I don’t think i am taking any class on stochastic calculus. The reason: i heard the person teaching this class is very theory oriented and probably won’t provide any applications. Bad. I am a simple person. I need to see some applications amidst the theory building. Instead I plan to take another standard algorithms class (i took one such grad class back at berkeley. but basically i just slogged thru the homework and forgot everything that wasn’t tested in hw/test. Perhaps about 20% of this class i might already know, but it would be a good review. I’m starting from ground zero essentially. We talk about fibonacci heaps which i didn’t study the last time. Amortized analysis is like magic. You come up with some brilliant potential function and the key idea is where is the actual cost incurred. Looking at , you sometimes overcount, sometimes undercount, but in the end, is all that matters and we can look at the amortized cost of every operation instead, and blithely ignore the actual cost . A link I find useful is www.cs.duke.edu/courses/fall05/cps230/L-11.pdf.

Another class I want to take is combinatorics: the probabilistic method. This is cool. I learn some basics as an undergrad but this class i hope is going to go beyond that. Simple tricks like the union bound () and linearity of expectation can be used to obtain powerful bounds. That’s all for now. Tomorrow I’ll be attending two classes on commutative algebra and computational algebra, one of which will use the book i am supposed to be reading (Cox, Little, O’Shea). Kleiman is teaching both courses. I hope he’s a thorough and considerate prof and I hope (too) that these can turn out to be useful for me in future. It’s about the connecting the dots, right Steve?

I have to make an effort to attend seminars too. CS seminars to find applications for what i’m learning. Math seminars to gather more tools. It ain’t easy trying to bridge these two sides.

Someone on livejournal posted this problem. Suppose you have a (hyper)sphere in dimensions (don’t worry, no one can really imagine anything dimensions higher than 3 or 4). Suppose you randomly pick two points on the surface of the sphere. What is the expected distance? Here is my “solution”.

Without loss of generality, fix one point to be the north pole. Note that though you do not need to imagine a (hyper)sphere, you do need to imagine a 3D sphere, and know spherical coordinates. Let denote the surface area of a sphere with radius , in dimensions. So , , and so on.

Before proceeding, you might find it helpful to review spherical coordinates, or even look at hyperspherical coordinates by checking out http://en.wikipedia.org/wiki/Hypersphere. Alright let us move on.

Let us work in dimensions. (You may like to substitute .)
Remember first point is the north pole. The main idea is to integrate circular strips from top to bottom. Each circle has radius where is radius of our sphere and is the “zenith angle” (http://en.wikipedia.org/wiki/Spherical_coordinates). Each of these strips have a certain probability. Its surface area is . is like the thickness of this strip. Convince yourself at least for . So the probability of picking a point on this strip is .

In fact, . For example, is equal to or .

Every point on each strip has equal distance from the north pole. The distance is . (just draw a line between the two points and bisect .) So, the expected distance is .

Let be surface area of unit sphere. Obviously, by rescaling, similarity etc, . Express the denominator as the integral over circular strips. Substitute in . Cancel these constants to obtain the expected distance as

. Great. At this point, you can crank up mathematica, type in the expression, and get ugly expressions and send . Very tempting but not nice. Let’s try something else .

If you plot for , you will notice that it becomes sharper and sharper at . Aha, it is becoming like a delta function multiplied by a constant . Therefore, the expected distance is becoming with which is .. We are more or less done.

This is the idea, but it is not rigorous. The rest of this post will make this as robust as titanium. However, it may also be a good point to stop reading.

I recall Stein and Shakarchi, chapter 3, section 2 on “good kernels and approximations to the identity”. This is what it says.

Suppose there is a family of functions such that

1. 
2. for some real number independent of
3. For every , as

Then for any bounded function , we have as , where is the convolution between and .

This is like , a Dirac delta function at the origin. Often, we write . To be mathematically correct, we should write as instead. Now, the integral becomes a convolution between and and the above “theorem” gives us the answer . Now back to our problem.

Scratch work: we want “equal” supported on . Let . Let . Now we know we have to define where is just some useless normalizing factor to make satisfy condition 1. Notice that is basically our over some constant, shifted to the left by . Let us verify further that is a good kernel/approximation.

Condition 2 is trivial since our function is nonnegative and condition 1 implies 2. Condition 3 is more tricky to check. Let’s work on it. Fix some . Let which is strictly less than 1.

