# Problems in Section 2 of GTM 167

Please post carefully solutions of following problems:

1. Show that the only automorphism of $\mathbb{Q}$ is the identity.

2. Show that the only automorphism of $\mathbb{R}$ is the identity. (Hint: If $\delta$ is an automorphism, show that $\delta |_{\mathbb{Q}}=\text{id}$ , and if $a>0$, then $\delta (a)>0$. It is an interesting fact that there are infinitely many automorphism of $\mathbb{C}$, even though $[\mathbb{C}:\mathbb{R}]=2$. Why is this fact not a contradiction to this problem? )

3.  Show that the six functions given in Example 2.21 extend to $\mathbb{Q}-$ automorphism of $\mathbb{Q}(\sqrt[3]{2},\omega)$. (Example 2.21: The extension $\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q}$ is Galois, where $\omega =e^{2\pi i/3}$. In fact, the field $\mathbb{Q}(\sqrt[3]{2},\omega)$ is the field generated over $\mathbb{Q}$ by three roots $\sqrt[3]{2},\omega \sqrt[3]{2},$ and $\omega^2 \sqrt[3]{2}$, of $x^3-2$, and since $\omega$ satisfies $x^2+x+1$ over $\mathbb{Q}$ and $\omega$ is not in $\mathbb{Q}(\sqrt[3]{2})$, we see that $[\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]=6$. It can be shown (see Problem 3) that six functions
$\text{id}:\sqrt[3]{2}\to\sqrt[3]{2},\omega\to\omega$
$f_1:\sqrt[3]{2}\to \omega \sqrt[3]{2},\omega\to\omega$
$f_2:\sqrt[3]{2}\to\sqrt[3]{2},\omega\to\omega^2$
$f_3:\sqrt[3]{2}\to \omega \sqrt[3]{2},\omega\to\omega^2$
$f_4:\sqrt[3]{2}\to \omega^2 \sqrt[3]{2},\omega\to\omega$
$f_5:\sqrt[3]{2}\to \omega^2 \sqrt[3]{2},\omega\to\omega^2$
extend to distinct automorphism of $\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q}$. Thus,
$|\text{Gal}(\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q}) |$

$= [\mathbb{Q}(\sqrt[3]{2},\omega):\mathbb{Q}]$
and so $\mathbb{Q}(\sqrt[3]{2},\omega)/\mathbb{Q}$ is Galois.)

4. Let $B$ be an integral domain with quotient field $F$. If $\delta : B\to B$ is a ring automorphism, show that $\delta$ induces a ring automorpism $\delta':F\to F$ defined by $\delta' (\dfrac{a}{b})=\dfrac{\delta (a)}{\delta (b)}$ if $a,b\in B$ with $b\not =0$.

5. Let $K=k(x_1,...,x_n)$ be the field of rational functions in $n$ variables over a field $k$. Show that the definition
$\delta\left(\dfrac{f(x_1,...,x_n)}{g(x_1,...,x_n)}\right)=\dfrac{f(x_{\delta (1)},...,x_{\delta (n)})}{g(x_{\delta (1)},...,x_{\delta (n)})}$
makes a permutation $\delta \in S_n$ into a field automorphism of $K$.
(Hint: The previous problem along with Problem 1.6 may help some.)

6.  Let $F$ be a field of characteristic not $2$, and let $K$ be an extension of $F$ with $[K:F]=2$. Show that $K=F(\sqrt{a})$ for some $a\in F$; that is, show that $K=F(\alpha)$ with $\alpha^2=a\in F$. Moreover, show that $K$ is Galois over $F$.

7. Let $F=\mathbb{F}_2$ and $K=F(\alpha)$, where $\alpha$ is a root of $1+x+x^2$. Show that the function $\delta :K\to K$ given by $\delta (a+b\alpha)=a+b+b\alpha$ for $a,b\in F$ is an $F-$ automorphism of $K$.

