structure, which is obvious.

Similarity and underlying field

As Jordan form needs the field to be extended to its algebraic closure, it raises a natural question: Let are two fields, and are two matrices with entries in . If are similar over , are they similar over ? The answer is yes.

Theorem 4 Let are two fields, and are two matrices with entries in . If are similar over , then they are similar over .

Proof for infinite field case

is equivalent to = 0. Thus we are asking if is solvable over , can it be solved over .

Using Kronecker product, rewrite the equation as

This shows that the solution space of (over either or ) can be spanned by matrices with entries in . Let be one such basis, and consider the multinomial

is not identically 0, because over , there exists some such that . If the field is infinite, then this implies that for some , , meaning that is solvable over .

There is a swft argument for the general case, as indicated by loup blanc in this post.

Proof 2

1. , are similar if and only if they have the same rational canonical form, as proved above.

2. Notice furthermore that the rational canonical form of in is the same as that in . Reason:

Suppose that we can find invertible such that is in Smith normal form. By uniqueness of normal form, and that , the Smith normal form in is the same. This implies that the rational canonical form are the same.

If are similar over , they have the same rational canonical form over . But their rational canonical form over are the same, so they are similar over .

Quick summary

1. Jordan form/rational canonical form of a linear map is merely the matrix form of the structure of as a -module.

2. The similarity problem is the same as whether the -module structure of are the same.

3. is exactly the relations matrix/kernel of the natural map . Thus it is significant in rational canonical form/Jordan form.

The polar decomposition is a generalization of the polar form of complex numbers .

Theorem 1 (Existence of polar decomposition) Let be an complex matrix. Then , where is a positive semi-definite Hermitian matrix, and is unitary. Furthermore, is uniquely determined by .

Lemma 1 There exists a unique positive semi-definite square root for a positive semi-definite Hermitian matrix .

Proof

Existence: By spectral theorem, there exists unitary such that is diagonal. Take the non-negative square root of each diagonal entry (remember that a Hermitian matrix admits real eigenvalues only), we get a diagonal matrix , such that . Then .

Uniqueness: Suppose that . BY spectral theorem, there exists an orthonormal eigenbasis for the left multiplication map by , i.e. for . Squaring, we get for all Since is uniquely determined by (its eigenvalues), are also uniquelydetermined by , asserting the uniqueness of positive semi-definite square root.

Proof of Theorem 1

Notice that is a Hermitian matrix. By spectral theorem, there exists an orthonomal eigenbasis for the left multiplication map by . Let . If is the complex inner product, we have

This means that is an orthogonal set. Extend this to an orthogonal basis, and make it orthonomal: , i.e. if , then .

Then it is trivial to see that the map can be decomposed to . The first operation is a unitary one because it is a change of orthonomal bases. The second operation is a mere scaling up, so it is positive semi-definite.

Remark

If we suppose that such a decomposition exists, then , i.e. is the positive semi-definite Hermitian square root of , thus is unique.

Another common canonical form is the rational canonical form. While Jordan form is more like a substitute for non-diagonalizable matrix and thus does not exist when the eigenvalues do not lie in the ground field, the rational canonical form exploits the properties of the characteristic/minimal polynomial.

Theorem 1 (Existence of rational canonical form, invariant factor version) Let be an matrix with coefficients in a field . Then it is similar to a direct sum of companion matrices such that .

Recall that a companion matrix is a matrix of the form

with the property that both the characteristic polynomial and the minimal polynomial are exactly .

Proof

Regard as the matrix of a linear map , where is a finite dimensional vector space over . Regard as an -module and apply the structure theorem for finitely generated modules over PID. Then we may write

such that are the invariant factors, up to units.

Finite dimensionality of implies that there is no free part in the decomposition. For each , it is clear that with respect to the basis ,
acting on takes the form of a companion matrix . Putting all these companion matrices together, we have shown the desired result.

Using rational canonical form, it is then clear that

1. The minimal polynomial of is , the characteristic polynomial of the largest cyclic block.
2. The characteristic polynomial of is .

