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.

Link download

http://imo.library.googlepages.com/GTM167_2.pdf

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

  1. 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.

  2. 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}.

  3. 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.😀

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