By symmetry, the numerator of the integral (see condition 3 and the definition of ) is , i.e. . The denominator also grows small with , but not as rapidly as we will show. Integrating by parts, we know that . Its lower bound turns out to be for some small constant .

For example, . Insert extra factors of the form to get a lower bound. Now, consecutive factors cancel off and we are left with some constant divided by 9.5. The same reasoning applies to any .

In short, the numerator dies faster than the denominator and condition 3 is satisfied.

Next, let . Working backwards using the “scratch work”, we see that the expected distance is just integrating , which is , and must converge to as . That is it.

Key ideas: Recognizing good kernels.

At the heart of this chapter is Fundamental Theorem, invariant factor form and elementary divisor form. Let us write this down nicely.

(invariant factor form) Let be a PID and be a finitely generated R-module. Then, can be written as the direct sum of finitely many cyclic modules. More precisely, , for some nonnegative integer , and (the annihilators of each cyclic module) are nonunits and satisfy .

Proof of this in Dummit and Foote relies on a “technical lemma”, aka “don’t read me because you will forget me immediately anyway”, which says:

Let be PID. Let be any submodule of a free R-module which has rank (max number of R-linearly independent elements). Then rank of is free of rank with , and there exists a basis of such that is a basis of and . (the ‘s are known as the invariant factors)

If we assume this result, the fundamental theorem falls easily. Let be basis for and be set of generators of of minimal size. Consider the map mapping to . This is surjective since generates , so . By the lemma, we can find another basis of such that is a basis for . This means . See that this is isomorphic to . (consider the map from to by “modding” the coefficients of by , for . The kernel is exactly .

If we look at each cyclic module, say , we can write (remember is PID and UFD). By Chinese Remainder Theorem, where is a prime in . For each cyclic module annihilated by an invariant factor, we can do this.

Now GROUP the cyclic modules annihilated by powers of the same prime together. Example: can be rewritten as where and . Notice that is annihilated by , so is an elementary divisor. is NOT a cyclic module, but it is generated by 2 elements, both of which are annihilated by .

This is the so-called “elementary divisor form” of the Fundamental theorem.

Now, let act on (or ) by . The “invariant factor form” of the Fundamental theorem states that where . The minimal polynomial is clearly . If it’s anything smaller, then elements in the “bigger” cyclic modules will not be annihilated by it.

Consider a cyclic module, annihilated by say . Say the cyclic module is generated by some , then consider the basis , i.e. . They are linearly independent, otherwise there is a smaller annihilator. They obviously spans the cyclic module. Moreover, and . Representing in this basis , we have the “companion matrix”. (google rational canonical form).

The discussion applies to only one cyclic module. Since is a sum of such modules, it follows that can be represented as a matrix with companion matrices on the diagonal.

To get the Jordan canonical form, do roughly the same thing. This time, is decomposed into , each annihilated by some . For Jordan form, we assume an algebraically closed ??, i.e. the only interesting primes are linear factors. Therefore, each is annihilated by some power of some linear factor . Pick that one element which is annihilated by (and no less). Call it .

Consider the set . This is a basis since is annihilated only by and no smaller polynomial. It has the right dimensions and must span. Let . Notice that . Representing in this basis gives us a Jordan block. Since is direct sum of such modules, is represented as a matrix with Jordan blocks on the diagonal. This is the Jordan canonical form.

Key ideas: decomposition into cyclic modules, and their annihilators

Mean Ergodic Theorem: Let be an isometry of the Hilbert space . Let be the invariant subspace of . Let be the orthogonal projection onto . Let . Then for each , converges to in norm, as .

(I think Prof Tao has like 4 different proofs of this theorem, but unfortunately I can only give the “slick but not particularly illuminating” proof by von Neumann.)

Define and . Note that are both closed subspaces because . (write ). Denote closure of as .

Lemma: and orthogonal complement of is .

Basic facts about isometry:
1) If is an isometry, for all .
I remember the way I did it is to write .

2) Therefore for all and conclude that .

Proof of lemma: If , then , i.e. . So . Now suppose . Check that now, . () Notice that by Cauchy-Schwarz, . But we actually have equality. This is only possible when , and obviously, .

For second part, means for all . Since for all , we have , which is equal to .

Main proof is short. Write where and . Now apply to .

since , the subspace invariant under . All we need to show that goes to 0. We can pick some arbitrarily close to . The idea is to bound the norm of and separately.

(telescoping sum). Check that this converges to 0.

. We can pick arbitrarily close to .