Similarly, we also have the elementary divisor version by using the “primary decomposition” in the structure theorem of finitely generated modules over PID.

Theorem 1′ (Existence of rational canonical form, primary decomposition version) Let be an matrix with coefficients in a field . Then it is similar to a direct sum of companion matrices such that each is a prime power.

So how can we compute the rational canonical form? The first question is how should the rational canonical form look like, and the second question is how to find one such basis.

To answer the first question, it suffices to find the invariant factors provided by the structure theorem. This is obtained by

• finding a presentation for the finitely generated module. In our case, we need to find a presentation for as an -module if we regard as the matrix of the linear map .
• Find the relations matrix, and apply Smith normal form to it. The resulting invariants are the invariant factors we want.

1. A presentation for

Let has an -basis for which is represented by the matrix . The most natural presentation is to consider the map defined by

We then need to find the relations matrix, i.e. the kernel of .

Claim 1 The kernel of is the -submodule generated by .

Proof

It is obvious that lies in .

For the other inclusion, let , i.e. . For each involved, we rewrite it as , and take the first term out. This way the degree of the polynomials in will decrease by 1. Repeat this process until all the s are eliminated. We are left with a linear expression in that equals 0. Therefore every coefficient in this last relation is 0.

Therefore equals the parts taken away. Since each part taken away is a multiple of , we have shown the other inclusion.

2. We now have the relations matrix , where is just an independent variable. Working out its Smith Normal Form give us the invariants needed.

Primary decomposition

If we factorize the invariants factors and put the prime powers separately (using Chinese remainder theorem), we would get the required primary decomposition.

Smith normal form is a generalization of Gaussian elimination to matrices with coefficients coming from PID. This can be used, for example, in finding a -basis for the kernel/image of a matrix of integers, or proving the often-cited structure theorem for finitely generated module over PID. The procedure is essentially the same as Gaussian elimination as well.

Theorem 1 (Existence of Smith Normal form) Let be an matrix with entries in a PID . Then there exists matrix and matrix , both invertible, such that is of the form

with .

Proof

The proof is similar to the Gaussian elimination process.

WLOG, assume that the first column is nonzero. Let us consider three operations:

1. Switching two rows

2. Adding some multiple of one row to another

3. For two relatively prime elements , we can find such that . For two rows 1 and 2, we may

• replace row 1 by row 1 + row 2
• replace row 2 by row 1 + row2

Only the last operation is different from the usual row operations. This is because multiplying a row by a constant involves scaling the determinant up by that constant. If we are working over a PID, that constant may not be a unit and thus is not what we want. The new operation is a replacement of scaling up two rows and replace one row by their sum, whose determinant is a unit.

Analogously we also have the column operations.

1. Assume that . WLOG, assume that the first column is nonzero (switching columns if needed), and actually the upper left element is nonzero (switching rows if needed). Take this as the pivot.

If the matrix is over a field, we can just eliminate every other entry in the first column. However in our case, division is not allowed, so the best we can do is somehow make the upper left entry the g.c.d. of all the entries in the first column, then eliminate.

Consider . If it is zero, leave it there. Otherwise, let the g.c.d. of and be . Then we can find such that. Apply the third type of row operation, we can replace by , and by a multiple of . Subtract the second row by a that multiple times the first row, we can make zero.

Repeat this, and we can get every element in the first column, except , to be zero. This finishes one round of elimination.

2. Repeat this again for the first row using column operations. A bad thing may happen: entries in the first column may become nonzero again. We can go back to step 1, and use row operations to clean up the first column. Of course the same question for the first row will arise. Does this loop terminate? The answer is yes, because after one round of elimination, the new divides the previous . Consider the chain of ideals generated by . It is an ascending chain, which stabilizes since a PID is Noetherian. That means eventually, stops changing – that only happens when we no longer need to use the third type of operation, i.e. every element in the row is divisible by . Using the second type of operation does NOT affect a cleaned-up row/column.

This means that by multiplying approriate invertible matrices on the left and on the right, is now of the form

3. Repeat for the lower right submatrix. Iterations will eventually give invertible matrices and such that is diagonal. By switching columns if needed, we can assume that all the nonzero columns are aligned to the left, i.e. it now takes the form

The final question is how to make , as needed.

