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.
Sunday, 31 May 2020
Friday, 29 May 2020
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 \((K_1 \cap K_2)[x]\).
(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 \((K_1 \cap K_2)[x]\).
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.
Friday, 22 May 2020
The ring \(\mathbb{Q}[x,y,z]/(xy-z^2)\) (D&F 9.1.12)
Let bars denote passage to \(\mathbb{Q}[x,y,z]/(xy-z^2)\). Prove that \(\bar{P} = (\bar{x}, \bar{z})\) is a prime ideal. Show that \(\bar{xy} \in \bar{P}^2\) but that no power of \(\bar{y}\) lies in \(\bar{P}\). (This shows \(\bar{P}\) is a prime ideal whose square is not a primary ideal.)
Let \(I = (xy-z^2)\). For any coset \(r+I\) (where \(r\) is any representative), define \(\varphi(r+I)\) as \(r\) but with all terms containing \(x\) and \(z\) removed. Note that this map is well defined as for any \(k \in \mathbb{Q}[x,y,z]\), we have \(\varphi(r+k(xy-z^2)) = \varphi(r)\). Furthermore, it can be seen that \(\varphi\) is a ring homomorphism to \(\mathbb{Q}[y]\), an integral domain. Thus \((x+I, z+I) = \ker \varphi\) is a prime ideal.
As for the second part, \(z^2+I\) = \((z^2+xy-z^2)+I\) = \(xy+I\). The latter statement is evident.
Let \(I = (xy-z^2)\). For any coset \(r+I\) (where \(r\) is any representative), define \(\varphi(r+I)\) as \(r\) but with all terms containing \(x\) and \(z\) removed. Note that this map is well defined as for any \(k \in \mathbb{Q}[x,y,z]\), we have \(\varphi(r+k(xy-z^2)) = \varphi(r)\). Furthermore, it can be seen that \(\varphi\) is a ring homomorphism to \(\mathbb{Q}[y]\), an integral domain. Thus \((x+I, z+I) = \ker \varphi\) is a prime ideal.
As for the second part, \(z^2+I\) = \((z^2+xy-z^2)+I\) = \(xy+I\). The latter statement is evident.
Monday, 18 May 2020
A maximal ideal wrt not containing a finitely generated ideal (D&F 7.4.35)
Let \(A = (a_1, \ldots, a_n)\) be a nonzero finitely generated ideal of \(R\). Prove that there is an ideal \(B\) which is maximal with respect to the property that it does not contain \(A\).
Let \(S\) be the set of ideals not containing \(A\). If \(C\) is a chain in \(S\), then define \(J = \cup_{I \in C} I\). Suppose that \(A \subseteq J\); then in particular each \(a_i\) is in some \(I_i \in C\). Thus the ideal \(I' = \cup_{1 \leq i \leq n} I_i\) contains each \(a_i\) and hence \(A\) itself. But this is a contradiction since \(I' \in C \subseteq S\). Thus \(J\) does not contain \(A\) and is an upper bound for \(C\). By Zorn's Lemma, there exists a maximal ideal \(B\) with respect to not containing \(A\).
Let \(S\) be the set of ideals not containing \(A\). If \(C\) is a chain in \(S\), then define \(J = \cup_{I \in C} I\). Suppose that \(A \subseteq J\); then in particular each \(a_i\) is in some \(I_i \in C\). Thus the ideal \(I' = \cup_{1 \leq i \leq n} I_i\) contains each \(a_i\) and hence \(A\) itself. But this is a contradiction since \(I' \in C \subseteq S\). Thus \(J\) does not contain \(A\) and is an upper bound for \(C\). By Zorn's Lemma, there exists a maximal ideal \(B\) with respect to not containing \(A\).
A minimal prime ideal (D&F 7.4.36)
Assume \(R\) is commutative. Prove that the set of prime ideals in \(R\) has a minimal element with respect to inclusion (possibly the zero ideal).
The prime ideals form a partially ordered set \(S\) by inclusion. If \(C\) is a chain in \(S\), then define \(P = \cap_{I \in C} I\). Suppose \(ab \in P\) and \(a,b \not\in P\). Then there is some \(I_a, I_b \in C\) such that \(a \not\in I_a\) and \(b \not\in I_b\). If \(I_a \subseteq I_b\), then \(b \not\in I_a\), contradicting the fact that \(I_a\) is a prime ideal (likewise if \(I_b \subseteq I_a\)). Thus \(P\) is a prime ideal and a lower bound for \(C\). By Zorn's Lemma, there exists a minimal prime ideal.
