Let , and let be a prime. Prove that mod .

Remember that mod for all . Say .

Then as desired.

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let , and let be a prime. Prove that mod .

Remember that mod for all . Say .

Then as desired.

Prove that for all .

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

Then as desired.

Prove that divides if and only if divdes .

Suppose ; say .

If , there is nothing to show. Suppose ; then , so that divides .

Now suppose divides . By the Division Algorithm in , we have and such that . Now . If , then (by the uniqueness part of the division algorithm in ) , so and . If , then , and again we have , so .

Let be a commutative ring and let . Recall that the derivative is defined to be . Prove that for all , , , and (for good measure) .

Let and . (For ease of exposition, pad or with zero terms so that they have the same nominal degree.)

Now .

We will now prove the ‘product rule’ in pieces. First, for monic monomials.

Lemma 0: If , then . Proof: .

Lemma 1: For all , we have . Proof: If , then as desired. Similarly for . Now .

Lemma 2: For all and , . Proof: We have as desired.

Lemma 3: For all , . Proof: We have as desired.

Now recall that if , then .

Lemma 4: For all , . Proof: We proceed by induction on . For and , the result is clear. (Note that .) Suppose the result holds for some . Now as desired.

Lemma 5: For all , . Proof: Let , , and , padding to make the degrees nominally . Then .

Lemma 6: For all , . Proof: We have as desired.

Compute the splitting field of over and its degree.

Note that factors as . (Using only the difference of squares and sum of cubes formulas familiar to middle schoolers.) Using the quadratic formula (again familiar to middle schoolers) we see that the roots of are and .

Now if is a 6th root of 1, then is a root of , where denotes the positive real 6th root of 4 (aka the positive cube root of 2). (Middle schoolers could verify that.) Evidently, then, the splitting field of is .

Now has degree 3 over , and has degree 2 over . (Use Eisenstein for both.) By Corollary 22 on page 529 of D&F, then, has degree (as a middle schooler could compute) over .

So we have proved not only that the splitting field of has degree 6 over , but, in a metamathematical twist, also that a middle schooler could prove this as well.

Compute the splitting field of over and its degree.

The roots of exist in (if we don’t know this yet, just assume some roots are in ). Let be such a root; then .

Evidently, . Comparing coefficients, we see that either , , or . If , then , a contradiction in . Likewise, if we get a contradiction. Thus . Substituting, we have , so , and so . There is a unique positive 4th root of , which we denote by ; so and , and hence . There are 4 such roots, and so we have completely factored . (WolframAlpha agrees.)

So the splitting field of is . Note that if and , then , and . Thus .

Note that is a root of , and that the reverse of is . Now is irreducible over the UFD , since it is Eisenstein at the irreducible element . So is irreducible over , the field of fractions of as a consequence of Gauss’ Lemma. As we showed previously, the reverse of , namely , is also irreducible over . So has degree 4 over , and thus has degree 8 over .

Show that is irreducible over . Use the fact (to be proven later) that is a root of to argue that the regular 7-gon is not constructible by straightedge and compass.

Using the rational root theorem, any rational roots of must be either 1 or -1. Certainly then has no rational roots. Since has degree 3, it is irreducible over if and only if it has no roots (in ), so that is irreducible. In particular, the roots of lie in degree 3 extensions of .

Suppose now that the regular 7-gon is constructible. The exterior angles of a regular 7-gon have measure , so in particular, if the regular 7-gon is constructible, then so is the point , so that the number is constructible. But we have seen that any constructible element must have degree a power of 2 over . Given that is a root of , this yields a contradiction.

Let be a field, an extension of of finite degree, and let . Show that if is the matrix of the linear transformation corresponding to ‘multiplication by ‘ (described here) then is a root of the characteristic polynomial of . Use this result to obtain monic polynomials of degree 3 satisfied by and .

Let be the -linear transformation described here. If is the characteristic polynomial of , then we have . On the other hand, , and so since is injective. So is a root of .

Consider the basis of over . Evidently, with respect to this basis, the matrix of is . As we showed previously, the characteristic polynomial of is . So satisfies . (Surprise!)

Similarly, has the matrix . Evidently, the characteristic polynomial of is . We can verify that actually satisfies this polynomial (WolframAlpha agrees.)

Let be an irreducible polynomial of degree over a field , and let be any polynomial in . Prove that every irreducible factor of the composite has degree divisible by .

Let be a root of , and let be the degree of over . Now , and moreover is a root of the irreducible polynomial . So has degree over . Graphically, we have the following diagram of fields.

By Theorem 14 in D&F, .

Let be a field and let be a polynomial over whose roots all lie in . Prove that all matrices over having characteristic polynomial are similar if and only if is squarefree.

Suppose has a repeated factor. Then there are at least two possible lists of elementary divisors whose product is ; in particular there exist two matrices with characteristic polynomial which are not similar. So if all matrices with characteristic polynomial are similar, then has no repeated factors.

Conversely, suppose has no repeated factors. then there is only one possible list of elementary divisors, and so any two matrices having characteristic polynomial are similar.