Let us focus on the upper left 2 by 2 block only. We show that after row/column operations we can make .

• Add the second column to the first.
• Using the third type of row operation, we make the upper left entry to be , the g.c.d. of . It is clear that the lower right entry is unchanged, and the other two entries are multiples of .
• Subtract the second column by the first with appropriate multiple such that the upper right entry is 0. Analogously make the lower left entry 0. The lower right entry may be changed, but it is definitely a multiple of the upper left one.

We may then repeat this process so that , and then and so forth. Repeat this and done.

Remark

1. This proof is essentially the same as Gaussian elimination.
2. The is needed to make the form canonical.

Example 1

Question

If we change the PID condition to a dimension condition, can we get anything similar? Maybe should try to see what happens.

In this post we focus on computing the Jordan form of a given matrix.

Proposition 1 (Calculating the number of Jordan blocks/sizes)

For each eigenvalue , the number of corresponding Jordan blocks is . More generally, let

= the number of Jordan blocks of size

Then , i.e.

This further shows that the number of Jordan blocks of size is

This is obvious by looking at the Jordan form. A nice way to keep track of all these numbers is to use Young tableaux, which will be done later.

The next question would be how to find a Jordan basis. A possible procedure is outlined here,

1. Fix an eigenvalue . Find the smallest such that stabilizes, i.e. . This is the size of the largest Jordan block.

2. Find a basis of . Pull this basis back to; this set is clearly linearly independent. Let this pullback set be . Notice that the set defined by

is also linearly independent. Moreover, it spans a -invariant subspace, where the matrix of with respect to this basis is blocks of Jordan block of size .

3. Consider . We already have a linearly independent set . Extend this to a basis, say we concatenate it by a set . The set

is again linearly independent.

4. Iterate, the final set would be a Jordan basis, since it is a set of linearly independent elements with the right dimension (why?) such that has Jordan form with respect to it.

Example 1 Find the Jordan form of .

The eigenvalues are clearly all 1. It is clear that , which means that there is only one Jordan block. Thus its Jordan form is .

To find a Jordan basis, notice that as there is only one Jordan block, the basis must be a cyclic one. Thus we only have to compute the image of .

Thus is one Jordan basis.

Example 2 Find the Jordan form of .

After computations, the characteristic polynomial is . Therefore it suffices to check the number of Jordan blocks for the eigenvalue .

is clearly of rank 1. So the kernel has rank 2, meaning that there are 2 Jordan blocks, i.e. is diagonalizable and is similar to .

To find a Jordan basis, we first consider the eigenvalue 1.

After performing Gaussian elimination, we get

Then clearly the kernel is spanned by .

For the eigenvalue -1,

thus after Gaussian elimination, it becomes

yielding a basis of .

Example 3 Find the Jordan form of .

After computations, the characteristic polynomial is . Therefore it suffices to check the number of Jordan blocks for the eigenvalue .

is clearly of rank 2. Thus the kernel is of 2 dimension, meaning that the Jordan form has to be

.

To find a Jordan basis, for the eigenvalue 2,

After some row reduction, we get

Therefore the kernel is spanned by .

For the eigenvalue 1,

1. By some computations, is spanned by .

The kernel contains . One can extend this to a basis by adding .

2. Compute

Then is the basis for the Jordan block of size 2.

3. Extend to a basis of by adding . Thus

spans .

Conclusion: A Jordan basis is .

Young tableaux

It is a convenient tool to track the number/size of Jordan blocks for each eigenvalue.

For example in the above case, we have exactly one Jordan block of size 3, we then give it a horizontal bar

If instead we have one Jordan block of size 2, and another of size 1, we give it a tableau

In general, each row represents one Jordan block, and the number of boxes represent the dimension. We list each block according to its size in a descending order. When we read this figure horizontally, we then obtain the numbers defined previously, i.e. the number of Jordan blocks with size .

