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 \(\mathbb{Q}\).

Product of primitive roots of unity (D&F 13.6.1)

Suppose \(m\) and \(n\) are relatively prime positive integers. Let \(\zeta_m\) and \(\zeta_n\) be primitive \(m\)th and \(n\)th roots of unity. Prove that \(\zeta_m \zeta_n\) is a primitive \(mn\)th root of unity.

This is essentially proving that if \(a\) and \(b\) are integers such that \((a,m) = (b,n) = 1\) then \((an+bm,mn) = 1\) (if the conclusion holds then \((an+bm-kmn,mn) = 1\) for any integer \(k\)). Note that any prime divisor \(d\) of \(an+bm\) cannot appear in the prime factorization of \(m\) or \(n\). For instance if \(d \mid m\) then it follows that \(d \mid an\), and since \((m,n) = 1\) we have \(d \mid a\), a contradiction. Thus the prime factorization of \(an+bm\) is disjoint from that of \(mn\), and the conclusion follows.

Composite and intersection of splitting fields (D&F 13.4.6)

Let \(K_1\) and \(K_2\) be finite extensions of \(F\) contained in the field \(K\), and assume both are splitting fields over \(F\).
(a) Prove that their composite \(K_1K_2\) is a splitting field over \(F\).
(b) Prove that \(K_1 \cap K_2\) is a splitting field over \(F\).

(a) \(K_1\) and \(K_2\) are splitting fields over sets of polynomials \(I\) and \(J\) respectively, whose roots make up sets \(A\) and \(B\). Note that \(K_1 = F(A)\) and \(K_2 = F(B)\). Let \(P\) be the set of all possible products of the polynomials; then \(K_1K_2\) contains \(A\) and \(B\). Furthermore, there cannot be a proper subfield \(F \subseteq K \subset K_1K_2\) with this property. Otherwise, \(K\) would also contain \(A\) and \(B\), and hence \(F(A)\) and \(F(B)\), contradicting the fact that \(K_1K_2\) is the smallest field containing \(F(A)\) and \(F(B)\).

(b) Let \(f(x)\) be an irreducible polynomial in \(F[x]\) with a root \(\alpha \in K_1 \cap K_2\). If \(f\) is linear, we are done. Otherwise, since \(f(x)\) splits completely in \(K_1[x]\) we can write \(f(x) = (x-\alpha)(x-\beta)g(x)\) for \(\beta \in K_1\). Since \(K_2(\alpha) = K_2 \cong K_2(\beta)\), it follows that \(\beta \in K_2\) as well. By induction we can prove similarly for irreducible factors of \(g(x)\), so that \(f(x)\) factors completely in \((K_1 \cap K_2)[x]\).

Elements of field extensions as matrices (D&F 13.2.19)

Let \(K\) be an extension of \(F\) of degree \(n\).

(a) For any \(\alpha \in K\) prove that \(\alpha\) acting by left multiplication on \(K\) is an \(F\)-linear transformation of \(K\).
(b) Prove that \(K\) is isomorphic to a subfield of the ring \(M_n(F)\), so \(M_n(F)\) contains an isomorphic copy of every extension of \(F\) of degree \(\le n\).

(a) Trivial.
(b) Each element \(\alpha \in K\) can be associated with a unique matrix \(M_\alpha \in M_n(F)\) as follows: let \(k_1, \cdots, k_n\) be a basis for \(K\) over \(F\). For each \(k_i\), we can write \(k_i \alpha = m_{i1}k_1 + \cdots m_{in}k_n\), where \(m_{ij} \in F\). Then let \((M_\alpha)_{ij} = m_{ij}\). Now we show that this is an isomorphism.

Let \(\alpha, \beta \in K\), \(m_{ij} = (M_\alpha)_{ij}\) and \(n_{ij} = (M_\beta)_{ij}\). Then \(k_i(\alpha+\beta) = \sum_{j} m_{ij}k_j + \sum_{j} n_{ij}k_j = \sum_{j} (m_{ij}+n_{ij})k_j = \sum_{j} (M_\alpha + M_\beta)_{ij}k_j\), so the map is additive.

Now for simplicity's sake assume that \(k_1 = 1\), so \(\beta = \sum_{j} n_{1j}k_j\) (hopefully it works). Thus \(k_i(\alpha\beta) = (\sum_{j} m_{ij}k_j)(\sum_{j} n_{1j}k_j) = \sum_{a,b} m_{ia}k_a(n_{1b}k_b) = \sum_{a,b} m_{ia}n_{ab}k_b\) \(= \sum_b (\sum_{a} m_{ia}n_{ab}) k_b = \sum_{b} (M_{\alpha}M_{\beta})_{ib}k_b\).

