Let be a root of unity. Show that the conjugates of are also roots of unity.

If is a root of 1, then is a root of for some . Thus the minimal polynomial of over is also a root of , so the conjugates of are roots of unity.

unnecessary lemmas. very sloppy. handwriting needs improvement.

Let be a root of unity. Show that the conjugates of are also roots of unity.

If is a root of 1, then is a root of for some . Thus the minimal polynomial of over is also a root of , so the conjugates of are roots of unity.

Let be a quadratic field, and let be an element with . Prove that there exists an algebraic integer such that , where denotes conjugate.

Let . If , then , and either or ; similarly if . So we assume that and . Now . Rearranging, we have . Choose some integers and such that is equal to this common ratio. Let ; certainly is an algebraic integer in . We claim that .

To that end, note that . From , we see that , and likewise since we have . So , and thus .

Let be an algebraic extension of , and let be an algebraic integer. Prove that if is a unit, then the conjugates of are also units.

By Lemma 7.3, the norm is , where (by definition) is the product of the conjugates of . In particular, each conjugate of divides 1 (in the extended sense, meaning that is an algebraic integer). Thus each conjugate of is a unit.

Alternately, if is a unit then for some algebraic integer . Letting , , and , and letting be the conjugates of , then , where and denote the conjugates of and , respectively.

If possible, exhibit and integral basis for whose elements are conjugate for .

Let be a squarefree integer, and let if mod 4 and if mod 4. Recall that the integers in have the form where . Let if mod 4 and if mod 4. Note that if mod 4 and if mod 4 (so that ), that if mod 4 and if mod 4, and that if mod 4 and if mod 4.

Note that . In particular, if then is the minimal polynomial of , so that the conjugates of are itself and .

Recall that the discriminant of is if mod 4 and if mod 4, and that any integral basis will have this discriminant.

Suppose now that is an integral basis of , with and . (Certainly if then cannot be a basis, much less an integral basis; if , then (for example) we cannot generate 1.) Note that . Then we must have

Note that if mod 4 and if mod 4. Evidently, this determinant is , and so we have .

If mod 4, then , and we have . This equation has no solutions in , and thus does not have an integral basis whose elements are conjugate.

If mod 4, then , and we have , so that . This gives a necessary condition on and if we wish to be an integral basis.

Recall that is an integral basis of . If is another integral basis, then there is a (unique) invertible matrix such that . Say , , and ; comparing coefficients, and recalling that (since mod 4), we have . Computing the determinant, we have . Combined with our necessary condition, . If , then , and if , then . Thus our only candidates for are and . Or, in terms of , and . It is easy to see that both and are integral bases.

To summarize, if mod 4, then is an integral basis whose elements are conjugate, where . If mod 4, then no such integral basis exists.

Let be squarefree. Find a basis for where is the conjugate of .

Let . Evidently, is a root of ; the other root is . If , then this polynomial has no rational roots, and so is irreducible over . In this case, is the minimal polynomial of , and so .

Note that if , then cannot be a basis for as it contains only one element. Thus we may assume that .

We claim that is a basis for so long as . To that end, suppose first that is a basis; if , then we have and . Now , with . So , a contradiction since is not rational. Thus .

Conversely, suppose . To show that is a basis, it suffices to show that it generates , as we know that this field has degree 2 over . To achieve this, we will show that and are linear combinations of and . Indeed, evidently we have and , as desired.

In particular, is a basis for over .

Let be an extension of contained in . Suppose is algebraic over with minimal polynomial and such that the complex conjugate of is also in . Prove that and are conjugates in .

Note that complex conjugation is an automorphism of . Thus . Since is a root of the minimal polynomial , and are conjugate in .

Let be a field and let be an algebraic extension of . Prove that the relation defined by if and only if and are conjugate over is an equivalence.

Recall that and are called conjugate precisely when they have the same minimal polynomial over . We can map to the set of all minimal polynomials of elements of by sending each element to its minimal polynomial; call this map . (Note that is well-defined since minimal polynomials are unique.) Now precisely when . Thus it is clear that is an equivalence relation.

Let be a field and let be a nonzero algebraic element over . Prove that the conjugates of are nonzero.

Let be the minimal polynomial of over . Recall that the conjugates of are precisely the roots of . If is linear, then the only conjugate of is itself. Suppose has degree at least 2. If one of the roots of is 0, then in fact ; note that is a root of and that has degree strictly smaller than that of , violating the minimalness of . So 0 is not a conjugate of .

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 .