The prime ideals form a partially ordered set \(S\) by inclusion. If \(C\) is a chain in \(S\), then define \(P = \cap_{I \in C} I\). Suppose \(ab \in P\) and \(a,b \not\in P\). Then there is some \(I_a, I_b \in C\) such that \(a \not\in I_a\) and \(b \not\in I_b\). If \(I_a \subseteq I_b\), then \(b \not\in I_a\), contradicting the fact that \(I_a\) is a prime ideal (likewise if \(I_b \subseteq I_a\)). Thus \(P\) is a prime ideal and a lower bound for \(C\). By Zorn's Lemma, there exists a minimal prime ideal.
Saturday, 16 May 2020
Order of conductor \(f\) in \(\mathbb{Q}(\sqrt{D})\) (D&F 7.1.23)
Let \(\mathcal{O}\) be the quadratic ring of integers in \(\mathbb{Q}(\sqrt{D})\). For any positive integer \(f\) prove that the set \(\mathcal{O}_f = \{a + bf\omega \mid a,b \in \mathbb{Z}\}\) is a subring of \(\mathcal{O}\) containing the identity, and that \(\left[\mathcal{O} : \mathcal{O}_f\right] = f\). Prove conversely that a subring of \(\mathcal{O}\) containing the identity and having finite index \(f\) (as additive group) is equal to \(\mathcal{O}_f\).
\(\implies\): Evident.
\(\impliedby\):Lemma. Let \(G = \mathbb{Z} \times \mathbb{Z}\) and \(H \le G\). If \(\left|G : H\right|\) is finite, then \(H \cong a\mathbb{Z} \times b\mathbb{Z}\) for positive integers \(a,b\).
Proof. Suppose there are nonzero integers \(m,n\) such that \((m,0),(0,n) \in H\). We can take \(m,n\) to be the smallest such positive integer, which implies \(H \cong m\mathbb{Z} \times n\mathbb{Z}\). Now suppose \((m,0) \not\in H\) for all \(m \in \mathbb{Z}-\{0\}\). Then any two cosets \((i,0) + H\) and \((j,0) + H\) with \(i,j \in \mathbb{Z}-\{0\}\), \(i \neq j\) are distinct, since \((i,0)-(j,0) \not\in H\). This contradicts the fact that \(\left|G : H\right|\) is finite. The same argument can be applied if \((0,n) \not\in H\) for all \(n \in \mathbb{Z}-\{0\}\). \(\blacksquare\)
As an additive group, \(\mathcal{O}\) is isomorphic to \(\mathbb{Z} \times \mathbb{Z}\). Any subgroup \(H\) of index \(f\) is of the form \(m\mathbb{Z} \times n\mathbb{Z}\), where \(m,n\) are the smallest positive integers appearing in their respective components and \(mn = f\). However by assumption \(H\) contains the multiplicative identity and so \(m = 1\), forcing \(H = \mathbb{Z} \times f\mathbb{Z}\). As a subring, \(H\) is precisely \(\mathcal{O}_f\).
Friday, 15 May 2020
Divisors in commutative rings
There is a general notion of divisibility and in particular, that of a greatest common divisor (GCD) for any commutative ring \(R\):
1. \(a \mid b\) if \(b = ra\) for some \(r \in R\).
2. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \(d \mid a\) and \(d \mid b\); (2) For any \(d'\) satisfying 1, \(d' \mid d\).
These statements can be expressed in terms of ideals:
1*. \(a \mid b\) if \((b) \subseteq (a)\).
2*. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \((a,b) \subseteq (d)\); (2) For any \(d'\) satisfying 1, \((d) \subseteq (d')\).
(Note that \((a,b)\) is the smallest ideal containing \((a)\) and \((b)\), so \((a),(b) \subseteq (d)\) automatically implies \((a,b) \subseteq (d)\).)
Based on the definition, we see that \((d)\) must be the unique smallest principal ideal containing \((a,b)\). This rules out situations where there are at least 2 minimal principal ideals containing \((a,b)\). Also note that there can exist multiple GCDs, provided they generate the same principal ideal. To take an extreme example, every real number is a GCD of every pair of real numbers (this is true for fields in general).
