My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.
1. Let \(R\) be a commutative ring with unity. Let \(U(R)\) be the set of units in \(R\).
(a) Show that \(U(R)\) is an abelian group under multiplication.
(b) Let \(R = \mathbb{Z}[x]/(x^2)\). Show that \(U(R) \cong \mathbb{Z} \times \mathbb{Z}_2\). Justify your answer carefully.
Proof (a). Given that \(R\) is a commutative ring with unity and \(U(R)\) is the set of units in \(R\). Hence, we know that \(1\in U(R)\). For any \(a\in U(R)\), we have \(a\cdot a^{-1}\in U(R)\) where \(a^{-1}\in U(R)\) since \(a\) is a unit. For any \(a, b\in U(R)\), we can know that \(ab\in U(R)\) since \(b^{-1}, a^{-1}\in U(R)\) and \(b^{-1}a^{-1}\in U(R)\). Therefore, we can know that \(U(R)\) is a group under multiplication. \(\blacksquare\)
Proof (b). Firstly, we can know that \((x^2) = \{f\cdot x^2 \mid f\in \mathbb{Z}[x]\}\). Hence, let \(f = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\), where \(a_i\in \mathbb{Z}\). Hence, we have \(f\cdot x^2 = a_nx^{n+2} + a_{n-1}x^{n+1} + \cdots + a_1x^3 + a_0x^2\). Thus, we can know that \(R = \mathbb{Z}[x]/(x^2) = \{ax + b\,\mid\, a, b\in \mathbb{Z}\}\). Suppose that \(ax + b\in U(R)\), then we can know that there exists \(cx + d\in U(R)\) such that \((ax + b)(cx + d) = 1\). Hence, we have \(acx^2 + (ad + bc)x + bd = (ad + bc)x + bd = 1\), which implies that \(ad + bc = 0\) and \(bd = 1\). Since \(b, d\in \mathbb{Z}\), we can know that \(b = d = -1\) or \(b = d = 1\) in order to make sure \(bd = 1\). Hence, we have \(a + c = 0\), which implies that \(a = -c\). Hence, we can know that \(U(R) = \{ax + 1, bx - 1\,\mid\, a, b\in \mathbb{Z}\}\). Let \(\varphi: U(R) \to \mathbb{Z}\times \mathbb{Z}_2\) be a map such that \(\varphi(ax + 1) = (a, 0)\) and \(\varphi(bx - 1) = (b, 1)\).
2. Recall that a splitting field for a polynomial \( f(x) \) over a field \( F \) is the smallest (in terms of degree) extension \( K \) of \( F \) such that \( f(x) \) factors completely in \( K[x] \). Let \( F = \mathbb{Z}_2 \).
(a) Prove that \( K = F[t]/(t^3 + t + 1) \) is a field of order 8.
(b) Prove that \( K \) is a splitting field for \( f(x) = x^7 - 1 \) over \( F \).
Proof (a). Firstly, we want to show that \(t^3 + t + 1\) is irreducible in \(F[t]\) by contradiction. Suppose that \(t^3 + t + 1 = (t + a)(t^2 + bt + c)\) for some \(a, b, c\in F\). Then we know that \(t^3 + t + 1\) has a root in \(F\), which is \(-a\). However, \(1^3 + 1 + 1 = 1\) and \(0^3 + 0 + 1 = 1\). Hence, we can know that \(t^3 + t + 1\) has no root in \(F\), which is a contradiction. Thus, we can know that \(t^3 + t + 1\) is irreducible in \(F[t]\). Hence, we have \(K = F[t]/(t^3 + t + 1)\) is a field. Hence, the element in \(K\) is of the form \(a + bt + ct^2\) where \(a, b, c\in F\). Each \(a, b, c\) has two choices, which implies that \(K\) has \(2^3 = 8\) elements. Therefore, we can know that \(K\) is a field of order 8. \(\blacksquare\)
Proof (b). Given that \(K\setminus\{0\}\) is a group of order \(7\) under multiplication. Thus, for any \(x\in K\setminus \{0\}\), we have \(x^7 = 1\). Since each element of \(K\setminus\{0\}\) is root of \(x^7 - 1\), we can conclude that \(K\) is a splitting field for \(f(x) = x^7 - 1\) over \(F\). \(\blacksquare\)
3. Let \( G \) be a group and \( \text{Aut}(G) \) be the group of automorphisms of \( G \). Recall that for any \( c \in G \), we have an automorphism of \( G \) given by \( f_c(x) = c^{-1}xc \) (such an automorphism is called an inner automorphism of \( G \)). It is not hard to see that \( \text{Inn}(G) \), the collection of inner automorphisms of \( G \), is a subgroup of \( \text{Aut}(G) \).
