Brian Bi
$\DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal}$

Miscellaneous problems for chapter 15

Problem 15.M.1 Obviously $$F(\beta) \subseteq F(\alpha)$$. The field $$F(\alpha)$$ is isomorphic to the field $$F(x)$$ of rational functions with coefficients in $$F$$, so write $$\beta = \frac{f(\alpha)}{g(\alpha)}$$ where $$f, g \in F[x]$$. Define $$h \in F(\alpha)[x]$$ such that $$h(x) = \beta g(x) - f(x)$$. Now, since $$g$$ is not the zero polynomial, it has some term with a nonzero coefficient, say, $$a x^k$$, where $$a \in F$$. Then $$\beta g(x)$$ will have a term of the form $$\beta a x^k$$. Since $$\beta \notin F$$, it is also the case that $$\beta a \notin F$$. The term $$\beta a x^k$$ in $$\beta g(x)$$ therefore can't be cancelled by any term in $$f(x)$$, so $$h$$ is not the zero polynomial. Since all of the coefficients of $$h$$ lie in $$F(\beta)$$, and $$h(\alpha) = 0$$, we conclude that $$\alpha$$ is algebraic over $$F(\beta)$$.

Problem 15.M.2 A solution can be found here.

Problem 15.M.3 Let $$L$$ be an extension of $$K$$ that contains a root of $$f$$, call it $$\alpha$$. Let $$M$$ be the subfield of $$L$$ defined by $$M = F(\alpha)$$. The subfield of $$L$$ generated by $$M$$ and $$K$$ together is, of course, $$K(\alpha)$$. Corollary 15.3.8 implies that $$[K(\alpha) : F] \le [K : F][M : F]$$ and that $$[K : F]$$ and $$[M : F]$$ both divide $$[K(\alpha) : F]$$. Now $$[K : F] = 2$$ and $$[M : F] = 6$$, so $$[K(\alpha) : F]$$ is either 6 or 12. If $$[K(\alpha) : F] = 6$$ then $$[K(\alpha) : K] = [K(\alpha) : F]/ [K : F] = 3$$. Likewise if $$[K(\alpha) : F] = 12$$ then $$[K(\alpha) : K] = 6$$. The degree of the irreducible factor of $$f$$ in $$K[x]$$ of which $$\alpha$$ is a root is therefore either 3 or 6. Both possibilities can be realized; for example, $$x^6 - 2$$ is irreducible over $$\Q$$ as well as over $$\Q(i)$$, but over $$\Q(\sqrt{2})$$ it splits into the product of irreducible cubics.

Problem 15.M.4

1. Let $$q = p$$. Partition $$\mathbb{F}_q^\times$$ into equivalence classes of elements that square to the same element. Since a polynomial $$x^2 - a$$ can have at most two roots in $$\mathbb{F}_q$$, each such equivalence class has size at most 2. If $$x^2 = a$$, then $$(-x)^2 = a$$ as well and $$x \ne -x$$ since the characteristic is odd. So in fact each equivalence class has size exactly 2. This implies that the number of such equivalence classes is half the size of $$\mathbb{F}_q^\times$$, so half the elements of $$\mathbb{F}_q^\times$$ are squares.

Fix $$\alpha \in \mathbb{F}_q^\times$$ where $$\alpha$$ is a nonsquare. The function $$f : \mathbb{F}_q^\times \to \mathbb{F}_q^\times$$ with $$f(\beta) = \alpha\beta$$ is a bijection. If $$\beta$$ is a square, then $$\alpha\beta$$ is a nonsquare, because if $$\alpha\beta$$ were a square then $$\alpha = (\alpha\beta)/\beta$$ would be a square too. So the $$(q-1)/2$$ distinct squares in $$\mathbb{F}_q^\times$$ are mapped to $$(q-1)/2$$ distinct nonsquares. This implies that the $$(q-1)/2$$ squares in $$\mathbb{F}_q^\times$$ are the images of at least $$(q-1)/2$$ elements of $$\mathbb{F}_q$$ that are nonsquares. This implies that every nonsquare is mapped to a square, which is what we wanted to prove.

2. In the above argument the only fact we used about $$\mathbb{F}_q$$ is that its characteristic is odd, therefore the above proof generalizes to all finite fields of odd order.

