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 \((K_1 \cap K_2)[x]\).
Friday, 29 May 2020
Thursday, 28 May 2020
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 \(k_i(\alpha\beta) = (\sum_{j} m_{ij}k_j)(\sum_{j} n_{1j}k_j) = \sum_{a,b} m_{ia}k_a(n_{1b}k_b) = \sum_{a,b} m_{ia}n_{ab}k_b\) \(= \sum_b (\sum_{a} m_{ia}n_{ab}) k_b = \sum_{b} (M_{\alpha}M_{\beta})_{ib}k_b\).
(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 \(F\).
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 \(F\).
Wednesday, 27 May 2020
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.
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.
Tuesday, 26 May 2020
Index of a subgroup's normalizer (FGT 1A.10)
Let \(H \subseteq G\).
(a) Show that \(\left|N_G(H) : H\right|\) is equal to the number of left cosets of \(H\) in \(G\) that are invariant under left multiplication of \(H\).
(b) Suppose that \(\left|H\right|\) is a power of the prime \(p\) and that \(\left|G : H\right|\) is divisible by \(p\). Show that \(\left|N_G(H) : H\right|\) is divisible by \(p\).
(a) \(h(aH) = aH\) for all \(h \in H\) \(\iff\) \(a^{-1}ha \in H\) \(\iff\) \(a^{-1}\), hence \(a\) in \(N_G(H)\) \(\iff\) \(aH \subseteq N_G(H)\)
(b) Let \(H\) act on the left cosets of \(H\) by left multiplication. Then the orbit of the coset \(1H\) has size 1. But all orbit sizes must be of some \(p\)-power (by the condition imposed on \(\left|H\right|\)) and sum up to \(\left|G : H\right|\) which is itself a \(p\)-multiple, so the number of size-1 orbits must be divisible by \(p\). The elements in these orbits are precisely the left cosets invariant under left multiplication by \(H\), so by (a) the conclusion follows.
(a) Show that \(\left|N_G(H) : H\right|\) is equal to the number of left cosets of \(H\) in \(G\) that are invariant under left multiplication of \(H\).
(b) Suppose that \(\left|H\right|\) is a power of the prime \(p\) and that \(\left|G : H\right|\) is divisible by \(p\). Show that \(\left|N_G(H) : H\right|\) is divisible by \(p\).
(a) \(h(aH) = aH\) for all \(h \in H\) \(\iff\) \(a^{-1}ha \in H\) \(\iff\) \(a^{-1}\), hence \(a\) in \(N_G(H)\) \(\iff\) \(aH \subseteq N_G(H)\)
(b) Let \(H\) act on the left cosets of \(H\) by left multiplication. Then the orbit of the coset \(1H\) has size 1. But all orbit sizes must be of some \(p\)-power (by the condition imposed on \(\left|H\right|\)) and sum up to \(\left|G : H\right|\) which is itself a \(p\)-multiple, so the number of size-1 orbits must be divisible by \(p\). The elements in these orbits are precisely the left cosets invariant under left multiplication by \(H\), so by (a) the conclusion follows.
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