Problems in Section 1 of GTM 167


Please post carefully solutions of the following ones:

1. Let K be a field extension of F. By defining scalar multiplication for \alpha\in F and a\in K by \alpha\cdot a=\alpha a, the multiplication in K, show that K is an F- vector space.

2. If K is a field extention of F, prove that [K:F]=1 iff K=F.

3. Let K be a field extension of F, and let a\in K. Show that the evaluation map \text{ev}_a:F[x]\to K given by \text{ev}_a(f(x))=f(a) is a ring and an F- vector space homomorphism. (Such a map is called an F- algebra homomorphism.)

4. Let K be a field extension of F and let a_1,a_2,...,a_n\in K. Then F[a_1,a_2,...,a_n]=\{f(a_1,a_2,...,a_n):f\in F[x_1,x_2,...,x_n]\} and F(a_1,a_2,...,a_n)=\{\dfrac{f(a_1,a_2,...,a_n)}{g(a_1,a_2,...,a_n)}:f,g\in F[x_1,x_2,...,x_n], \\ g(a_1,a_2,...,a_n)\not =0\},

so F(a_1,a_2,...,a_n) is the quotient field of F[a_1,a_2,...,a_n].

5. Show that \mathbb{Q}(\sqrt{5},\sqrt{7})=\mathbb{Q}(\sqrt{5}+\sqrt{7}).

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

(a) Let A be a ring containing a field F. If a_1,a_2,...,a_n\in A, show that there is a unique ring homomorphism \varphi : F[x_1,x_2,...,x_n]\to A with \varphi (x_i)=a_i for each i.

(b) Moreover, suppose that B is a ring containing F, together with a function f:\{x_1,x_2,...,x_n\}\to B, satisfying the following property: For any ring A containing F and elements a_1,a_2,...,a_n\in A, there is a unique ring homomorphism \varphi :B\to A with \varphi (f(x_i))=a_i. Show that B is isomorphic to F[x_1,x_2,...,x_n].

7. Let A be a ring. If A is also an F- vector space and \alpha (ab)=(\alpha a)b=a(\alpha b) forall \alpha\in F and a,b\in A, then A is said to be an F- algebra. If A is an F- algebra , show that A contains an isomorphic copy of F. Also show that if K is a field extension of F, then K is an F- algebra.

8. Let K=F(a) be a finite extension of F. For \alpha\in K, let L_{\alpha} be the map from K to K defined by L_{\alpha}(x)=\alpha x. Show that L_{\alpha} is an F- linear transformation. Also show that \det (xI-L_{\alpha}) is the minimal polynomial \min (F,a) of a. For which \alpha\in K is \det (xI-L_{\alpha})=\min (F,\alpha) ?



9. If K is an extension of F such that [K:F] is prime, show that there are no intermediate fields between K and F.



10. If K is a field extension of F and if a\in K such that [F(a):F] is odd, show that F(a)=F(a^2). Given an example to show that this can be false if the degree of F(a) over F is even.



11. If K is an algebraic extension of F and if R is a subring of K with F\subseteq R\subseteq K, show that R is a field.



12. Show that \mathbb{Q}(\sqrt{2}) and \mathbb{Q}(\sqrt{3}) are not isomorphic as fields but are isomorphic as vector spaces over \mathbb{Q}.



13. If L_1=F(a_1,a_2,...,a_n) and L_2=F(b_1,b_2,...,b_m), show that the composite L_1L_2 is equal to F(a_1,a_2,...,a_n,b_1,b_2,...,b_m).


14. If L_1 and L_2 are field extensions of F that are cotained in a common field, show that L_1L_2 is a finite extension of F iff both L_1 and L_2 are finite extensions of F.


15. If L_1 and L_2 are field extensions of F that are cotained in a common field, show that L_1L_2 is algebraic over F iff both L_1 and L_2 are algebraic over F.


16. Let \mathbb{A} be the algebraic closure of \mathbb{Q} in \mathbb{C}. Prove that [\mathbb{A}:\mathbb{Q}]=\infty.


17. Let K be a finite extension of F. If L_1 and L_2 are subfields of K containing F, show that [L_1L_2:F]\leq [L_1:F]\cdot [L_2:F]. If \gcd ([L_1:F], [L_2:F])=1, prove that [L_1L_2:F]= [L_1:F]\cdot [L_2:F].