This concludes the proof.

I’m too lazy to study the pointwise ergodic theorem (by Birkoff?). Shall just quote its interesting corollary.

Let be an ergodic measure-preserving transformation. For an integrable function , we have

converges to for a.e. as .

The LHS can be interpreted as the “time average” of . RHS is like the “space average”. is like modifying every time step.

Recall some definitions. (assumed to be a measure preserving map, i.e. ) is ergodic if for any measurable set which is invariant under (i.e. and differs by measure 0), either or has measure 0. (in probability space, this just says any invariant set under has either measure 0 or 1)

We shall skip the proof.

An interesting exercise. Consider the map defined on as . Show that is measure preserving, mixing (hence ergodic). Prove as a consequence that for “almost every” number , is “normal”.

Let us define what we mean by “normal”. Suppose we do a “m-adic” expansion of , i.e. write in base , i.e. . Then is normal if for every between 0 and , we have as .

This seems quite profound. Almost every number in is “normal”, i.e. every digit is repeated equal number of times!

Let’s do this exercise. Clearly, is measure-preserving. Its inverse is times smaller but repeated times. Showing that it is mixing amounts to generalizing the proof for the doubling map on page 305.

Finally, apply the “space average=time average” corollary. Define to be characteristic function on the interval (in m-adic form). Then . Notice that only when . Hence the time average is the LHS. Simple application of a deep theorem gives us an interesting result.

Key ideas: Ergodic measure-preserving maps, properties of an isometry.

Localizing a ring by its subset really just means constructing a ring of “fractions” such that elements of becomes units. This ring of fractions shall be denoted as .

Here’s an example from Wikipedia. Let be the ring of functions defined on some algebraic variety . Suppose we are interested in studying this variety “locally” at a point . We will let be set of functions not zero at point and “localizes” with . The resulting ring contains additional information about the behavior of near .

Reference as usual is Dummit and Foote (chapter 15.4).

The formal construction of is as follows. Assume to be multiplicatively closed. Define a relation on : iff for some .

This is an equivalence relation. Let denote the equivalence class of . Define addition and multiplication like fractions. e.g. and . Note that for every , is a unit in . Its inverse is simply .

Define the map as for the rest of the post. Two initial observations.

1) If contains no zero divisors, then is an injection. Easy to check. implies that there exists some such that , implies that has zero divisors. Example: if we localize by , then the map is an injection. The ring of fractions here is just .

2) iff . Proof: LHS iff , i.e. there exists such that , i.e. . So almost all the time, we assume does not contain 0.

Do not bogged down by the formalism! Really just think of as fractions where denominators are taken from .

Now, we introduce two operators (contraction) and (extension). Given an ideal , . NOT just . It is best to interpret as . (anything in is of the form where and and and ). The contraction is simply . We can think of as .

An important fact is that is a bijective correspondence between prime ideals of disjoint from and prime ideals of . So if is a prime ideal of , disjoint from , then is a prime ideal in and moreover, . On the other hand, if is a prime ideal of , then is a prime ideal in , disjoint from and also, .

We shall skip the routine but glorious details. This is proposition 38, page 709.

Note that for any ideal , we always have . And for any ideal , . i.e. these are elements of such that there are some such that . i.e. we are adding elements of divided by elements of . Typically is larger than . If they are equal, we say is saturated. Giving denominators to it does not add any new elements.

Let us get to the algorithm. Say we want to know whether is a prime ideal in . To do so, we try to show iteratively/inductively that is prime. Think of as , i.e. let . Assume by induction that is prime. Let . This is an integral domain. Let be the quotient field of , i.e. the localization of by . Note that (or ) is an Euclidean domain (smell something nice). Let us draw the following very important picture:

Let us put proposition 39 in this context. For the forward direction, it says that if is prime, then has to be prime. Moreover, let , is a prime ideal in and is saturated, i.e. .

The other direction says that if these conditions are true, then is a prime ideal. Inductively, the condition about being an integral domain is already satisfied. So it suffices to check that is prime AND if is saturated. The first conditon is easy since is an Euclidean domain, PID, and every prime ideal is maximal and we just need to test if the generator is irreducible or not, or equal to 0$. Checking whether is saturated sounds more hairy.

Again, I’m not going into proof details.. Say . As we mention earlier, we need to check if is 0 or irreducible in . That aside, we need to check if . Proposition 40 comes to the rescue. Before that, note that we can assume that . This is because and we can clear the denominators of all the coefficients to make it a polynomial in . Let be the leading coefficient of . Let be the localization of with respect to powers of , i.e. introduce powers of as denominators.