Generally, we do not require \((a,b) = (d)\). For example, in the polynomial ring \(\mathbb{Z}[x]\), \((2,x)\) is a maximal non-principal ideal (because the associated quotient is a field). Thus 1 is the GCD of 2 and \(x\) even though \((2,x) \subset (1)\).
On the other hand, there is a hierarchy of rings satisfying \((a,b) = (d)\) for all \(a,b \in R\):
Euclidean domains \(\subset\) PIDs \(\subset\) Bezout domains \(\subset\) Commutative rings
Bezout domains are defined precisely to satisfy this condition. However, one must note that not all Bezout domains are principal ideal domains. For instance, they can have infinitely generated non-principal ideals.
PIDs satisfy this condition trivially; in fact all PIDs are Bezout domains. In particular, Euclidean domains enable one to algorithmically compute \((d)\) and express \(d\) as an \(R\)-linear combination of \(a\) and \(b\), via the Euclidean algorithm.
1. \(a \mid b\) if \(b = ra\) for some \(r \in R\).
2. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \(d \mid a\) and \(d \mid b\); (2) For any \(d'\) satisfying 1, \(d' \mid d\).
These statements can be expressed in terms of ideals:
1*. \(a \mid b\) if \((b) \subseteq (a)\).
2*. A GCD \(d\) of \(a\) and \(b\) must satisfy the following: (1) \((a,b) \subseteq (d)\); (2) For any \(d'\) satisfying 1, \((d) \subseteq (d')\).
(Note that \((a,b)\) is the smallest ideal containing \((a)\) and \((b)\), so \((a),(b) \subseteq (d)\) automatically implies \((a,b) \subseteq (d)\).)
Based on the definition, we see that \((d)\) must be the unique smallest principal ideal containing \((a,b)\). This rules out situations where there are at least 2 minimal principal ideals containing \((a,b)\). Also note that there can exist multiple GCDs, provided they generate the same principal ideal. To take an extreme example, every real number is a GCD of every pair of real numbers (this is true for fields in general).
Generally, we do not require \((a,b) = (d)\). For example, in the polynomial ring \(\mathbb{Z}[x]\), \((2,x)\) is a maximal non-principal ideal (because the associated quotient is a field). Thus 1 is the GCD of 2 and \(x\) even though \((2,x) \subset (1)\).
On the other hand, there is a hierarchy of rings satisfying \((a,b) = (d)\) for all \(a,b \in R\):
Euclidean domains \(\subset\) PIDs \(\subset\) Bezout domains \(\subset\) Commutative rings
Bezout domains are defined precisely to satisfy this condition. However, one must note that not all Bezout domains are principal ideal domains. For instance, they can have infinitely generated non-principal ideals.
PIDs satisfy this condition trivially; in fact all PIDs are Bezout domains. In particular, Euclidean domains enable one to algorithmically compute \((d)\) and express \(d\) as an \(R\)-linear combination of \(a\) and \(b\), via the Euclidean algorithm.
Tuesday, 12 May 2020
Group with order \(pm\), with \(p > m\) and \(p\) prime (FGT 1A.9)
Suppose \(\left|G\right| = pm\), where \(p > m\) and \(p\) is prime. Show that \(G\) has a unique subgroup of order \(p\).
Assume there are \(k\) distinct order \(p\) subgroups. Since they have trivial intersection, the number of order \(p\) elements is \(k(p-1)\). By 1A.8, we have \(k(p-1) \equiv -1\) mod \(p\), or \(k(p-1) = lp-1\) for some \(l \in \mathbb{Z}^+\).
The sequences \(\{k(p-1)\}\) and \(\{lp-1\}\) form an arithmetic progression with step size \(p-1\) and \(p\), and clearly they meet at \(p-1\) when \(k = l = 1\). Since \(p\) and \(p-1\) are relative prime, the sequences next meet at \(1(p-1)+p(p-1) = (p+1)(p-1)\), so the next smallest solution has \(k = p+1\). But then there are a total of \((p+1)(p-1)+1 = p^2 > pm\) distinct elements, a contradiction.