18. Show that [\mathbb{Q}(\sqrt[4]{2},\sqrt{3}):\mathbb{Q}]=8.


19. Given an example of field extensions L_1,L_2 of F for which [L_1L_2:F]<[L_1:F]\cdot [L_2:F].


20. Given an example of a field extension K/F with [K:F]=3 but with K\not =F(\sqrt[3]{b}) for any b\in F.


21. Let a\in\mathbb{C} be a root of x^n-b, where b\in\mathbb{C}. Show that x^n-b factors as \prod_{i=0}^{n-1}(x-\omega^ia), where \omega =e^{\dfrac{2\pi i}{n}}.



22. (a)Let F be field , and let f(x)\in F[x]. If f(x)=\sum_{i}a_ix^i and \alpha\in F, let f(x+\alpha)=\sum_{i}a_i(x+\alpha)^i. Prove that f is irreducible over F iff f(x+\alpha) is irreducible over F for any \alpha\in F.

(b)Show that x^{p-1}+...+x+1 is irreducible over \mathbb{Q} if p is a prime.

(Hint: Replace x by x+1 and use the Eisenstein criterion.)


23. Recall that the characteristic of a ring R with identity is the smallest positive integer n for which n\cdot 1=0, if such an n exists, or else the characteristis is \text{0}. Let R be a ring with identity. Define \varphi :\mathbb{Z}\to R by \varphi (n)=n\cdot 1, where 1 is the identity of R. Show that \varphi is a ring homomorphism and that \ker (\varphi)=m\mathbb{Z} for a unique nonnegative integer m, and show that m is the characteristis of R.

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


25. If R is an integral domain, show that either \text{char} (R)=0 or \text{char} (R) is prime.



26. Let R be a commutative ring with identity. The prime subring of R is the intersection of all subrings of R. Show that this intersection is a subring of R that is contained inside all subrings of R. Moreover, show that the prime subring of R is equal to \{n\cdot 1:n\in\mathbb{Z}\}, where 1 is the multiplicative identity of R.



27. Let F be a field. If \text{char} (F)=p>0, show that the prime subring of F is isomorphic to the field \mathbb{F}_p, and if \text{char} (F)=0, then the prime subring is isomorphic to \mathbb{Z}.



28. Let F be a field. The prime subfield of F is the intersection of all subfileds of F. Show that this subfield is the quotient field of the prime subring of F, that it is contained inside all subfileds of F , and that it is isomorphic to \mathbb{F}_p or \mathbb{Q} depending on whether the characteristis of F is p>0 or \text{0}.



Those are all problems in Section 1 of GTM 167.



Here is version pdf

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


21 thoughts on “Problems in Section 1 of GTM 167”

  1. Solution of problem 5:
    We’ll prove that: If n,m\in\mathbb{N} then \mathbb{Q}(\sqrt {n} + \sqrt {m}) = \mathbb{Q}(\sqrt {n},\sqrt {m}).

    Because \sqrt{m},\sqrt{n}\in\mathbb{Q}(\sqrt{m},\sqrt{n})

    we have \sqrt{m}+ \sqrt{n}\in\mathbb{Q}(\sqrt{m},\sqrt{n}),

    so \mathbb{Q}(\sqrt{m} + \sqrt{n})\subseteq\mathbb{Q}(\sqrt{m},\sqrt{n}).(1)

    \sqrt {m} + \sqrt {n}\in \mathbb{Q}(\sqrt {m} + \sqrt{n})

    and \sqrt{m} - \sqrt{n} = \dfrac{m - n}{\sqrt {m} + \sqrt{n}}\in \mathbb{Q}(\sqrt{m} + \sqrt{n})

    therefore \sqrt {m},\sqrt {n}\in \mathbb{Q}(\sqrt{m} + \sqrt{n}),
    so
    \mathbb{Q}(\sqrt{m},\sqrt{n})\subseteq \mathbb{Q}(\sqrt{m} + \sqrt{n}). (2)

