Let be a primitive th root of unity and let . Prove that is a primitive th root of unity.

Certainly . Now if , we have , and so divides . So is the order of , and thus is a primitive th root of unity.

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let be a primitive th root of unity and let . Prove that is a primitive th root of unity.

Certainly . Now if , we have , and so divides . So is the order of , and thus is a primitive th root of unity.

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 .

Let be an ideal in the ring of integers of . Find an algebraic integer such that , where .

We saw in the text that the class number of is 2. Note that , as indeed . Let . By our proof of Theorem 10.6, .

Let be the ring of integers in an algebraic number field of class number . Let be an ideal. Show that if is minimal such that is principal, then .

This is precisely the order of in the class group of . The result then follows by Lagrange’s Theorem.

Let and be ideals in a commutative unital ring . Show that if and then . Characterize the ideals which divide .

If , then for some . So . Likewise, if then . Thus if and , then .

Suppose is an ideal such that for some ideal . Now , so that . That is, the only unit in the semigroup of ideals in under ideal multiplication is itself.

Let be the ring of integers in an algebraic number field , and let . Prove that if , then in .

Recall that means that for some ideal . Now , so that , and thus for some . Thus in .

Prove that every algebraic integer divides a rational integer. (Here we mean divisibility in the extended sense that if is an algebraic integer, not necessarily in a given field.)

Let be an algebraic integer with conjugates . (Recall that the are also algebraic integers.) Let denote the norm of .

By Theorem 7.1 in TAN, is a rational integer. By definition, .

Let be an algebraic extension of , and let , where the are the conjugates of . Let , and denote by and the th conjugate of and (respectively). Show that if in , then in .

Suppose is an algebraic integer in . In particular, , so that for some polynomial .

Now is a root of some irreducible monic polynomial with rational integer coefficients. That is, . In particular, is a root of , so that the conjugates are also roots of . So for each , and thus is a root of a monic polynomial with rational integer coefficients. So is an algebraic integer in (in fact in ).

Let be an algebraic extension of and let . Recall that if are algebraic integers, we say that if the quotient is an algebraic integer. Prove that in if and only if in .

If in , then is a rational integer. Certainly is an algebraic integer; so in .

Now suppose in . Since is an algebraic integer and a rational number, is a rational integer. So in .