Assume there are \(k\) distinct order \(p\) subgroups. Since they have trivial intersection, the number of order \(p\) elements is \(k(p-1)\). By 1A.8, we have \(k(p-1) \equiv -1\) mod \(p\), or \(k(p-1) = lp-1\) for some \(l \in \mathbb{Z}^+\).
The sequences \(\{k(p-1)\}\) and \(\{lp-1\}\) form an arithmetic progression with step size \(p-1\) and \(p\), and clearly they meet at \(p-1\) when \(k = l = 1\). Since \(p\) and \(p-1\) are relative prime, the sequences next meet at \(1(p-1)+p(p-1) = (p+1)(p-1)\), so the next smallest solution has \(k = p+1\). But then there are a total of \((p+1)(p-1)+1 = p^2 > pm\) distinct elements, a contradiction.
McKay's proof of Cauchy's theorem (FGT 1A.8)
Let \(G\) be a finite group, let \(n > 0\) be an integer. Let \(A\) be the set of \(n\)-tuples \((x_1, \ldots, x_n)\) of \(G\) such that \(x_1x_2 \ldots x_n = 1\).
(a) Show that \(\mathbb{Z}/n\mathbb{Z}\) acts on \(A\) according to the formula \(\overline{k} \cdot (x_1, \ldots, x_n) = (x_{1+k}, \ldots, x_{n+k})\), where indices are taken mod \(n\).
(b) Suppose \(n = p\) is a prime dividing \(\left|G\right|\). Show that \(p\) divides the number of size 1 orbits, and deduce that the number of elements of order \(p\) in \(G\) is congruent to -1 mod \(p\).
(a) Notice that if \(x_1 \ldots x_n = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\) then \(x_{1+k} \ldots x_{n+k} = (x_{1+k} \ldots x_n)(x_{n+1} \ldots x_{n+k}) = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\).
(b) The size of any orbit must divide \(\left|\mathbb{Z}/p\mathbb{Z}\right| = p\), hence the only possibilities are 1 and \(p\). Let the number of size 1 and size \(p\) orbits be \(n_1\) and \(n_p\). The number of possible \(p\)-tuples is easily checked to be \(\left|G\right|^{p-1}\), so we have \(n_1 + pn_p = \left|G\right|^{p-1}\). Since \(p \mid pn_p\) and \(p \mid \left|G\right|^{p-1}\), it follows that \(p \mid n_1\).
If \(\{(x_1, \ldots, x_n)\}\) is a size 1 orbit, then we have \(x_1 = \ldots = x_n\) and \({x_1}^p = 1\). Thus \(x_1\) is either 1 or an order \(p\) element. Eliminating the former possibility yields the congruence relation.
(a) Show that \(\mathbb{Z}/n\mathbb{Z}\) acts on \(A\) according to the formula \(\overline{k} \cdot (x_1, \ldots, x_n) = (x_{1+k}, \ldots, x_{n+k})\), where indices are taken mod \(n\).
(b) Suppose \(n = p\) is a prime dividing \(\left|G\right|\). Show that \(p\) divides the number of size 1 orbits, and deduce that the number of elements of order \(p\) in \(G\) is congruent to -1 mod \(p\).
(a) Notice that if \(x_1 \ldots x_n = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\) then \(x_{1+k} \ldots x_{n+k} = (x_{1+k} \ldots x_n)(x_{n+1} \ldots x_{n+k}) = (x_1 \ldots x_k)(x_{k+1} \ldots x_n) = 1\).
(b) The size of any orbit must divide \(\left|\mathbb{Z}/p\mathbb{Z}\right| = p\), hence the only possibilities are 1 and \(p\). Let the number of size 1 and size \(p\) orbits be \(n_1\) and \(n_p\). The number of possible \(p\)-tuples is easily checked to be \(\left|G\right|^{p-1}\), so we have \(n_1 + pn_p = \left|G\right|^{p-1}\). Since \(p \mid pn_p\) and \(p \mid \left|G\right|^{p-1}\), it follows that \(p \mid n_1\).
If \(\{(x_1, \ldots, x_n)\}\) is a size 1 orbit, then we have \(x_1 = \ldots = x_n\) and \({x_1}^p = 1\). Thus \(x_1\) is either 1 or an order \(p\) element. Eliminating the former possibility yields the congruence relation.