Prop 40 says that . Let be ideal generated by and in (weird). Then .

We just need to check if . Exercise 3 says that this is the same as checking the unmodded version, i.e. whether the ideal generated by and in (instead of ) is equal to , which can be achieved by working with the reduced Groebner basis of .

Although we didn’t go into details, we gain an idea of the main principles behind the algorithm. By localization, we can turn rings into fields, work in Euclidean domains, PIDs, and then translate back.

Key ideas: extension and contraction of ideals, bijective correspondence between prime ideals, prime in implies that is prime, i.e. is integral domain which can be extended/localized to a field, and image of in is saturated.

First, we define what is an integral extension. Note that we are dealing with commutative rings. Say is a subring of . An element is integral over if is the root of a monic polynomial in . We say the ring is integral over , i.e. integral extension, if every is integral over .

Next, we define what is meant by “algebraically independent”. Let be a field and be in some -algebra. We say they are algebraically independent (over ) if there is no nonzero polynomial in such that . Another way of saying this is that is isomorphic to by the map . (kernel is trivial)

The Noether’s Normalization Lemma says: Let be a field and be a finitely generated -algebra. Then we can find such that these elements are algebraically independent (over ) and is integral over and .

The proof looks rather technical. We shall skip this, despite its importance..

Weak Form of Hilbert’s Nullstellensatz (WFHN): Let be an algebraically closed field. Then is a maximal ideal in the polynomial ring iff for some . Equivalently, there is a correspondence between points in and maximal ideals in .

Remark: Note that . To see this, consider the map from to by mapping . Clearly, . So LHS has to be a maximal ideal. The RHS is obviously a subset of LHS. Moreover, the RHS is a maximal ideal. If we add any polynomial to RHS, polynomial division gives us an element in . This expands the ideal to the whole ring. In short, RHS is a subset of LHS, but both are maximal. So they have to be the same.

Remark: Let us see why the two statements are equivalent. Suppose WFHN holds. Pick a point . Its ideal is . By WFHN, this is a maximal ideal. If we apply to this ideal, we get back the same point.

Now, suppose we pick a maximal ideal. By WFHN, it is of the form . The variety of is a single point . Apply to this single point to get back . So we got a bijective correspondence between maximal ideals and points.

Now consider the other direction. Assume the correspondence. We already know that is a maximal ideal. Suppose we have a maximal ideal . By the bijective correspondence, . But (by the correspondence) is a single point and its ideal is of the form .

Finally proof of WFHN: One direction we have already explained, that is is a maximal ideal. Consider the other direction. Let be a maximal ideal. Let be a field finitely generated by . Apply Noether’s Normalization Lemma to say is integral over some polynomial ring . Since is a field, must be a field too!

(in general, if is integral over and is a field, then is also a field. reason: given . is a field. So there is some . But is integral over . So there is some monic polynomial satisfied by , say . Multiply everything by to get .)

A polynomial ring is in general not a field, unless . So is integral over . Both are fields, so is an algebraic extension of . But is algebraically closed, i.e. , i.e. for some , i.e. . Hence, . Both are maximal ideals and are thus equal.

Important remark: The significance of the weak form is.. If , then has to be the whole polynomial ring. If not, for some maximal ideal . must therefore contain , i.e. contain the point . Contradiction.

In English, the only ideal which has no common roots is the whole polynomial ring.

Now the stronger form says: . Moreover, there is now a correspondence between affine algebraic sets and radical ideals.

Proof is rather slick. Clearly, . Focus on the other direction. By Hilbert Basis Theorem, assume . Let . Introduce new variable . Consider the ideal .

Claim is that . Reason: Pick any point . By definition of , . So, . But . So . But implies that , contradiction.

By WFHN’s important remark, we know is the whole ring, i.e. . We can write for some . Now replace with and multiply both sides by extremely high powers of , say , to wipe out in the denominator of the ‘s. We will obtain where is with replaced with , multiplied by where is huge so that . Thus, .

As for the correspondence, it’s easy once we have the main result. We already know that . So all we need to show is that when is a radical ideal. The Nullstellensatz says .

Some of the tricks are just too clever.

