Indicate whether each of the following if-then statements is true, and give a brief explanation to support your answer. If the statement is false, and there is a reasonable addition to the hypotheses that would make it true, please indicate that as well.
If \(f\) is an irreducible 4th degree polynomial with rational coefficients then the roots of \(f\) are constructible numbers.
If \(g\) is an irreducible 6th degree polynomial with rational coefficients then the roots of \(g\) are constructible numbers.
If \(E\) and \(F\) are finite fields and \(F\subset E\) then \(E/F\) is Galois.
If \(E/F\) and \(K/F\) are both Galois extensions then \(EK/F\) is a Galois extension.
If \(E\) and \(F\) are finite fields and \(|F|\) divides \(|E|\) then \(E\) has a subfield isomorphic to \(F\).
Solution
The answers are false, false, true, true, false. We use the quadratic-tower criterion for straightedge-and-compass constructibility [@DF04]. In particular, a constructible algebraic number lies in a finite tower of quadratic extensions of \(\mathbf Q\), and its degree divides \(2^m\) for some \(m\). Finitely many constructible numbers lie in a common such tower: adjoin their quadratic generators successively, noting that each new extension has degree at most \(2\). Their sums and products are therefore constructible as well.
1. Part (a) is false. An explicit counterexample is\[ f(x)=x^4-4x^2+x+1. \]
1.1. Modulo \(2\), this polynomial becomes \(x^4+x+1\). It has no root in\(\mathbf F_2\). The only monic irreducible quadratic over \(\mathbf F_2\) is\(x^2+x+1\), and modulo that quadratic one has \(x^3=1\), so\(x^4+x+1\equiv1\). Thus the quartic has neither a linear nor a quadraticfactor over \(\mathbf F_2\), proving irreducibility there and hence over\(\mathbf Q\) by Gauss’s lemma [@DF04]. It also has four real roots:its values at \(-3,-2,0,1,2\) are respectively \(43,-1,1,-1,3\), giving aroot in each of the four intervening intervals.
1.2. Write its roots as \(\alpha_1,\ldots,\alpha_4\) and put\[ \beta_1=\alpha_1\alpha_2+\alpha_3\alpha_4,\quad \beta_2=\alpha_1\alpha_3+\alpha_2\alpha_4,\quad \beta_3=\alpha_1\alpha_4+\alpha_2\alpha_3. \]If \(e_i\) are the elementary symmetric functions of the \(\alpha_j\),Vieta’s identities give \((e_1,e_2,e_3,e_4)=(0,-4,-1,1)\). Expanding thethree displayed expressions gives\[ \begin{aligned} \sum_i\beta_i&=e_2=-4,\\ \sum_{i<j}\beta_i\beta_j&=e_1e_3-4e_4=-4,\\ \beta_1\beta_2\beta_3&=e_3^2+e_1^2e_4-4e_2e_4=17. \end{aligned} \]Consequently each \(\beta_i\) is a root of\[ h(y)=y^3+4y^2-4y-17. \]Modulo \(3\) this is \(y^3+y^2+2y+1\), whose values at \(0,1,2\) are\(1,2,2\). A cubic without a root over its coefficient field isirreducible. Hence \(h\) is irreducible over \(\mathbf Q\), and each\(\beta_i\) has degree \(3\).
1.3. If all four \(\alpha_i\) were constructible, their products and sumswould make the \(\beta_i\) constructible. Their degree \(3\) contradictsthe quadratic-tower criterion. Thus an irreducible quartic need not haveconstructible roots, even when all its roots are real.
A sufficient additional hypothesis is that the quartic is biquadratic: \(f(x)=a x^4+b x^2+c\) with \(a,b,c\in\mathbf Q\) and \(a\ne0\). Then every root is obtained by the two successive square-root operations \[ x=\pm\sqrt{\frac{-b\pm\sqrt{b^2-4ac}}{2a}}, \] so all roots are constructible (with complex numbers interpreted as points in the plane).
2. Part (b) is false, and in fact no root of an irreducible sexticover \(\mathbf Q\) is constructible.
For every root \(\alpha\), irreducibility gives \([\mathbf Q(\alpha):\mathbf Q]=6\). If \(\alpha\) lay in a quadratic tower of total degree \(2^m\), the tower law would give \(6\mid2^m\), which is impossible. Thus no additional hypothesis compatible with irreducibility and degree \(6\) makes this assertion true.
3. Part (c) is true.
Let \(|F|=q\) and \([E:F]=d\), so \(|E|=q^d\). Every element of \(E\) satisfies \(x^{q^d}=x\): for nonzero elements this follows from the order \(q^d-1\) of the multiplicative group, and it also holds at zero. Hence \(E\) is the splitting field over \(F\) of \(x^{q^d}-x\). The derivative of this polynomial is \(-1\), so its roots are distinct. Thus the extension is normal, and the minimal polynomial of every element divides a separable polynomial and is itself separable. Therefore \(E/F\) is Galois.
4. Part (d) is true, with \(E\) and \(K\) viewed in a common algebraicclosure \(\overline F\).
The compositum \(EK\) is algebraic and separable over \(F\). Indeed, each of its elements belongs to a field generated by finitely many separable elements of \(E\cup K\), and a finite extension generated by separable elements is separable [@DF04]. Every \(F\)-embedding \(\sigma:EK\to\overline F\) restricts to embeddings of \(E\) and \(K\). Their normality gives \(\sigma(E)=E\) and \(\sigma(K)=K\), so \[ \sigma(EK)=\sigma(E)\sigma(K)=EK. \] Thus \(EK/F\) is normal and separable, hence Galois. The same argument applies when \(E/F\) and \(K/F\) have infinite degree.
5. Part (e) is false. For \(|F|=p^a\) and\(|E|=p^b\), the field \(E\) has a subfield isomorphic to \(F\) if and onlyif \(a\mid b\).
5.1. Take \(F=\mathbf F_4\) and \(E=\mathbf F_8\). For explicit models use\(\mathbf F_2[x]/(x^2+x+1)\) and \(\mathbf F_2[y]/(y^3+y+1)\); thepolynomials have no root in \(\mathbf F_2\) and are therefore irreduciblein degrees \(2\) and \(3\). Although \(4\mid8\), a field embedding \(F\to E\)would embed the multiplicative group of order \(3\) into one of order\(7\), contradicting Lagrange’s theorem.
5.2. More generally, an embedding of \(F\) into \(E\) identifies their primefields. The tower law then gives\[ b=[E:\mathbf F_p]=[E:F]\,[F:\mathbf F_p]=[E:F]a, \]so \(a\mid b\) is necessary.
Conversely, suppose \(b=ar\). In characteristic \(p\) the identity \[ X^{p^{ar}}-X=\sum_{j=0}^{r-1}(X^{p^a}-X)^{p^{aj}} \] shows that \(X^{p^a}-X\) divides \(X^{p^b}-X\). The latter splits into distinct linear factors over \(E\), so the former has exactly \(p^a\) distinct roots in \(E\). Its root set is closed under addition, subtraction, multiplication, and inverses of nonzero elements, since it is the fixed set of the \(p^a\)-power Frobenius map. It is therefore a subfield of order \(p^a\). It is isomorphic to \(F\): both are splitting fields of \(X^{p^a}-X\) over \(\mathbf F_p\), and splitting fields are unique up to an isomorphism fixing the coefficient field [@DF04]. Thus the additional exponent-divisibility hypothesis is both necessary and sufficient.