8. Suppose that $a\in\mathbb{C}$ is algebraic over $\mathbb{Q}$ with $p(x)=\min (\mathbb{Q},a)$, and let $b$ be any root in $\mathbb{C}$ of $p$. Show that the map $\delta :\mathbb{Q} (a)\to\mathbb{C}$ given by $\delta (f(a))=f(b)$ is a well-defined $\mathbb{Q}-$ homomorphism.

9. Show that the complex numbers $i\sqrt{3}$ and $1+i\sqrt{3}$ are roots of $f(x)=x^4-2x^3+7x^2-6x+12$. Let $K$ be the field generated by $\mathbb{Q}$ and the roots of $f$. Is there an automorphism $\delta$ of $K$ with $\delta (i\sqrt{3})=1+i\sqrt{3}$?

10. Determine whether the following fields are Galois over $\mathbb{Q}$.
(a)$\mathbb{Q}$, where $\omega=\exp (2\pi i/3)$.
(b)$\mathbb{Q}(\sqrt[4]{2})$.
(c)$\mathbb{Q} (\sqrt{5},\sqrt{7})$.(Hint: The previous section has a problem that might be relevant.)

11. Prove or disprove the following assertion and its converse : If $F\subseteq L\subseteq K$ are fields with $K/L$ and $L/F$ Galois, then $K/F$ 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 $S$ is a set , a relation $\leq$ on $S$ is called a partial order on $S$ provided that $a\leq a\forall a\in S$ ; if $a\leq b$ and $b\leq a$ , then $a=b$; and if $a\leq b$ and $b\leq c$, then $a\leq c$. Let $S$ and $T$ be sets with partial orders $\leq_S$ and $\leq_T$, respectively. Suppose that there are functions $f: S\to T$ and $g: T\to S$ such that : (i) if $s_1\leq_Ss_2$, then $f(s_2)\leq_T f(s_1)$, (ii) if $t_1\leq_T t_2$, then $g(t_2)\leq_S g(t_1)$, and (iii) $s\leq_S g(f(s))$ and $t\leq_T g(g(t))$ for all $s\in S$ and $t\in T$. Prove that there is a $1-1$ order reversing correspondence between the image of $g$ and image of $f$, given by $s \mapsto f(s)$, whose inverse is $t\mapsto g(t)$.

13. Let $k$ be a field, and let $K=k(x)$ be the rational function field in one variable over $k$. Let $\sigma$ and $\tau$ be the automorphisms of $K$ defined by $\sigma (f(x)/g(x))=f(1/x)/g(1/x)$ and $\tau (f(x)/g(x))=f(1-x)/g(1-x)$, respectively. Dertemine the fixed field $F$ of $\{\sigma,\tau\}$ and determine $\text{Gal}(K/F)$. Find an $h\in F$ so that $F=k(h)$.

14. Let $k$ be a field, and let $K=k(x)$ be the rational function field in one variable over $k$. If $u\in K$, show that $K=k(u)$ iff $u=(ax+b)/(cx+d)$ for some $a,b,c,d\in k$ with $\det\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\not = 0.$ (Hint: See the example before Proposition 1.15) .

15. Use the previous problem to show that any invertible $2\times 2$ matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$ determines an element of $\text{Gal}(k(x)/k)$ with $x\mapsto (ax+b)/(cx+d)$. Moreover, show that every element of $\text{Gal}(k(x)/k)$ is given by such a formula. Show that the map from the set of invertible $2\times 2$ matrixs over $k$ to $\text{Gal}(k(x)/k)$ given by $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\mapsto \varphi$, where $\varphi (x)=(ax+b)/(cx+d)$, is a group homomorphism. Determine the kernel to show that $\text{Gal}(k(x)/k)\cong \text{PGL}_2(k)$, the group of invertible $2\times 2$ matrices over $k$ modulo the scalar matrices.