3. Suppose $$F$$ is a finite field of even order and therefore characteristic 2. Suppose $$a \in F$$ has a square root $$b$$. Then $$(x - b)^2 = x^2 - 2b + b^2 = x^2 + a = x^2 - a$$ in $$F[x]$$, so $$a$$ has only one square root in $$F$$. So the function $$x \mapsto x^2$$ is injective on $$F$$, and since $$F$$ is finite, it is also surjective.

4. Suppose $$\varphi : R_1 \to R_2$$ is a homomorphism of rings, and we have elements $$a \in R_1, b \in R_2$$ such that $$b^2 = \phi(a)$$. Then there is a homomorphism of rings $$\varphi' : R_1[x]/(x^2 - a) \to R_2$$ that agrees with $$\varphi$$ on $$R_1$$ and such that $$\varphi'(x) = b$$.

Take the substitution homomorphism $$\Phi : R_1[x] \to R_2$$ that agrees with $$\varphi$$ on $$R_1$$ and such that $$\Phi(x) = b$$. The ideal $$(x^2 - a)$$ is contained in the kernel of $$\Phi$$ since $$\Phi(x^2 - a) = b^2 - \Phi(a) = b^2 - b^2 = 0$$. The fundamental homomorphism theorem implies that $$\Phi$$ factors through the quotient map $$\pi : R_1[x] \to R_1[x]/(x^2 - a)$$, yielding the desired homomorphism $$\phi' : R_1[x]/(x^2 - a) \to R_2$$. (Evidently, the Lemma can be made more general, but for our purposes we will only use this version.)

Let $$f$$ be the minimal polynomial for $$\sqrt{2} + \sqrt{3}$$ over $$\Q$$. We showed in Exercise 15.8.2 that $$\sqrt{2} + \sqrt{3}$$ is a primitive element of $$\Q(\sqrt{2}, \sqrt{3})$$, so $$f$$ has degree 4. In fact $$f(x) = x^4 - 10x^2 + 1$$.

Fix $$p$$ and let $$K = \mathbb{F}_{p^2}$$. Now $$K$$ contains square roots of 2 and 3, which we will denote by $$a$$ and $$b$$, respectively. The homomorphism from $$\Z$$ to $$\mathbb{F}_p$$ can be naturally extended to a homomorphism $$\varphi : \Z \to K$$, and the Lemma allows us to extend this to a homomorphism from $$\Z[\sqrt{2}]$$ to $$K$$ and then again from $$\Z[\sqrt{2}, \sqrt{3}]$$ to $$K$$. Denote this latter homomorphism by $$\varphi'$$, so that $$\varphi'(\sqrt{2}) = a$$ and $$\varphi'(\sqrt{3}) = b$$ where $$a^2 = 2, b^2 = 3$$. Since $$f(\sqrt{2} + \sqrt{3}) = 0$$, and the coefficients of $$f$$ lie in $$\Z$$, applying $$\varphi'$$ to both sides implies that $$f(a + b) = 0$$, where the coefficients of $$f$$ are interpreted modulo $$p$$. However, $$a + b$$ lies in the quadratic field extension $$K/\mathbb{F}_p$$, so its degree over $$\mathbb{F}_p$$ is at most 2, implying that $$f$$ splits over $$\mathbb{F}_p$$.

Remark: In our solution we have used the fact that every element of $$\mathbb{F}_p$$ has a square root in $$\mathbb{F}_{p^2}$$, which is a consequence of Theorem 15.7.3 and was proven in an alternative fashion in Exercise 15.7.9. However, here is an alternative solution that uses the previous parts of this problem in lieu of the facts about finite fields from Section 15.7. If $$p = 2$$, then part (c) implies that instead of taking $$K = \mathbb{F}_{p^2}$$ we can simply take $$K = \mathbb{F}_p$$ itself, so that $$a + b$$ has degree 1, and $$f$$ splits completely modulo $$p$$. Otherwise, $$p$$ is odd, and instead of taking $$K = \mathbb{F}_{p^2}$$, we may take $$K$$ to be produced by two successive extensions, where one adjoins a square root of 2 and one adjoins a square root of 3. If at least one of 2 and 3 is a quadratic residue modulo $$p$$, then one of these extensions can be taken to be trivial, so $$[K : \mathbb{F}_p] \le 2$$. If neither is a quadratic residue, then part (a) implies that 6 is a quadratic residue modulo $$p$$. Say $$c^2 = 6$$ where $$c \in \mathbb{F}_p$$; then the other square root of 6 is just $$-c$$. As $$(ab)^2 = 6$$, this implies that $$ab = \pm c$$. So this implies that $$\mathbb{F}_p(a, b)$$ in fact has degree only 2, so again $$f$$ splits.