Therefore if we merely wish to find the Jordan form, we may, for each eigenvalue ,

1. Calculate until it stabilizes.

2. Calculate , where is the number of Jordan blocks with size .

3. Draw boxes in the -th column, aligned to the top.

4. We then obtain a Young tableau, and we can just read off the number of Jordan blocks (the number of rows), and their sizes (the number of boxes in each row).

Remark

The last two examples are taken from the last page of http://www.math.uga.edu/~roy/rev.lin.alg.pdf .

Random words/My thoughts/Motivation(?!)

Very often we encounter functions that satisfies “local property”, e.g. continuous, differentiable, smooth functions on manifolds. An idea is that an object can be described by the functions we “allow” on them, e.g. a topological manifold is a manifold with allowed functions being continuous maps, whereas a smooth manifold is a topological manifold with allowed functions being “smooth” maps. Of course, smoothness is not something defined on a topological manifold, but we can try to extract all the smooth functions (defined over any open set), and look at them as a whole. For this purpose, we would need some terminology that allows us to track functions defined on an open set, and allows us to restrict a function to a smaller open set.

This is where presheaves and sheaves come in. A presheaf is something that tries to track data (aka functions) while allowing restrictions. While “restriction” in this sense can be arbitrary, a sheaf imposes more conditions in order to make the “global” functions depend on “local behaviour”.

Definition 1 (Presheaf) Let be a topological space. A presheaf on is a collection of abelian groups for every open set of .  We also have the restriction morphisms for open sets . They are subjected to the following axioms,

1. 
2. Restriction on itself is identity: identity
3. Composition of restriction: for open sets .

Remark

1. The definition above defines a presheaf of abelian groups. We can of course deal with other categories and consider respective morphisms. However, according to the wikipedia page the first condition is dropped. It is somehow embedded into the definition of sheafs and only when we deal with concrete categories, i.e. a category with a faithful functor to the category of sets.
2. Consider the category whose objects are the open subsets of , and the morphisms are the inclusion maps. A presheaf can then be defined as a contravariant functor from this category to the category of abelian groups.
3. Some terminologies:
   Suppose is open, and . is then called a section of over .
   If , then the restriction of , i.e. is often denoted by .

Definition 2(Sheaf) A sheaf on is a presheaf subjecting to two more conditions:

1. (Identity axiom) Let be open with an open cover , and let . If for each , then .
2. (Gluing axiom) Let be open with an open cover , and let . If for any we have , then there exists such that for all in the index set.

Remark

1. The gluing axiom means that if a bunch of elements are compatible, then one can glue them. The identity axiom then says that this patching is unique.
2. The identity axiom has to be modified in general category, because a zero object may not exist. We change it to if s and t are locally the same, then they are the same.

Stalk, sheafification, effects on stalk.

Maps, kernel, cokerenel, image, quotient, pushforward, pullback

Examples: diff/cts functions, constant sheaf, sheaf hom, skyscraper sheaves

Constructions: direct sum/product, direct limit/inverse limit, extension by 0, gluing.

Notions: support of sheaves, flasque sheaves.

In this post we will work over algebraically closed fields.

The diagonalization problem asks whether a linear transformation admits an eigenbasis. Eequivalently it asks when an matrix is similar to a diagonal matrix, i.e. for some invertible matrix and a diagonal matrix .

Let the characteristic polynomial of be , where are the distinct eigenvalues of and are the respective algebraic multiplicities.

Theorem 1 Let is an endomorphism for finite dimensional vector space (over an algebraically closed field). Let its characteristic polynomial be . Then .

Proof

1. The sum part

Notice that , , …, are relatively prime. Therefore we can find polynomials such that

For any vector , notice that , by Cayley-Hamilton theorem. Analogously we have other inclusions. Therefore by applying (1) to , we have

This gives a required decomposition.

2. The direct part

WLOG suppose that

where , is the shortest possible relation for which .(). Applying on both sides gives

(3) and (4) are linearly independent equations, because are all distinct. Therefore it is possible to eliminate one of the vectors involved, e.g. to eliminate , and still have other coefficients to be nonzero. This contradicts the minimality assumption, so must be linearly independent.

Corollary 1 is diagonalizable if and only if for each , .