Field \(\subseteq\) ring \(\subseteq\) field (D&F 13.2.16)

Let \(K/F\) be an algebraic extension and let \(R\) be a ring contained in \(K\) and containing \(F\). Show that \(R\) is a subfield of \(K\) containing \(F\).

By virtue of \(R\) being a subring of \(K\), many properties are automatically satisfied, such as the ring axioms, commutativity and 'integral-domain-ness'. The only remaining property to check is closure under inverses.

Let \(\alpha\) be a nonzero element of \(R\). Since \(\alpha\) is algebraic, its degree over \(F\) is finite. Each element of \(F(\alpha)\) is the form \(a_0 + a_1\alpha + \cdots + a_{n-1}\alpha^{n-1}\), where \(n\) is the degree of \(\alpha\) over \(F\). Thus \(F(\alpha) \subseteq R\) and so \(R\) contains \(\alpha^{-1}\) since \(F(\alpha)\) is a field.

Irreducible factors of composite polynomials (D&F 13.2.17)

Let \(f(x) \in F[x]\) be an irreducible degree-\(n\) polynomial, and \(g(x) \in F[x]\). Prove that every irreducible factor of the composite polynomial \(f(g(x))\) has degree divisible by \(n\).

If \(\alpha\) be some root of \(f(x)\), then the field extension \(F(\alpha)/F\) has degree \(n\). If \(\beta\) is a root of \(f(g(x))\), then \(\beta\) must also be a root of the polynomial \(g(x)-\alpha \in F(\alpha)[x]\). Thus the degree of the extension \(F(\alpha,\beta)/F\) is divisible by \(n\).

Now observe the following chain: \(F \subseteq_a F(\beta) \subseteq_b F(\beta)(\alpha)\). Note that \(\beta\) is a root of some irreducible factor \(p(x)\) of \(f(g(x))\), thus \(a = \text{deg}\ p\). Conversely for every irreducible factor \(p(x)\) we can find a root \(\beta\) such that \([F(\beta) : F] = \text{deg}\ p\). Suppose the degree of some irreducible factor \(p(x)\) is not divisible by \(n\). Then \(b\) divides \(n\), since \(ab\) divides \(n\) (as shown in the previous paragraph). Furthermore, \(b\) is the degree of the minimal polynomial of \(\alpha\) in \(F(\beta)[x]\).

The above statement implies that if we can exhibit a degree \(n\) polynomial in \(F(\beta)[x]\) with \(\alpha\) as a root, then it must be a unit-multiple of the minimal polynomial, meaning it is irreducible. But \(f(x)\) is precisely such a polynomial. However, since \(g(\beta) = \alpha\) is a root of \(f(x)\), \(f(x)\) has a linear factor \((x-g(\beta))\), contradicting its irreducibility (it is critical that we are working in \(F(\beta)[x]\) and not just \(F[x]\)). Thus \(b\) cannot divide \(n\) and the conclusion follows.

Condition for a biquadratic extension (D&F 13.2.8)

Let \(F\) be a field of characteristic \(\neq 2\). Let \(D_1\) and \(D_2\) be elements of \(F\), neither of which is a square of \(F\). Prove that \(F(\sqrt{D_1}, \sqrt{D_2})\) is of degree 4 over \(F\) if \(D_1D_2\) is not a square in \(F\) and is of degree 2 over \(F\) otherwise (when \(F(\sqrt{D_1}, \sqrt{D_2})\) is of degree 4 over \(F\) it is called a biquadratic extension of \(F\)).

Remark: note that the condition \(\text{chr}(F) \neq 2\) is required, to exclude e.g. \(\mathbb{F}_2\), all of whose elements are squares.

Let \(F' = F(\sqrt{D_1D_2})\) and \(F'' = F(\sqrt{D_1},\sqrt{D_2})\). Consider the chain \(F \subseteq_{n} F' \subseteq_{m} F''\), where \(n\) and \(m\) denote the respective degrees. Note the second inclusion is valid since each element of \(F'\) can be written in the form \(a+b\sqrt{D_1D_2}\) for \(a,b \in F\). Furthermore, \(m > 1\) because \(\sqrt{D_1}\) cannot be expressed in that form (otherwise \(\sqrt{D_2}^{-1} \in F\) which is a contradiction).