Problem 15.M.5(a) In fact a slightly stronger statement holds; we will prove the stated claim for $$GL_2(\Q)$$. Suppose $$M \in GL_2(\Q)$$ has finite order. $$M$$ is similar to some matrix $$J \in GL_2(\C)$$ which is in Jordan form. If $$J$$ is not diagonal, then it takes the form $J = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}$ It is easy to show using induction that $J^n = \begin{pmatrix} \lambda^n & n\lambda^{n-1} \\ 0 & \lambda^n \end{pmatrix}$ so this will never be the identity matrix for any positive integer $$n$$. Since $$M^n$$ is similar to $$J^n$$, this contradicts $$M$$ having finite order. Therefore, $$J$$ is diagonal. Denote the two eigenvalues by $$\lambda_1, \lambda_2 \in \C$$. Each eigenvalue is a root of unity. Also, each has degree at most 2 over $$\Q$$. We found in Exercise 15.3.5 that a primitive $$n$$th root of unity has degree 1 or 2 over $$\Q$$ only when $$n \in \{1, 2, 3, 4, 6\}$$. If $$\lambda_1$$ is a primitive $$n_1$$th root of unity and $$\lambda_2$$ is a primitive $$n_2$$th root of unity, then the order of $$M$$ is $$\lcm(n_1, n_2)$$. As long as neither of $$n_1, n_2$$ is 4, the LCM will be 1, 2, 3, or 6. If $$n_1 = 4$$, then $$\lambda_1 = \pm i$$ and since the characteristic polynomial of $$M$$ is real, this implies $$\lambda_2 = \overline{\lambda_1}$$, so $$n_2 = 4$$ as well, and the order of $$M$$ is 4.

Elements of $$GL_2(\Z)$$ do indeed exist with each of these five orders. It is easy to construct them since we know what we want the eigenvalues to be, we know their minimal polynomials, and this allows us to construct matrices with those polynomials as the characteristic polynomials. Orders 1 and 2 are obvious. For order 3, we would like the characteristic polynomial to be $$\lambda^2 + \lambda + 1$$ so the trace should be $$-1$$ and the determinant $$+1$$. An integer matrix with these properties is $M_3 = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}$ For order 6, the characteristic polynomial should be $$\lambda^2 - \lambda + 1$$ so this time we want the trace to be 1, so we can use $M_6 = \begin{pmatrix} 0 & 1 \\ -1 & 1 \end{pmatrix}$ For order 4, the characteristic polynomial should be $$\lambda^2 + 1$$ so we need the trace to be 0 and the determinant 1: $M_4 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$

Problem 15.M.6

1. The field $$\C(f(t))$$ is the field of rational functions of $$f(t)$$. If $$\C(f(t)) = \C(t)$$, then $$t \in \C(f(t))$$, so there is some rational function $$g$$ with $$g(f(t)) = t$$. This implies that $$f$$ should be injective. A difficulty is that the values $$c \in \C$$ such that $$|f^{-1}(c)| > 1$$ may be points at which $$g$$ is undefined. Let $$S$$ denote the set of points at which $$g$$ is undefined. Since $$S$$ is finite, and $$f$$ is not a constant function, $$f^{-1}(S)$$ is finite. So we may delete the points in $$f^{-1}(S)$$ to yield $$T'$$. Now we can be sure that $$f$$ is injective when restricted to $$T'$$.

We will show that $$f$$ is also surjective, but first, it is important to note that although Artin defines $$T'$$ to deleting a finite set of points may not be sufficient for $$f$$ to be a bijection from $$T'$$ to $$T'$$. For example, if $$f(t) = t/(t+1)$$, then $$f$$ never attains the value 1, so 1 must be deleted from $$T'$$. But $$f$$ is injective, so if $$1 \notin T'$$, then $$f(1) = 1/2 \notin T'$$. Deleting $$1/2$$ from $$T'$$ implies that we must also delete $$f(1/2) = 1/3$$, and so on, so in fact we need to delete the countably infinite set of points $$\{1, 1/2, 1/3, \ldots\}$$. To avoid this difficulty, we will pretend that a bijective map $$T' \to T'$$ really means a bijective map $$T' \to T''$$ where $$T'$$ and $$T''$$ are both complements of finite sets but may be different from each other.

