Please post carefully solutions of following problems:

1. Show that the only automorphism of is the identity.

2. Show that the only automorphism of is the identity. (Hint: If is an automorphism, show that , and if , then . It is an interesting fact that there are infinitely many automorphism of , even though . Why is this fact not a contradiction to this problem? )

3. Show that the six functions given in Example 2.21 extend to automorphism of . (Example 2.21: The extension is Galois, where . In fact, the field is the field generated over by three roots and , of , and since satisfies over and is not in , we see that . It can be shown (see Problem 3) that six functions

extend to distinct automorphism of . Thus,

and so is Galois.)

4. Let be an integral domain with quotient field . If is a ring automorphism, show that induces a ring automorpism defined by if with .

5. Let be the field of rational functions in variables over a field . Show that the definition

makes a permutation into a field automorphism of .

(Hint: The previous problem along with Problem 1.6 may help some.)

6. Let be a field of characteristic not , and let be an extension of with . Show that for some ; that is, show that with . Moreover, show that is Galois over .

7. Let and , where is a root of . Show that the function given by for is an automorphism of .

8. Suppose that is algebraic over with , and let be any root in of . Show that the map given by is a well-defined homomorphism.

9. Show that the complex numbers and are roots of . Let be the field generated by and the roots of . Is there an automorphism of with ?

10. Determine whether the following fields are Galois over .

(a), where .

(b).

(c).(Hint: The previous section has a problem that might be relevant.)

11. Prove or disprove the following assertion and its converse : If are fields with and Galois, then is Galois.

12. Galois connections.

The relationship given in Corollary 2.10 between the set of intermediate fields of a Galois extension and the set of subgroups of its Galois group apprears in other situations, so we study it here. We first need a definition. If is a set , a relation on is called a partial order on provided that ; if and , then ; and if and , then . Let and be sets with partial orders and , respectively. Suppose that there are functions and such that : (i) if , then , (ii) if , then , and (iii) and for all and . Prove that there is a order reversing correspondence between the image of and image of , given by , whose inverse is .

13. Let be a field, and let be the rational function field in one variable over . Let and be the automorphisms of defined by and , respectively. Dertemine the fixed field of and determine . Find an so that .

14. Let be a field, and let be the rational function field in one variable over . If , show that iff for some with (Hint: See the example before Proposition 1.15) .

15. Use the previous problem to show that any invertible matrix determines an element of with . Moreover, show that every element of is given by such a formula. Show that the map from the set of invertible matrixs over to given by , where , is a group homomorphism. Determine the kernel to show that , the group of invertible matrices over modulo the scalar matrices.

16. Let and let be the matrix given by rotating the plane around the origin by . Using the previous problem, show that determines a subgroup of of order . Let be the fixed field. Show that is Galois, find a so that , find the minimal polynomial , and find all the roots of this polynomial.

17. Let and let be the rayional function field in one variable over . Define by . Show that has finite order in . Determine this order, find a so that is the fixed field of , determine the minimal polynomial over of , and find all the roots of this minimal polynomial.

18. Let be a field of characteristic , and let . Let . Show that is fixed by the automorphism of defined by for any . Show that is the fixed field of .

19. Prove that , as we claimed in Example 2.22. ().

Those are all problems in Section 2 of GTM 167.

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Solution of problem 19. Induction on .

Solution of problem 1. Assume that is an automorphism of , we have and . By induction and we have , from here . In , set then and therefore .

So in , set we have . From here and we have .

Final, And we are done.

Solution of problem 2. Assume that is an automorphism of , from problem 1, . For all : .

We have is increasing on In fact, .

Now, for every , assume that and . Therefore , or , setting we have and we’re done.

Why is this fact not a contradiction to this problem? Because

.

Solution of problem 3. Routine because form of elements of .

Solution of problem 4. Trivial by definition of automorphism. 😀