Please post carefully solutions of the following ones:

1. Let be a field extension of . By defining scalar multiplication for and by , the multiplication in , show that is an vector space.

2. If is a field extention of , prove that iff .

3. Let be a field extension of , and let . Show that the evaluation map given by is a ring and an vector space homomorphism. (Such a map is called an algebra homomorphism.)

4. Let be a field extension of and let . Then and ,

so is the quotient field of .

5. Show that .

6. Verify the following universal mapping property for polynomial rings:

(a) Let be a ring containing a field . If , show that there is a unique ring homomorphism with for each .

(b) Moreover, suppose that is a ring containing , together with a function , satisfying the following property: For any ring containing and elements , there is a unique ring homomorphism with . Show that is isomorphic to .

7. Let be a ring. If is also an vector space and forall and , then is said to be an algebra. If is an algebra , show that contains an isomorphic copy of . Also show that if is a field extension of , then is an algebra.

8. Let be a finite extension of . For , let be the map from to defined by . Show that is an linear transformation. Also show that is the minimal polynomial of . For which is ?

9. If is an extension of such that is prime, show that there are no intermediate fields between and .

10. If is a field extension of and if such that is odd, show that . Given an example to show that this can be false if the degree of over is even.

11. If is an algebraic extension of and if is a subring of with , show that is a field.

12. Show that and are not isomorphic as fields but are isomorphic as vector spaces over .

13. If and , show that the composite is equal to .

14. If and are field extensions of that are cotained in a common field, show that is a finite extension of iff both and are finite extensions of .

15. If and are field extensions of that are cotained in a common field, show that is algebraic over iff both and are algebraic over .

16. Let be the algebraic closure of in . Prove that .

17. Let be a finite extension of . If and are subfields of containing , show that . If , prove that .

18. Show that .

19. Given an example of field extensions of for which .

20. Given an example of a field extension with but with for any .

21. Let be a root of , where . Show that factors as , where .

22. (a)Let be field , and let . If and , let . Prove that is irreducible over iff is irreducible over for any .

(b)Show that is irreducible over if is a prime.

(Hint: Replace by and use the Eisenstein criterion.)

23. Recall that the characteristic of a ring with identity is the smallest positive integer for which , if such an exists, or else the characteristis is . Let be a ring with identity. Define by , where is the identity of . Show that is a ring homomorphism and that for a unique nonnegative integer , and show that is the characteristis of .

24. For any positive integer , given an example of a ring of characteristis .

25. If is an integral domain, show that either or is prime.

26. Let be a commutative ring with identity. The prime subring of is the intersection of all subrings of . Show that this intersection is a subring of that is contained inside all subrings of . Moreover, show that the prime subring of is equal to , where is the multiplicative identity of .

27. Let be a field. If , show that the prime subring of is isomorphic to the field , and if , then the prime subring is isomorphic to .

28. Let be a field. The prime subfield of is the intersection of all subfileds of . Show that this subfield is the quotient field of the prime subring of , that it is contained inside all subfileds of , and that it is isomorphic to or depending on whether the characteristis of is or .

Those are all problems in Section 1 of GTM 167.

Here is version pdf

http://imo.library.googlepages.com/GTM167_1.pdf

Solution of problem 5:

We’ll prove that: If then .

Because

we have ,

so .(1)

and

therefore ,

so

. (2)

From (1) and (2) we are done!

Bai 2 😀 (anh chi lam may bai de thoi con bai kho phan chu va ku T.Moc)

gia su co so cua la voi thoa ma nen vay nen dpcm

Bac con thieu chieu nguoc lai, nhung no la don gian! 😀

Solution of problem 1: We can check conditions in GTM 31, and we’re done! 😛

Loi giai cua bai 9: Neu co ton tai mot truong nhu the, ki hieu la . The thi , vi la so nguyen to nen hoac , hay boi bai 2 ta co hoac . 😀

Solution of problem 11:

It is easy see that is an integral domain. We need only prove that if then . Because , we have is a root of a polynomial

, therefore .

Solution of problem 16. For every positive integer the polynomial is irreducible over by Eisenstein criterion, therefore , so . Final , , and we’re done.

Solution of problem 24. An example . 😀

P.S: Thanks, Toyo.

Solution of problem 25.

If then . Assume that and , then , so or (because is an integral domain), contradiction with .

Solution of problem 26. Assume that are all subrings of and . Easy see by definition of a subring that, is a subring of . Now, since we have , so . But by is a subring of we have . And we’re done.

Thanks for your help.

Problems solved: 1,2,5,9,11,16,24,25,26.

Solution of problem 10. Since we see that . Therefore, we need only prove that , from hypothesis we obtain with , so , therefore and . An Example .

Note that :

Problem 27 is a trivial corollary of Problem 26! 😀

Solution of problem 13. Easy see that is smallest subfield containing and of , here is an extension field of and , therefore we’re done.

Solution of problem 14.

If is a finite extension of then by we have and are finite extensions of .

Now, assume that and are finite extensions of . Then and are algebraic and finitely generated over , write and , here are algabraic over . From problem 13 we have , but are algabraic over , we’re done!

Solution of problem 15.

If is algebraic over then by we have and are algebraic over .

Now, assume that and are algebraic over . Then we have is algebraic over (because is algebraic over , therefore over ) and is algebraic over therefore we’re done!

Thank you very much!

Now, problems solved: 1,2,5,9,10,11,13,14,15,16,24,25,26,27.

Loi giai cua bai 12.

Chung khong dang cau nhu cac truong vi trong khong co phan tu nao thoa man , nhung trong thi co. Chung dang cau nhu cac khong gian vecto vi chung la cac khong gian vec to co chieu huu han va bang nhau.

P.S: Bac Minh va bac Thinh dau roi? 😀

Now, problems solved: 1,2,5,9,10,11,12,13,14,15,16,24,25,26,27.

Solution of problem 17(Minh and -oo-)

Assume that , then and

, therefore , so

.

If then , so we have identity.

Solution of problem 19 (ZetaX)

Take and any nontrivial finite extensions.