Let be a field extension with prime degree . Show that any subfield of containing is either or .

Let . Then by Theorem 14 in D&F. Since the degree of a (finite) field extension is an integer, either (so ) or (so ).

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let be a field extension with prime degree . Show that any subfield of containing is either or .

Let . Then by Theorem 14 in D&F. Since the degree of a (finite) field extension is an integer, either (so ) or (so ).

Let be a principal ideal domain, let be nonzero, and let . Given a prime , say , where does not divide . Prove that is module-isomorphic to if and to if .

We begin with some lemmas.

Lemma 1: Let be ideals of a ring , and consider as an -module. Then . Proof: If , then . If , then . Thus .

Lemma 2: Let be a principal ideal domain and let be nonzero with . Note that . Prove that . Proof: Let be given by . (This is well-defined since is a domain, and is clearly an -module homomorphism.) Certainly is surjective. Now if , then , so that . Conversely, if , then for some , and so . By the First Isomorphism Theorem, .

Lemma 3: Let be a principal ideal domain and such that . If , then . Proof: Say . Note that for some , so that . Now , so that .

If , then , so that . By Lemma 1, we have by the Third Isomorphism Theorem. Using Lemma 2, this is isomorphic to .

If with (say) , then , using Lemma 3 and the Third Isomorphism Theorem.

Let be a principal ideal domain, let be a torsion (left, unital) -module, and let be prime. Prove that if for some nonzero , then .

[Note: I’m not sure why D&F assume that is torsion. This proof doesn’t use that fact as far as I can tell. Read with caution.]

Recall that is an ideal (by this previous exercise). Since is a principal ideal domain, we have for some .

Consider the cyclic submodule generated by . Again, is an ideal, so for some . By our assumption, , so that for some . Since is prime, either or . If , then we have for some . Now , so that ; in particular, is a unit, and so . But then , a contradiction. So , and thus .

Since , we have , and so , as desired.

Let and let be the ring of integers in . Factor in , where is a ramified rational prime.

By Theorem 9.6, if does not divide the discriminant of (using Theorem 6.11), then is not ramified. Now , so that if is ramified in , it is either 2 or 5.

We claim that . Indeed, the direction is clear, and we have . We claim also that is maximal. To this end, let , and say mod 2 where . Evidently, mod ; if mod 2 then mod , and if mod 2 then mod . Now suppose ; then . Comparing coefficients mod 2, we have mod 2, a contradiction. So , and thus is a field. Hence is maximal, and is the prime factorization of .

Certianly . We claim that is prime. To see this, note that . If , then mod , where . Suppose . Then for some . Comparing coefficients, we have mod 5. In particular, mod , where are distinct. Thus is a field, so that is maximal. Thus is the prime factorization of .

Let be a quadratic extension of and let be a rational prime. Prove that, as an algebraic integer in , has at most two irreducible factors.

Recall that the conjugates of for are itself with multiplicity 2. So the norm of over is . Since (by Lemma 7.1) the norm of an algebraic integer is a rational integer and the norm is multiplicative, has at most two irreducible integer factors in .

Prove that every prime integer congruent to 1 mod 4 is a sum of two squares. Show that every product of two such primes is also a sum of two squares.

Let be such a prime. We know that is not irreducible in ; thus there exist nonunits such that . Note that neither of and can have norm 1. Since , letting , we see that as desired.

Now suppose are integer primes congruent to 1 mod 4. As above, we have and , where and have norm and as needed. Now . In particular, if , then .

Let be irreducible which is not real and not an associate of . Prove that if and , then .

Since , we have . Note that conjugation preserves products and has order 2; thus . Since and is irreducible, either or . In the first case, since is not real, by this previous exercise, is an associate of , a contradiction. So . Thus divides .

Let be an irreducible Gaussian integer with . Show that if is a factor of its conjugate , then is an associate of .

Let and suppose . This equality yields the two equations and , which can be rearranged as and . Now , so that $latex . Hence .

If , then , so that , a contradiction. Similarly, if then and we have . If , then , so that . If , then , and we have . In either case, . If , then has a nontrivial factorization (namely ) and so is not irreducible. Thus , and so . These are precisely the associates of .

Prove that if is irreducible in , then so is .

Note that conjugation preserves multiplication in ; for all and . As a consequence, if is irreducible then is as well.

Note that . Thus, if is irreducible, then so is .

Let be a natural number. Prove that if , then is prime.

Suppose is composite; say , where the are pairwise coprime and greater than 1. (Say each is the highest power of some prime dividing .) If , each of the is represented in the set . Thus we have . If on the other hand is a prime power, and , then and are in the set , so that . If and , then , and . If , then .

So must be prime.