\(D_1D_2\) not being a square in \(F\) means \(n = 2\), forcing \(nm = 4\) as \(nm \le 4\) and \(m > 1\). On the other hand, \(nm \neq 2\) forces \(nm = 4\) (as \(nm = 1\) is not a valid possibility), so \(n = 2\) and \(D_1D_2\) is not a square in \(F\).

Algebraic extensions: The very basics (part 1)

Let \(F\) be a field. The degree of a field extension \(E/F\), denoted \([E : F]\), is the dimension of \(E\) as a vector space over \(F\). For example \([\mathbb{C} : \mathbb{R}] = 2\), where a basis is \(\{1,i\}\). Note that it is not necessarily finite, and finding an explicit basis is not always possible.

There is a particular type of field extension whose aforementioned properties are known. Consider an irreducible polynomial \(p(x) \in F[x]\). Then we can construct a field \(F'\) containing a root \(\alpha\), and an isomorphic copy of \(F\). In particular, we let \(F'\) be the quotient \(F[x]/(p(x))\), whose elements can be represented by polynomials of degree \(< \text{deg}
\ p\).

Note that the condition that \(p(x)\) is irreducible is required for this quotient to be a field. Furthermore, it forces the \(p(x)\) to be of minimal degree. (If irreducible polynomials \(a(x),b(x)\) have the same root \(\alpha\), but \(a(x)\) is of minimal degree while \(b(x)\) is not, we obtain a contradiction from dividing \(b(x)\) by \(a(x)\) and considering the remainder.)

If we impose the further condition that \(p(x)\) be monic, then we obtain the minimal polynomial \(m_{\alpha,F}(x)\) which is unique. In particular, \([F' : F] = \deg m_{\alpha,F}\) (where \(F\) is identified with its isomorphic copy). Also, the element \(\bar{x} \in F'\) is the desired root of \(p(x)\), and \(F'\) has basis \(\{1,\bar{x},\cdots,\bar{x}^{n-1}\}\), where \(n = \deg m_{\alpha,F}\). Furthermore, the image of \(F\) under the projection is precisely the isomorphic copy of \(F\).

As an example, the field \(F = \mathbb{R}[x]/(x^2+1)\) contains a root \(\alpha\) to the polynomial \(x^2+1\). The degree of \(F\) over \(\mathbb{R}\) is 2, and every element of \(F\) can be represented in the form \(a+b\alpha\), where \(a,b \in \mathbb{R}\). \(F\) contains a copy of \(\mathbb{R}\), namely the set of elements with \(b=0\). The rules of complex addition and multiplication can then be derived via usual addition and multiplication of polynomials, mod \(\alpha^2+1\).

If we know a priori some field extension \(E/F\) containing a root \(\alpha\) to \(p(x)\), then we use \(F(\alpha)\) to denote the smallest subfield containing \(F\) and \(\alpha\). One key isomorphism which arises is \(F(\alpha) \cong F[x]/(p(x))\), so it does not matter whether we construct a field containing a root, or simply take an existing one. Taking the abovementioned example, we have \(\mathbb{C} \cong \mathbb{R}(i) \cong \mathbb{R}[x]/(x^2+1)\).

Now, suppose there are 2 distinct roots \(\alpha,\beta\) of \(p(x) \in F[x]\) in some field \(F'\). Then \(F(\alpha) \cong F[x]/(p(x)) \cong F(\beta)\), i.e. the roots are algebraically indistinguishable. Furthermore, there is a canonical isomorphism \(F(\alpha) \rightarrow F(\beta)\) mapping \(a_0+a_1\alpha+\cdots+a_n\alpha^n\) to \(a_0+a_1\beta+\cdots+a_n\beta^n\). For example, the map \(a+bi \rightarrow a-bi\) is an isomorphism.

This isomorphism can be generalized to the case where the fields \(F\) and \(F'\) where \(\alpha\) and \(\beta\) are adjoined to are isomorphic. Let \(\varphi : F \rightarrow F'\) be an isomorphism. Then \(\varphi\) extends naturally to an isomorphism \(F[x] \rightarrow F'[x]\) and thus we have \(F[x]/(p(x)) \cong F'[x]/(p(x))\). So we can conclude that \(F(\alpha) \cong F'(\beta)\). This fact is useful later on in proving the uniqueness of splitting fields up to isomorphism.