Monday, 11 May 2020
Burnside's lemma (FGT 1A.6)
Let \(\chi\) be the permutation character associated with the action of a group \(G\) on a set \(A\). Prove that \(\sum_{g \in G} \chi(g) = \sum_{a \in A} \left|G_a\right| = n\left|G\right|\), where \(n\) is the number of orbits.
The action of \(G\) on \(A\) can be represented by a table where the rows are elements of \(G\) and the columns are elements of \(A\). For example, here is the action of \(D_6\) on the three vertices of a triangle:
If we highlight all cells \((g,a)\) where \(g \cdot a = a\), then we get the following table:
Notice that the sum \(\sum_{g \in G} \chi(g)\) corresponds to be a horizontal summation of the highlighted squares, while \(\sum_{a \in A} \left|G_a\right|\) corresponds to a vertical summation.
Lastly, \(\sum_{a \in A} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \frac{\left|G\right|}{\left|\mathcal{O}\right|} = n\left|G\right|\).
The action of \(G\) on \(A\) can be represented by a table where the rows are elements of \(G\) and the columns are elements of \(A\). For example, here is the action of \(D_6\) on the three vertices of a triangle:
If we highlight all cells \((g,a)\) where \(g \cdot a = a\), then we get the following table:
Notice that the sum \(\sum_{g \in G} \chi(g)\) corresponds to be a horizontal summation of the highlighted squares, while \(\sum_{a \in A} \left|G_a\right|\) corresponds to a vertical summation.
Lastly, \(\sum_{a \in A} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \left|G_a\right| = \sum_{\mathcal{O}} \sum_{a \in \mathcal{O}} \frac{\left|G\right|}{\left|\mathcal{O}\right|} = n\left|G\right|\).
Friday, 8 May 2020
Ring projection from \(\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\)
Let \(n,m \in \mathbb{Z}^+\) with \(n \mid m\). Prove that the natural surjective ring projection \(\mathbb{Z}/m\mathbb{Z} \rightarrow \mathbb{Z}/n\mathbb{Z}\) is also surjective on the units: \((\mathbb{Z}/m\mathbb{Z})^\times \rightarrow (\mathbb{Z}/n\mathbb{Z})^\times\).
Let \(\varphi\) be the ring projection. Suppose that \(m = p^a\) and \(n = p^b\) where \(p\) is a prime and \(a,b \in \mathbb{Z}^+ \cup \{0\}\). Clearly if \((x,n) = 1\), then \((x,m) = 1\) as well. Thus \(\varphi\) is surjective on the units.
Now suppose \(m,n\) are any positive integers; let \(m = {p_1}^{a_1} \ldots {p_n}^{a_n}\) and \(n = {p_1}^{b_1} \ldots {p_n}^{b_n}\) be their prime factorization (where \(b_i\) is set to 0 where needed). By the Chinese Remainder Theorem we have \((\mathbb{Z}/m\mathbb{Z})^\times \cong (\mathbb{Z}/{p_1}^{a_1}\mathbb{Z})^\times \times \ldots \times (\mathbb{Z}/{p_n}^{a_n}\mathbb{Z})^\times\) (likewise for \((\mathbb{Z}/n\mathbb{Z})^\times\)). Since the natural projections corresponding to each factor is surjective on the units as shown in the previous paragraph, \(\varphi\) is surjective.
Let \(\varphi\) be the ring projection. Suppose that \(m = p^a\) and \(n = p^b\) where \(p\) is a prime and \(a,b \in \mathbb{Z}^+ \cup \{0\}\). Clearly if \((x,n) = 1\), then \((x,m) = 1\) as well. Thus \(\varphi\) is surjective on the units.
Now suppose \(m,n\) are any positive integers; let \(m = {p_1}^{a_1} \ldots {p_n}^{a_n}\) and \(n = {p_1}^{b_1} \ldots {p_n}^{b_n}\) be their prime factorization (where \(b_i\) is set to 0 where needed). By the Chinese Remainder Theorem we have \((\mathbb{Z}/m\mathbb{Z})^\times \cong (\mathbb{Z}/{p_1}^{a_1}\mathbb{Z})^\times \times \ldots \times (\mathbb{Z}/{p_n}^{a_n}\mathbb{Z})^\times\) (likewise for \((\mathbb{Z}/n\mathbb{Z})^\times\)). Since the natural projections corresponding to each factor is surjective on the units as shown in the previous paragraph, \(\varphi\) is surjective.
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