Write $$f(t) = p(t)/q(t)$$ where $$p, q \in \C[t]$$ and $$p, q$$ have no common nonconstant factor. Suppose $$c \in \C$$. We can solve the equation $$f(t) = c$$ by writing it in the form $$p(t) - cq(t) = 0$$. Since $$f$$ is not a constant function, the polynomial $$p - cq$$ is not identically zero, and $$p - cq$$ can only be a constant for at most one value of $$c$$. For all other values, $$p - cq$$ is a nonconstant polynomial, therefore there exists $$t_0 \in \C$$ such that $$p(t_0) - cq(t_0) = 0$$. Note that $$q(t_0)$$ cannot be zero, because if it were, then $$p(t_0)$$ would also be zero and $$t - t_0$$ would be a common factor of $$p(t)$$ and $$q(t)$$. Since $$q(t_0) \ne 0$$, $$t_0$$ also solves the equation $$p(t)/q(t) = c$$. So $$f : T' \to T''$$ will be surjective once we have removed a finite number of points from $$T$$, namely the images of the points we removed from $$T$$ to yield $$T'$$, together with at most one $$c$$ for which $$p - cq$$ is a constant. The map $$f : T' \to T''$$ is then bijective.

2. Suppose $$p, q \in \C[t], q \ne 0, \deg q < \deg p$$, and $$\deg p \ge 2$$. Then there are at most two values of $$c \in \C$$ such that $$p(t) - cq(t)$$ is a perfect power, that is, takes the form $$a(t - r)^n$$, where $$n = \deg p$$ and $$a, r \in \C$$.

By induction on $$n = \deg p$$. For the base case, suppose $$n = 2$$. If $$\deg q = 1$$, then the discriminant of $$p - cq$$ is a quadratic function of $$c$$, so it vanishes for at most two values of $$c$$; these are precisely the values for which $$p(t) - cq(t) = a(t - r)^2$$. Otherwise, if $$\deg q = 0$$, then the discriminant is a linear function of $$c$$ and there is at most one $$c$$ for which it vanishes. Now for the inductive case, first suppose $$\deg q = 0$$. In a perfect power of the form $$a(t - r)^n$$, the leading and second-to-leading coefficients (the coefficients of degree $$n$$ and $$n-1$$) completely determine the other coefficients, so the constant term of $$p - cq$$ will have the right value for $$p - cq$$ to be a perfect power only for at most one $$c$$. If on the other hand $$\deg q \ge 1$$, then the polynomials $$p', q'$$ satisfy the inductive hypothesis and there are at most two values of $$c$$ for which $$p' - cq'$$ is a perfect power. For all other values of $$c$$, if $$p' - cq' = (p - cq)'$$ is not a perfect power, then $$p - cq$$ also cannot be, as the derivative of a perfect power is always a perfect power.

Now let $$f$$ be a rational function, $$f(t) = p(t)/q(t)$$, written in lowest terms. Suppose that $$f$$ is not of the indicated form; either $$\deg p \ge 2$$ or $$\deg q \ge 2$$. We will show that $$f$$ fails to be injective at infinitely many points of its image.

• Case 1: $$\deg p > \deg q$$. The idea is that $$f(t) = p(t)/q(t) = c$$, for some arbitrary $$c \in \C$$, will be satisfied precisely when $$p(t) - cq(t) = 0$$. (Note that the latter equation cannot be satisfied when $$q(t) = 0$$ since that would also imply $$p(t) = 0$$, which would imply that $$f$$ was not written in lowest terms.) This latter equation is polynomial and of degree at least 2, so it will have at least two different roots. We apply the Lemma to ensure that this will indeed be the case infinitely often; only a finite number of $$c$$ yield a polynomial that has only a single root repeated $$\deg p$$ times. For the infinite number of other $$c$$ in the image of $$f$$, the latter fails to be injective there.
• Case 2: $$\deg p < \deg q$$. Let $$g = 1/f$$. Case 1 establishes that $$g$$ fails to be injective at infinitely many points of its image. One of these points might be 0, but there are still infinitely many that aren't. So $$f$$ also fails to be injective at infinitely many points of its image.
• Case 3: $$\deg p = \deg q$$. Write $$p/q = a + r/q$$ where $$\deg r < \deg q$$ and $$a$$ is a constant. Then $$r/q$$ falls into case 2, and fails to be injective at infinitely many points of its image. The same is then true of $$p/q$$ since it is just a shifted version of $$r/q$$.

