Show that mod .

Note that by the binomial theorem. The coefficient of is . Over , we have , and now the coefficient of is . In particular, we have mod .

unnecessary lemmas. very sloppy. handwriting needs improvement.

Show that mod .

Note that by the binomial theorem. The coefficient of is . Over , we have , and now the coefficient of is . In particular, we have mod .

Prove that for all .

Let . Remember that the elements of are precisely the roots of ; in particular, for all .

Then as desired.

Prove that for any prime and any nonzero , is irreducible and separable over .

Note that , so that and are relatively prime. So is separable.

Now let be a root of . Using the Frobenius endomorphism, , so that is also a root. By induction, is a root for all , and since has degree , these are all of the roots.

Now , and in particular the minimal polynomials of and have the same degree over – say . Since is the product of the minimal polynomials of its roots, we have for some . Since is prime, we have either (so that , a contradiction) or , so that itself is the minimal polynommial of , hence is irreducible.

Fix an integer . Prove that for all , divides if and only if divides . Conclude that if and only if .

If , then by this previous exercise, divides in , and so divides in .

Conversely, suppose divides . Using the Division Algorithm, say . Now . If , then we have , so that and . If , then , and again we have , so that as desired.

Recall that is precisely the roots (in some splitting field) of . Now if and only if divides by the above argument, if and only if divides by a previous exercise, if and only if divides , if and only if .

Find all the irreducible polynomials of degree 1, 2, or 4 over , and verify that their product is .

Note that there are (monic) polynomials of degree , as each non-leading coefficient can be either 0 or 1.

The polynomials of degree 1, and , are both irreducible.

There are 4 polynomials of degree 2, one of which is irreducible.

- is reducible
- is reducible
- is reducible
- has no roots, and so is irreducible.

Before we address the degree 4 polynomials, we prove a lemma.

Lemma 1: If is a degree 4 polynomial over with constant term 1 and factors as a product of quadratics, then the linear and cubic terms of are equal. Proof: We have (as we assume) . Now , so that . So , as desired.

Lemma 2: is irreducible over . Proof: Clearly this has no roots. By Lemma 1, if factors into two quadratics, then the factors’ linear terms, and , satisfy , which is impossible over .

There are 16 polynomials of degree 4.

- is reducible
- is reducible
- is reducible
- is reducible
- is reducible
- clearly has no roots, and by the lemma, has no quadratic factors. So is irreducible.
- is reducible
- clearly has no roots, and by Lemma 1, has no quadratic factors. So is irreducible.
- is reducible
- is reducible
- is reducible
- has 1 as a root, so is reducible
- is reducible
- has 1 as a root, so is reducible
- is reducible
- is irreducible

So there are six irreducible polynomials of degree 1, 2, or 4 over :

It is easy (if tedious) to verify that the product of these polynomials is . (WolframAlpha agrees.)

Let and . Construct finite fields of order 4, 8, 9, and 27 elements. For the fields with 4 and 9 elements, give the multiplication tables and show that the nonzero elements form a cyclic group.

Note that to construct fields of the given sizes, it suffices to show that and are each irreducible over the fields and . (More precisely, that these polynomials are irreducible when we inject the coefficients into the respective fields.)

It is easy to see that neither nor has a root in either or in , so these are indeed irreducible. Then , , , and are fields of order 4, 8, 9, and 27 by Proposition 11 in D&F, and specifically are isomorphic as fields to the extension of or by adjoining a root of or .

The multiplication table of is as follows.

Evidently, is generated by .

The multiplication table of is as follows.

Evidently, is generated by .

Let be a finite field of characteristic . Prove that for some .

Suppose has a nonzero element of additive order . That is, is minimal such that . Note that , since annihilates every element of (being the characteristic of ). If , then by the division algorithm in we have for some and with . If , then , violating the minimalness of . If , then . Since is prime, either (so ) or (so ), which both give a contradiction. Thus .

So every element of has additive order , and thus by Cauchy’s theorem is a power of .

Let be a prime. Compute the Jordan canonical form over of the matrix whose diagonal entries are 0 and whose off-diagonal entries are 1. ()

Let . The computation we carried out here shows that over (indeed, over any ring with 1). So the minimal polynomial of divides . Note that and are nonzero (any off-diagonal entry is 1), so the minimal polynomial of is precisely .

Let be an -module via as usual. Now the invariant factors of divide , and we have .

We claim that has dimension 1 over . Indeed, if , then .

Using Lemmas 2 and 3 from this previous exercise, we have . Note that any mapping which realizes an isomorphism as -modules is necessarily also an -vector space isomorphism; since has dimension 1, the right hand side of this isomorphism also has dimension 1 over . Now one of the left summands is , corresponding to the minimal polynomial. Thus the remaining summands must be trivial; that is, there are no invariant factors which are not divisible by , and every invariant factor other than the minimal polynomial is exactly .

So the elementary divisors of are ( times) and . The corresponding Jordan canonical form is where if , if , and otherwise.

Let be a prime. Compute the Jordan canonical form over the finite field of the matrix whose every entry is 1. (.)

Let . Now the computation in this previous exercise holds over (indeed, over any ring with 1), and so we have .

If is not divisible by , then is a unit in . If is an eigenvector with eigenvalue , then evidently (as we saw in the exercise referenced above) . So the eigenspace of has dimension 1, the characteristic polynomial of is , and the Jordan canonical form of is where and otherwise.

Suppose divides . Since , the minimal polynomial of is now . We claim that the remaining invariant factors of are . There is probably a slick linear-algebraic way to argue this, but I don’t see it. Instead I will show explicitly that is similar to a matrix in Jordan canonical form. (If anyone can offer a simplification of this proof I’d be grateful!)

If , then the minimal polynomial of is the only invariant factor, so the Jordan form of is . Suppose .

Let have dimension , and let be the matrix whose every entry is 1. Note that and . Let and . Evidently , so that . Moreover, we have , and .

Now let have dimension . Let and . Evidently, , so that . Now , and we have . If we squint just right, this matrix is in Jordan canonical form, and is similar to .

Let be a prime. Show that the following matrices are similar in : (the companion matrix of ) and (the Jordan block with eigenvalue 1).

Recall that the characteristic polynomial of is (as we showed here), and that over we have (as can be deduced from this precious exercise).

We claim that in fact is the minimal polynomial of . Recall (from this previous exercise) that (the symmetric group on objects) is embedded in by letting a permutation act index-wise on the elements of an ordered basis (i.e. ). Now is in the image of this representation, and in fact is the matrix representing the permutation whose cycle decomposition is . In particular, is not the identity transformation, so the minimal polynomial of is indeed .

So the elementary divisors of are just , and so is similar to the Jordan block .