This is because being able to find an eigenbasis means exactly that .

Proposition 2 .

This will be clear using Jordan form.

Corollary 2 is diagonalizable if and only if for each , .

is usually called the geometric multiplicity of the eigenvalue .  Thus the above corollary may be rephrased as

Corollary 2′ is diagonalizable if and only if for each , the algebraic multiplicity coincides with the geometric multiplicity.

Proposition 2 also gives the following quick corollary.

Corollary 3 The algebraic multiplicity is always no less than the geometric multiplicity.

So the diagonalization problem boils down to whether . This may not hold in many situations. The natural question is then: can we choose a nice basis from each such that when is diagonalizable, the selected basis is an eigenbasis?

one answer to this qusetion is the Jordan form. The Jordan canonical form picks a basis such that the matrix constitutes Jordan blocks of the form

Theorem 2 (Existence of Jordan form) Let is an endomorphism for finite dimensional vector space (over an algebraically closed field). Let its characteristic polynomial be . Then there exists a basis of so that the matrix representation of is a direct sum of Jordan blocks.

Furthermore, the number of Jordan blocks and the size of each of them is an invariant for .

Proof

Since we have already shown , it suffices to choose a nice basis in each .

Let . Then is a -invariant map. Therefore, we may regard as an -module. Since is a principal ideal domain, we may invoke the structure theorem for finitely generated modules over PID to see that

where each is a cyclic subspace. Take as an example. Let be its generator, and suppose that is the smallest number such that . Then it is immediate that forms a basis for . It is then routine to show that with respect tothis basis, takes exactly the form of a Jordan block.

Similarly, one can choose basis for the other to get a required basis for for each eigenvalue . Put them together and we have found a basis with the desired matrix representation.

Notice that the number of Jordan blocks is determined by the number of distinct eigenvalues and the number of cyclic modules in the decomposition. The size of Jordan blocks is determined by the invariants associated to the decomposition into cyclic modules. The uniqueness thus follows.

Although the above proof cannot help us to do any computation, it has some theoretical implications because the Jordan form is easier to manipulate.

Proposition 3 (Minimal Polynomial)

For each eigenvalue , let be the size of the largest Jordan block corresponding to . Then the minimal polynomial for is

Remark

1. In fact one can directly apply the structure theorem for finitely generated modules over PID to get Jordan form directly. (skipping the proof of theorem 1) To me, a constructive proof is better though.
2. The algebraically closed field condition can be replaced by “the characteristic polynomial splits”, because all we used is that eigenvalues exist.

It is written in the wikipedia page that Cayley-Hamilton actually holds in any commutative ring with unity as well. Here I will present a proof which should be the same as the one in wikipedia.

In the old proof, we considered the vector space , and our argument used the concept of basis. We cannot copy our proof to the new case because the basis of modules does not behave that nicely. Anyway we can at least consider the matrix as a -module homomorphism from to itself, i.e. we are trying to regard as a -module.

Second proof

By definition, we have that , this means that

Multiplying both sides by the classical adjoint of the square matrix on the left. This immediately implies that

i.e. is the zero endomorphism, which is precisely what we want.

Generalization

This proof can be easily adapted to the following theorem that can be found in Chapter 2 of Atiyah and MacDonald’s book.

Theorem 1 Let be a commutative ring with unity, be a finitely generated module, and be an ideal. Let be a -module homomorphism such that
. Suppose that can be generated by elements. Then satisfies a monic polynomial of degree with coefficients, except the leading one, in .

Applying the above theorem to the identity map, we get

Corollary 1 Let be a commutative ring with unity, be a finitely generated module, and be an ideal such that . Then there exists such that and .

Recall that the Jacobson radical of a commutative ring is the intersection of all its maximal ideals. In particular, if then must be a unit. We then obtain the following version of Nakayama’s lemma,

Corollary 2 (Nakayama’s lemma) Let be a commutative ring with unity, be a finitely generated module, and be an ideal contained in such that . Then .

Reference

Atiyah, M. F.; MacDonald, I. G. (1969), Introduction to Commutative Algebra