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

  2. Loi giai cua bai 9: Neu co ton tai mot truong nhu the, ki hieu la R. The thi [K:F]=[K:R][R:F] , vi [K:F] la so nguyen to nen [K:R]=1 hoac [R:F]=1, hay boi bai 2 ta co K=R hoac R=F.😀

  3. Solution of problem 11:
    It is easy see that R is an integral domain. We need only prove that if r\in R^* then r^{-1}\in R. Because r^{-1}\in K, we have r^{-1} is a root of a polynomial
    a_nx^n+a_{n-1}x^{n-1}+...+a_0\in F[x], therefore r^{-1}=-\dfrac{1}{a_n}(a_{n-1}+...+a_0r^{n-1})\in R.

  4. Solution of problem 16. For every positive integer n the polynomial x^n-2 is irreducible over \mathbb{Q} by Eisenstein criterion, therefore x^n-2=\min (\mathbb{Q},\sqrt[n]{2}), so [\mathbb{Q} (\sqrt[n]{2}):\mathbb{Q}]=n. Final , [\mathbb{A}:\mathbb{Q}]\geq [\mathbb{Q} (\sqrt[n]{2}):\mathbb{Q}]=n\forall n\geq 1, and we’re done.

  5. Solution of problem 26. Assume that a_i,i\in I are all subrings of R and a=\cap_{i\in I}a_i. Easy see by definition of a subring that, a is a subring of R. Now, since 1\in a_i\forall i we have 1\in a, so \{n\cdot 1:n\in\mathbb{Z}\}\subseteq a. But by \{n\cdot 1:n\in\mathbb{Z}\} is a subring of R we have a\subseteq \{n\cdot 1:n\in\mathbb{Z}\}. And we’re done.

  6. Solution of problem 10. Since a^2\in F(a) we see that F(a^2)\subseteq F(a). Therefore, we need only prove that F(a)\subseteq F(a^2), from hypothesis we obtain a^{2n+1}+b_{2n}a^{2n}+...+b_1a+b_0=0 with b_i\in F, so a(a^{2n}+b_{2n-1}a^{2n-2}+...)=-(b_{2n}a^{2n}+b_{2n-2}a^{2n-2}+...), therefore a\in F(a^2) and F(a)\subseteq F(a^2). An Example K=\mathbb{R},F=\mathbb{Q},a=\sqrt{2}.

    Note that :\min (F,a)=x^{2n+1}+b_{2n}x^{2n}+...

  7. Solution of problem 13. Easy see that F(a_1,...,a_n,b_1,...,b_m) is smallest subfield containing L_1 and L_2 of K, here K is an extension field of L_1 and L_2, therefore we’re done.

  8. Solution of problem 14.
    If L_1L_2 is a finite extension of F then by F\subseteq L_1,L_2\subseteq L_1L_2 we have L_1 and L_2 are finite extensions of F.

    Now, assume that L_1 and L_2 are finite extensions of F. Then L_1 and L_2 are algebraic and finitely generated over F, write L_1=F(a_1,...,a_n) and L_2=F(b_1,...,b_m), here a_1,...,a_n,b_1,...,b_m are algabraic over F. From problem 13 we have L_1L_2=F(a_1,...,a_n,b_1,...,b_m), but a_1,...,a_n,b_1,...,b_m are algabraic over F, we’re done!

  9. Solution of problem 15.
    If L_1L_2 is algebraic over F then by F\subseteq L_1,L_2\subseteq L_1L_2 we have L_1 and L_2 are algebraic over F.

    Now, assume that L_1 and L_2 are algebraic over F. Then we have L_1L_2=L_1(L_2) is algebraic over L_1 (because L_2 is algebraic over F, therefore over L_1) and L_1 is algebraic over F therefore we’re done!

  10. Loi giai cua bai 12.
    Chung khong dang cau nhu cac truong vi trong \mathbb{Q} (\sqrt{3}) khong co phan tu x nao thoa man x^2=2, nhung trong \mathbb{Q} (\sqrt{2}) thi co. Chung dang cau nhu cac \mathbb{Q}- khong gian vecto vi chung la cac \mathbb{Q}- khong gian vec to co chieu huu han va bang nhau.

    P.S: Bac Minh va bac Thinh dau roi?😀

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s