We have shown that when either the numerator or denominator of $$f$$ has degree greater than 1, $$f$$ fails to be injective at infinitely many points of its image. Part (a) implies that such $$f$$ cannot generate $$\C(t)$$. So now we consider the remaining rational functions, namely those of the form $$f(t) = (at + b)/(ct + d)$$. If $$ad - bc = 0$$, then $$f$$ is actually a constant function equal to either $$a/c$$ or $$b/d$$ (whichever one is defined) so it does not generate $$\C(t)$$. If $$ad - bc \ne 0$$, then $$f$$ has a rational inverse function $$f^{-1}(t) = \frac{-dt + b}{ct - a}$$, so $$t \in \C(f)$$, so $$f$$ generates $$\C(t)$$.

3. Suppose $$\varphi : \C(x) \to \C(x)$$ is an automorphism of fields that is the identity on $$\C$$. The image of $$\varphi$$ is generated by $$\varphi(x)$$, so by part (b), $$\varphi(x)$$ must be a Möbius transformation. Conversely, for each Möbius transformation $$g(x)$$, the field $$\C(g(x))$$ will be the same as $$\C(x)$$ by the result of part (b), and the extensions $$\C(x)/\C$$ and $$\C(g(x))/\C$$ will be isomorphic since $$x$$ and $$g(x)$$ are both transcendental over $$\C$$, so there is indeed an automorphism of $$\C(x)$$ sending $$x$$ to $$g(x)$$ and the constants to themselves. The group of automorphisms of $$\C(x)$$ that fix the constants is therefore the group of Möbius transformations under composition. This is the complex projective linear group of order 2, $$PGL(2, \C)$$.

Problem 15.M.7 The problem does not ask us to show that $$\varphi : SL_2(\Z) \to SL_2(\mathbb{F}_p)$$ is actually a homomorphism, so we will just take it as a given that it is. Let $E_1 = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad E_2 = \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \qquad F = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ where the entries are interpreted modulo $$p$$. Evidently these three matrices lie in the image of $$\varphi$$. It suffices to show that they generate $$SL_2(\mathbb{F}_p)$$. We can use $$E_1$$ and $$E_2$$ to perform a limited form of row reduction in $$SL_2(\mathbb{F}_p)$$ in which either row may be added to the other; the result will also lie in $$SL_2(\mathbb{F}_p)$$. Starting with the matrix $$M \in SL_2(\mathbb{F}_p)$$ with $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ Either $$a$$ or $$c$$ must be nonzero. Suppose first that $$a$$ is nonzero. We can perform successive row operations as follows: $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \rightarrow \begin{pmatrix} a & b \\ 1 & e \end{pmatrix} \rightarrow \begin{pmatrix} 0 & f \\ 1 & e \end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & e \end{pmatrix} \rightarrow \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ In the first step, we add row 1 to row 2 $$a^{-1}(1-c)$$ times in order to make the bottom-left entry 1. In the second step, we add row 2 to row 1 $$-a$$ times to zero out the top-left entry. In the resulting matrix, $$f$$ must be equal to $$-1$$ since the determinant must be 1. Adding the first row to the second row $$e$$ times then gives the matrix $$F$$ as the result.

If $$a$$ is zero then $$c$$ must be nonzero, and we have a very similar reduction sequence $\begin{pmatrix} a & b \\ c & d \end{pmatrix} \rightarrow \begin{pmatrix} 1 & e \\ c & d \end{pmatrix} \rightarrow \begin{pmatrix} 1 & e \\ 0 & f \end{pmatrix} = \begin{pmatrix} 1 & e \\ 0 & 1 \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ ending in the identity matrix, which of course lies in $$\im\varphi$$. So every matrix in $$SL_2(\mathbb{F}_p)$$ can be generated by $$E_1, E_2, F$$, therefore $$\varphi$$ is surjective.

Remark: I don't know of a field-theoretic solution to this problem. Feel free to email me to contribute one.