# 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

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

1. trungtuan says:

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

gia su co so cua $K/F$ la $\{e\}$ voi $\forall k \in K \exists f\in F$ thoa $k = fe$ ma $1_F = f_1.e$ nen $e = f_1^{-1} \in F$ vay nen $k = fe = ff^{-1}_1 \in F$ dpcm

3. trungtuan says:

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

4. trungtuan says:

Solution of problem 1: We can check conditions $A_1,A_2,A_3,A_4$ in GTM 31, and we’re done! 😛

5. trungtuan says:

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 $R=F$. 😀

6. Toyo says:

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 $r^{-1}=-\dfrac{1}{a_n}(a_{n-1}+...+a_0r^{n-1})\in R$.

7. Toyo says:

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.

8. trungtuan says:

Solution of problem 24. An example $\mathbb{Z}_n$. 😀
P.S: Thanks, Toyo.

9. trungtuan says:

Solution of problem 25.
If $\text{char}(R)\not =0$ then $\text{char} (R)=n\in\mathbb{N}-\{1\}$. Assume that $n=pq$ and $p,q>1$, then $(p\cdot 1)\cdot (q\cdot 1)=n\cdot 1=0$, so $p\cdot 1=0$ or $q\cdot 1=0$ (because $R$ is an integral domain), contradiction with $n=\text{char} (R)$.

10. Toyo says:

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.

11. trungtuan says:

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

12. Toyo says:

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

13. Tuan says:

Problem 27 is a trivial corollary of Problem 26! 😀

14. Toyo says:

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.

15. Toyo says:

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!

16. Toyo says:

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!

17. trungtuan says:

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

18. Tuan says:

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? 😀

19. trungtuan says:

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

20. trungtuan says:

Solution of problem 17(Minh and -oo-)
Assume that $\ [L_2:F]=n$ , then $L_2=\sum_{i=1}^nF\cdot e_i$ and
$\ L_1L_2=\sum_{i=1}^nL_1\cdot e_i$, therefore $\ [L_1L_2:L_1] \leq n$, so
$\ [L_1L_2:F]=[L_1L_2:L_1]\cdot \ [L_1:F]\leq n\cdot \ [L_1:F] =\\ \ [L_1:F] \cdot \ [L_2:F]$.

If $\gcd ([L_1:F],[L_2:F])=1$ then $\ [L_1:F]\cdot [L_2:F]|\ [L_1L_2:F]$, so we have identity.$\Box$

21. trungtuan says:

Solution of problem 19 (ZetaX)
Take $L_1 = L_2$ and any nontrivial finite extensions.