Non-simplicity of \(\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)\) (D&F 14.4.6)

Prove that \(\mathbb{F}_p(x,y)/\mathbb{F}_p(x^p,y^p)\) is not a simple extension by explicitly exhibiting an infinite number of infinite subfields.

Let \(F = \mathbb{F}_p(x^p,y^p)\), and consider the extensions \(F(x^{pn+1}+y)\), where \(n \in \mathbb{Z}^+ \cup \{0\}\). They are degree \(p\) over \(F\) (due to the lone \(y\) term) and so are proper subfields of the degree \(p^2\) extension \(\mathbb{F}_p(x,y)\).

Suppose \(F(x^{pi+1}+y) = F(x^{p(i+k)+1}+y)\) where \(i \ge 0, k > 0\) (denote the field \(K\)). Then \(x^{p(i+k)+1}-x^{pi+1} = x^{pi+1}(x^{pk}-1) \in K\), and dividing by \(x^{pi}(x^{pk}-1)\) (an element of \(F\)) gives \(x \in K\). In particular, \(x^{pi+1} \in K\) so \(y \in K\), a clear contradiction. Hence the extensions \(F(x^{pn+1}+y)\) are all distinct, and there are infinitely many of them.

Galois group of \(\mathbb{Q}(\sqrt{2+\sqrt{2}})\) (D&F 14.2.14)

Show that \(\mathbb{Q}(\sqrt{2+\sqrt{2}})\) is a cyclic quartic field.

Let \(a_+ = \sqrt{2+\sqrt{2}}\) and \(a_- = \sqrt{2-\sqrt{2}}\). It can be seen that \(a_+\) is a root of the minimal polynomial \(p(x) = (x^2-2)^2-2 = x^4-4x^2+2\) (being irreducible by Eisenstein), and solving for all roots of \(p(x)\) gives \(\pm a_+, \pm a_-\). Also note that since \({a_+}^2-2 = \sqrt{2}\) and \({a_+}{a_-} = \sqrt{2}\ (*)\), it follows that \(a_- \in \mathbb{Q}(a_+)\), and so \(\mathbb{Q}(a_+)\) contains all roots of \(p(x)\). Since \(\mathbb{Q}(a_+)\) has degree 4 over \(\mathbb{Q}\), it is in fact the splitting field of \(p(x)\) and is thus Galois.

We need only find 4 automorphisms in \(\text{Gal}(\mathbb{Q}(a_+)/\mathbb{Q})\) and show that they form a cyclic group. Note that the automorphisms are fully determined by their action on \(a_+\) (due to the starred expression above). Let \(\varphi\) be such an automorphism; then \(\varphi(a_+)\varphi(a_-) = \varphi(a_+)^2-2\), so \(\varphi(a_-) = (\varphi(a_+)^2-2)/\varphi(a_+)\). If \(\varphi(a_+) = a_+\), then clearly \(\varphi(a_-) = a_-\). If \(\varphi(a_+) = a_-\), then \(\varphi(a_-) = -a_-\). If \(\varphi(a_+) = a_-\), then a manual calculation gives \(\varphi(a_-) = -a_+\). Lastly if \(\varphi(a_+) = -a_-\), then we similarly get \(\varphi(a_-) = a_+\).

Associate \(a_+, -a_+, a_-, -a_-\) with the numbers \(1,2,3,4\) in order. Then the automorphisms listed above correspond to the permutations 1, \((1\ 2)(3\ 4)\), \((1\ 3\ 2\ 4)\) and \((1\ 4\ 2\ 3)\). It can be checked that \((1\ 3\ 2\ 4)\) generates this set and thus the conclusion follows.

Galois group of \(x^4-14x^2+9 \in \mathbb{Q}[x]\) (D&F 14.2.12)

Determine the Galois group of the splitting field over \(\mathbb{Q}\) of \(x^4-14x^2+9\).

We can solve for the roots explicitly: they are \(\pm\sqrt{7 \pm \sqrt{40}}\). For convenience, let \(a_+ = \sqrt{7+\sqrt{40}}\) and \(a_- = \sqrt{7-\sqrt{40}}\), so that the roots are \(\pm a_+, \pm a_-\). Since \(x^4-14x^2+9\) is separable, the splitting field is Galois over \(\mathbb{Q}\), i.e. there are exactly 4 automorphisms fixing \(\mathbb{Q}\).