Key ideas: maximal ideals corresponds to points, radical ideas corresponds to affine algebraic sets/varieties, polynomial magic

Question 10: Let be cyclotomic field for some prime and let . Let be any -th root of unity. Prove that (the trace) is or depending on whether is or is not a primitive -th root of unity.

Proof: This is quite obvious. If is a primitive root of unity, the sum is over all the roots of unity, except 1. This has to be since . If is not a primitive root of unity, i.e. , then the sum will give since . therefore corresponds to

Question 11. The setup: Let be the cyclotomic field, for some odd prime . Let . Say is generated by . Let be a subgroup of index 2 in cyclic group . Define and .

Let’s think about this set-up: and . Let denote the map that sends to . Recall there is an isomorphism between and by sending to . turns out to be . The proof lies in elementary number theory. iff or . So the set has exactly elements. Also, it is a group since the product of squares is a square.

Part (a): Show that and and that and .

Proof: For first part, just apply the definitions. The second part follows from our remark earlier about being a set of squares. The coset is disjoint from , and has to contain nonsquares.

Part (b): Prove that and where those bracketed terms are the Lagrange resolvents of .

Proof: What is the Lagrange resolvent in our context? . So . By same argument as in question 10, this sum is . Showing is just as straightforward.

Part (c): Let (the classical Gauss sum). Prove that .

Proof: This part is nice. Remember that in . So .

Part (d): Prove that if and if . Conclude that . Recall that complex conjugation is the automorphism on . Conclude that if is a square mod , i.e. , and if is not a square mod , i.e. .

Proof: It’s late now but I will persevere! is of index 2 and must be a normal subgroup. Recall that . If we apply to , clearly the first term is still a sum over . As for the second term, we will be summing over where . Usual trick here. Write (noting that is a normal subgroup). Hence, .

Actually this is stupid. We are dealing with a cyclic group. So . So clearly, if , then is some odd power of and thus and . In this case, .

We have so far shown that is the subgroup which fixes . By the fundamental theorem of Galois theory, is the index of , i.e. .

If is a square , then and must fix , i.e. . Otherwise, it must send to as what we have just shown.

Part (e): Prove that .

Proof: Use the hint. Conjugate in part (c), we have . The product is therefore . We let and sum in a different manner: . If , then the inner term is 1 and the inner sum is . Otherwise, that inner term is some primitive root of unity and the inner sum is (by question 10). Thus, the sum becomes .

Part (f): Conclude that and that is the unique quadratic subfield of .

Proof: We know from part (d) that or . Split into 2 cases. If , then the exponent on RHS is even and from RHS: . If , then from RHS: .

Part (d) shows that is a quadratic subfield. In addition, it is unique. Given any other quadratic subfield, we can let be the corresponding subgroup in and apply the results above to show that its fixed field is .

Key ideas: working with , changing order of summing

First, a reminder about direct products/sums of modules. If where ‘s are submodules of , we mean that every element in is written uniquely as where , or equivalently the map defined as is an isomorphism. So direct sum or product, it doesn’t matter.

Next, recall what are free modules. When we say is a free R-module, we have to specify a subset such that every element of can be written uniquely as a linear combination of elements from , i.e. if , then there exist uniquely in and in such that . We call a basis for . (Think linear algebra.) If is commutative, then all bases have the same size, called the rank. If not commutative, the bases may be of different size. An important property of free modules is the “universal property”:

Given any R-module and any map , we can always extend linearly to a R-module homomorphism between and . (Think linear algebra again.)

Note that if is finite, say , then any free module on is isomorphic to as expected.

Now, we get to the main topic: exact sequences. Given , we say the sequence is exact at if . Notice that is exact at iff is injective, and that is exact at iff is surjective. In this chapter, we often deal with “short exact sequences”: . When we see such a sequence, we should intuitively think of being an extension of (by ) because .

Notice that there is always an easy extension of a module by , by letting . In a certain sense, got “split” by and . Let us define the notion of a “split sequence” now. If is “split”, we mean that up to isomorphism, . What’s happening in more precise terms, is that where by , where is restricted to . We can say is a “split extension” of by .

Another way to think of a split sequence is that there is a R-module homomorphism such that . Let us see why this is equivalent. If is provided, then we let . If is given, we define .

Before the definition of a projective module, we need to consider and understand the following scenario.

Suppose we have a short exact sequence and some R-module . What is the relationship between and and ? Given , we can easily get a by letting . We denote this obvious map between and as . Since is injective, has to be injective. Reason is that if and is injective, then . i.e. is exact.