16. Let $k=\mathbb{R}$ and let $A$ be the matrix $\left(\begin{matrix}-1/2&-\sqrt{3}/2\\\sqrt{3}/{2}&-1/2\end{matrix}\right)$ given by rotating the plane around the origin by $120^\circ$. Using the previous problem, show that $A$ determines a subgroup of $\text{Gal}(k(x)/k)$ of order $3$. Let $F$ be the fixed field. Show that $k(x)/F$ is Galois, find a $u$ so that $F=k(u)$, find the minimal polynomial $\min (F,x)$, and find all the roots of this polynomial.

17. Let $k=\mathbb{F}_p$ and let $k(x)$ be the rayional function field in one variable over $k$. Define $\varphi : k(x)\to k(x)$ by $\varphi (x)=x+1$. Show that $\varphi$ has finite order in $\text{Gal}(k(x)/k)$. Determine this order, find a $u$ so that $k(u)$ is the fixed field of $\varphi$, determine the minimal polynomial over $k(u)$ of $x$, and find all the roots of this minimal polynomial.

18. Let $k$ be a field of characteristic $p>0$, and let $a\in k$. Let $f(x)=x^p-a^{p-1}x$. Show that $f$ is fixed by the automorphism $\varphi$ of $k(x)$ defined by $\varphi (f(x)/g(x))=f(x+a)/g(x+a)$ for any $f(x),g(x)\in k[x]$. Show that $k(f)$ is the fixed field of $\varphi$.

19. Prove that $(t-x_1)\cdots (t-x_n)=t^n-s_1t^{n-1}+\cdots + (-1)^ns_n$, as we claimed in Example 2.22. ($s_1=x_1+\cdots + x_n, s_2=x_1x_2+x_1x_3+\cdots + x_{n-1}x_n,\cdots,\\ s_n=x_1\cdots x_n$).

Those are all problems in Section 2 of GTM 167.

## 4 thoughts on “Problems in Section 2 of GTM 167”

1. Toyo says:

Solution of problem 19. Induction on $n$.

2. Tuan says:

Solution of problem 1. Assume that $f$ is an automorphism of $\mathbb{Q}$, we have $f(x+y)=f(x)+f(y)\forall x,y\in\mathbb{Q}(1)$ and $f(xy)=f(x)f(y)\forall x,y\in\mathbb{Q}(2)$. By induction and $(1)$ we have $f(n)=nf(1)\forall n\in\mathbb{Z}$, from here $f(1)\not =0$. In $(2)$, set $x=y=1$ then $f(1)=1$ and therefore $f(n)=n\forall n\in\mathbb{Z}(3)$.

So in $(2)$, set $y=x^{-1}$ we have $f(x)f(1/x)=1\forall x\in\mathbb{Q}-\{0\}$. From here and $(3)$ we have $f(1/n)=1/n\forall n\in\mathbb{Z}-\{0\}$.

Final, $f(p/q)=f(p)f(1/q)=p/q\forall p/q\in\mathbb{Q}.$ And we are done.

3. Tuan says:

Solution of problem 2. Assume that $f$ is an automorphism of $\mathbb{R}$, from problem 1, $f(q)=q\forall q\in\mathbb{Q}$. For all $t>0$ : $f(t)=(f(\sqrt{t}))^2>0$.

We have $f$ is increasing on $\mathbb{R}$ In fact, $f(x)=f(y)+f(x-y)>f(y)\forall x>y$.

Now, for every $x\in\mathbb{R}$, assume that $r_n\leq x\leq s_n\forall n,$ and $r_n,s_n\in\mathbb{Q}, \lim r_n=\lim s_n=x$. Therefore $f(r_n)\leq f(x)\leq f(s_n)\forall n$, or $r_n\leq f(x)\leq s_n\forall n$, setting $n\to\infty$ we have $x\leq f(x)\leq x$ and we’re done.

Why is this fact not a contradiction to this problem? Because
$\text{automorphism}\not\rightarrow\mathbb{R}-\text{automorphism}$.

4. trungtuan says:

Solution of problem 3. Routine because form of elements of $\mathbb{Q}(\sqrt[3]{2},\omega)$ .

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