Prove that for all .

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

Then as desired.

unnecessary lemmas. very sloppy. handwriting needs improvement.

Prove that for all .

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

Then as desired.

Let be an imperfect field of characteristic . Prove that there exist irreducible inseparable polynomials over . Conclude that there exist inseparable finite extensions of .

Let be an element which is not a th power in , and let be a root of . In particular, . Now consider . Suppose is an irreducible factor of over . Now cannot be linear, since . But any factor of of degree at least 2 has a multiple root (since all the roots are ). So is an irreducible inseparable polynomial over .

Then is an inseparable extension of .

Prove that . Conclude that . Deduce Wilson’s Theorem: if is an odd prime, then mod .

Recall that by definition, and that is merely the root having minimal polynomial . So . Comparing constant coefficients, we have , so that $latex , and hence .

Restrict now to the field with odd. Then , and . Thus mod .

(I’d like to point out that this is a *really* roundabout way to prove Wilson’s Theorem. The easy(ier) way is to note that in , every element but -1 is distinct from its inverse.)

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 .

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 .

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 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.

Let be finite dimensional splitting fields over a field and contained in a field . Show that and are also finite dimensional splitting fields over (and contained in ).

Say and are splitting fields over for the finite sets .

We claim that is a splitting field over for . To see this, note that each polynomial in splits over . Now if is a field over which split, then the polynomials in split over , so that . Likewise , and so . So is a splitting field for over .

Now suppose is irreducible over with a root in . This root is in the splitting field , so splits over by this previous exercise. Likewise, splits over . Thus splits over . Again using this exercise, is a splitting field over .

Let be a finite extension of a field . Prove that is a splitting field over if and only if every irreducible polynomial over with a root in splits over .

We begin with some lemmas.

Lemma 1: Let be a finite extension of , and let be algebraic over . If is a splitting field for a set , then is a splitting field for considered as polynomials over . Proof: Certainly the roots of elements of are in . Now suppose is a splitting field for over ; in particular, contains and the roots of the polynomials in , so . Moreover , so . Hence is a splitting field for over .

Suppose is a splitting field for some (finite) set . Let be irreducible over , and suppose and are roots of with . By Theorem 8 on page 519 of D&F, extends to an isomorphism . Now is the splitting field for , and likewise is the splitting field for over . By Theorem 27 on page 541 of D&F, the isomorphism extends to an isomorphism . We can visualize this using the following diagram.

Since , , and so , and we have . That is, if an irreducible polynomial over has any roots in , then it splits completely over .

Conversely, suppose is finite over and that every irreducible polynomial with a root in has all roots in . Since is a finite extension, it is algebraic (see Theorem 17 on page 526 of D&F). Say , and let be the minimal polynomial of over . By our hypothesis, each splits over . Moreover, any field over which all the split must contain the . So in fact is the splitting field over of the set of polynomials .