The relationship between and is less clear. It turns out that the map is not always surjective! Given some , we cannot always “lift” it to the extension , i.e. find a such that , i.e. .

Nevertheless, it is known that is exact. The left part is done, so we just need to show the exactness at , i.e. . This looks hairy and a little routine, so we shall skip.

Some category theory at this point We can think of as a covariant functor between R-modules and abelian groups. If is a R-module, then this functor will map it to . And if are R-modules and , we can define a group homomorphism , defined by . Our discussion earlier shows that is a “left exact functor”.

SO, what are projective modules? We say is a projective R-module if: for any exact sequence ( can be anything), is also exact. We know this is not always true in general by the previous discussion. But to a projective module, this is true for any exact sequences. This is equivalent (by previous discussion) to saying that: for any R-modules such that is exact, we can lift any to some such that , i.e. is surjective. We call this definition 1.

We will end this post by establishing two more equivalent definitions of a projective module. Particularly, we are curious why are such modules called “projective”?

Definition 2 (of being projective): If is a quotient of , then is isomorphic to a direct summand of , i.e. every short exact sequence splits.

Definition 3: is a direct summand of a free R-module.

First we do .

Say we are given some short exact sequence . From definition 1, let , i.e. lift the identity map in to a map in . So there exists some such that . Think of as in the section about split sequences.

Second, we do .

We can always construct a free module on the generators of . Say is a set of generators for . Define as a free module on . Define as the unique “linear extension” of inclusion map from to (recall the “universal property of free module ). Hence we have an short exact sequence , i.e. is the quotient of a free module. Apply definition 2.

Lastly, we do .

Assume .

Suppose where is a free R-module on set . Let . We want to lift it to some .

The main trick here I would say is “when you lift a bigger thing, you lift a smaller thing on it together”. The bigger thing is where is a natural projection from to . We want to lift it to some . The “lifting” means that . So how is the smaller thing lifted together?

We can define as (the bigger thing lifted) restricted to . will be the smaller thing is lifted to. Let us check this. Let . Then .

It remains to “lift the bigger thing”, i.e. define such that for every , .

To do so, we exploit the “universal property” of to define . We will define a function such that and then extend it to get by the “universal property” of . Getting is easy. maps into and is surjective: given any , there is a such that . Define to be and we are done.

Definition 2 is like saying that given any module which projects onto has as a direct summand (up to isomorphism). That sort of explains why these modules are called “projective”.

Another way of stating definition 1 is: is projective iff the functor is exact, i.e. take short exact sequences to short exact sequences.

Key ideas: universal property, split exact sequences, “lift big to lift small”

any finite division ring is commutative (i.e. a field)

Step 1: Let be center of . Show that is a field containing for some prime . If , then has order for some integer .

We want to show that is a finite field. Remember is a division ring, so every element of already has a multiplicative inverse. Moreover, the inverse is also in because: Let and . Since , we have . So and is a finite field with some characteristic . It acts on like a field acting on a vector space. The last statement follows readily from counting. ( has to be some power of )

The nonzero elements of form a multiplicative group. For any , show that the elements of which commute with form a division ring which contains . Show that this division ring is of order and if is not an element of .

The proof is similar to the previous. This is the centralizer of and obviously contains . By the vector space argument, division ring has size . If is not in , then its centralizer cannot equal and thus .

Show that the class equation for group is where are representatives of distinct conjugacy classes in not contained in the center of . Conclude from step 2 that for each , for some .

Let us revise the class equation. Assume that acts on itself by group conjugation. That where is orbit of . LHS is . First term on RHS is . And second term is using the stabilizer-orbit relationship.

The previous step says that the centralizer has order . Its multiplicative group has zero removed.

Prove that since is an integer (it’s the size of ), then divides . Conclude that divides and hence that the integer divides for .

For the first part: Let and where . Write . The first term is divisible by by considering geometric series. That means divides which is smaller. Contradiction.

For the second part: Recall that is the cyclotomic polynomial with roots and . This excludes . So (whose roots are all powers of ) is divisible by .

Prove that step 3 and 4 imply that divides . Prove that (complex absolute value) for any root of unity . Conclude that , i.e. that is a field, i.e. commutative.

Look at the class equation from step 3. divides LHS as well as every term in the summation. Hence it divides .

Since is some integer, it lies on the real line and its closest point on unit circle is 1. The inequality is clear. So cannot divide unless .