Observe that for any automorphism \(\varphi\), we have \(\varphi(a_+)\varphi(a_-) = \varphi(\sqrt{49-40}) = 3\). Hence the action of \(\varphi\) on \(a_+\) determines its action on \(a_-\), and hence the remaining roots. In particular, the automorphisms induced by mapping \(a_+\) to each of the 4 roots exhaust the Galois group of the splitting field. It can then be verified that the group of automorphisms is isomorphic to \(V_4\).

Simple extensions in splitting fields with Galois group \(S_4\) (D&F 14.2.11)

Suppose \(f(x) \in \mathbb{Z}[x]\) is an irreducible quartic whose splitting field has Galois group \(S_4\) over \(\mathbb{Q}\). Let \(\theta\) be a root of \(f(x)\) and set \(K = Q(\theta)\). Prove that \(K\) is an extension of \(\mathbb{Q}\) of degree 4 which has no proper subfields.

Let the splitting field be \(L\). We have \(\text{Gal}(L/K) \cong S_3\) as one root is fixed, but the other 3 roots can be freely permuted. If \(M \subset K\) is a proper subfield, it must have degree 2 and hence \(\text{Gal}(L/M) \cong A_4\), since \(A_4\) is the unique index 2 subgroup in \(S_4\). However, we show that this leads to a contradiction as \(A_4\) does not contain an isomorphic copy. Simply, note that there are 3 elements of order 2 in both \(S_3\) and \(A_4\), hence any embedding of \(S_3\) and \(A_4\) gives a one-to-one correspondence between these. But only the order 2 elements of \(A_4\), together with the identity, form a subgroup (namely \(Z_4\)), meaning there is no such embedding.

Automorphisms of a rational function field (D&F 14.1.8)

Prove that the automorphisms of the rational function field \(k(t)\) which fix \(k\) are precisely the fractional linear transformations determined by \(t \rightarrow \frac{at+b}{ct+d}\) for \(a,b,c,d \in k\), \(ad-bc \neq 0\).

It is easy to verify that fractional linear transformations are automorphisms of \(k(t)\) fixing \(k\). Conversely, let \(\varphi \in \text{Aut}(k(t)/k)\). Then \(k(\varphi(t)) = \varphi(k(t)) = k(t)\) (1st equality follows from the definition of a homomorphism, 2nd equality follows from surjectivity), so that \([k(t) : k(\varphi(t))] = 1\). If \(\varphi(t) = p(t)/q(t)\) for \(p(t),q(t) \in k[t]\), where \(p(t),q(t)\) are relatively prime and \(q(t) \neq 0\), then by 13.2.18 the LHS is equal to \(\max (\deg p, \deg q)\). Thus \(p(t) = at+b\) and \(q(t) =  ct+d\) for \(a,b,c,d \in k\). Lastly, if \(ad-bc = 0\), then \(a/c = x = b/d\) for some \(x \in k\) and so \(p(t) = xq(t)\), contradicting the fact that they are relatively prime. Notice in particular that we cannot have \(a = c = 0\).

Algebraic extensions: Degrees (part 2)

We start by mentioning the following result: If a field extension \(K/F\) has degree \(n\), then any \(\alpha \in K\) satisfies a polynomial with degree \(\le n\). Observe that if \(K = F(\alpha)\) then equality always holds, but this need not always be the case. For example, the extension \(\mathbb{Q}(\sqrt[4]{2})/\mathbb{Q}\) is of degree 4, but the element \(\sqrt{2}\) satisfies a degree 2 (< 4) polynomial.

The next major result is that \(F \subseteq K \subseteq L\) then \([L:F] = [L:K][K:F]\). Intuitively, a basis for the vector space of \(L\) over \(F\) is \(\alpha_i \beta_j\), where \(\alpha_i\) and \(\beta_j\) run through bases of \(L\) over \(K\) and \(K\) over \(F\) respectively (one can check this by manual expansion). A special case is where \(K = F(\alpha)\) and \(L = F(\alpha,\beta) = F(\alpha)(\beta)\). The degree \(d\) of \(\beta\) over \(F(\alpha)\) is \(\le\) its degree over \(F\), and an \(F\)-basis for \(L\) is \(\alpha^i \beta^j\), where \(1 \le i \le [K:F]\) and \(1 \le j \le d\).

This leads to the following result: \(K/F\) is finite iff it is generated by finitely many algebraic elements \(\alpha_i\), and its degree is \(\le \prod_{i} \text{deg}\ \alpha_i\). This is useful in proving that the set of algebraic elements of \(K/F\) form a subfield (in particular an algebraic extension of \(F\)). For instance, the set of algebraic elements over \(\mathbb{Q}\) in \(\mathbb{C}\) form an infinite extension of \(\mathbb{Q}\).