(a) Show that \( G/Z(G) \cong \text{Inn}(G) \). Here \( Z(G) \) is the center of \( G \).
(b) Show that if \( \text{Aut}(G) \) is cyclic, then \( G \) is abelian.
(c) Find all prime numbers \( p \) such that there is a finite group \( G \) with \( \text{Aut}(G) \cong \mathbb{Z}_p \). (Hint: show that if \( G \) is not \( \mathbb{Z}_2 \), then \( \text{Aut}(G) \) must have an element of order 2).
Proof (a) Firstly, we define a map \(\varphi: G\to \text{Inn}(G)\) such that \[ \varphi(g) = f_g(x) = g^{-1}xg. \] Then, we can see that \(\varphi\) is well-defined since if \(g = h\in G\), then we will have \(g^{-1}xg = h^{-1}xh\). Now, we want to show that \(varphi\) is a group homomorphism. Suppose that \(g, h\in G\), then we have \[ \varphi(h)\cdot \varphi(g) = \varphi(g)\circ\varphi(h) = f_g\circ f_h(x) = f_g(h^{-1}xh) = g^{-1}h^{-1}xhg = (hg)^{-1}x(hg) = f_{hg}(x) = \varphi(hg). \] Hence, we can know that \(\varphi\) is a group homomorphism. Since for any \(f_g(x)\in \text{Inn}(G)\), we have \(\varphi(g) = f_g(x)\), which implies that \(varphi\) is surjective. Now, we want to figure the kernel of \(\varphi\). In \(\text{Inn}(G)\), we can know that the identity of \(\text{Inn}(G)\) is \(f_e(x) = e^{-1}xe = x\). Hence, for any \(g\in \ker(\varphi)\), we have \(f_g(x) = g^{-1}xg = x\), which implies that \(g\in Z(G)\). In other direction, for any \(z\in Z(G)\), we have \(f_z(x) = z^{-1}xz = x\), which implies that \(z\in \ker(\varphi)\). Thus, we can know that \(\ker(\varphi) = Z(G)\). By the \(\textbf{First Isomorphism Theorem}\), we can know that \(G/Z(G) \cong \text{Inn}(G)\). \(\blacksquare\)
Proof (b). We know that if \(G\) is a cyclic group, any subgroup of a cyclic group is cyclic. Since \(\text{Inn}(G)\) is a subgroup of \(\text{Aut}(G)\) and \(\text{Aut}(G)\) is cyclic, we can know that \(\text{Inn}(G)\) is cyclic. And we already showed that \(\text{Inn}(G)\} \cong G/Z(G)\). Hence, we can know that \(G/Z(G)\) is cyclic. Since \(G/Z(G)\) is cyclic, we can know that \(G/Z(G) = \langle aZ(G)\rangle\) for some \(a\in G\). Hence, for any \(g\in G\), we can know that \(gZ(G) = (aZ(G))^n = a^nZ(G)\) for some \(n\in \mathbb{Z}\). Thus, we have \(Z(G) = (a^{-1})^ngZ(G) = a^{-n}gZ(G)\), which implies that there exists \(z\in Z(G)\) such that \(a^{-n}g = z\). Hence, we have \(g = za^n = a^nz\) for some \(z\in Z(G)\) and \(n\in \mathbb{Z}\). Now, suppose that \(g = a^nz_1\) and \(h = a^mz_2\) for some \(n, m\in \mathbb{Z}\) and \(z_1, z_2\in Z(G)\). Then we have \[ gh = a^nz_1a^mz_2 = (a^nz_1a^m)z_2 = z_2a^nz_1a^m = z_2a^na^mz_1 = z_2a^{n+m}z_1 = z_2a^ma^nz_1 = a^mz_2a^nz_1 = hg. \] Therefore, we can know that \(G\) is abelian. \(\blacksquare\)
Solution (c). Suppose that \(G\) is a finite group and there exists a prime \(p\) such that \(\text{Aut}(G) = \mathbb{Z}_p\). We know that \(\mathbb{Z}_p\) is a cyclic group. Based on part (b), we can know that \(G\) is an abelian group. Assume that \(G\neq \mathbb{Z}_2\). In that case, we can know there exists \(g\in G\) such that \(g^{-1}\neq g\). Now define \(\varphi: G\to G\) such that \(\varphi(g) = g^{-1}\). Firstly, we can see that \(\varphi\) is well-defined since if \(g = h\), we will have \(g^{-1} = h^{-1}\). Now, we want to show that \(\varphi\) is a group homomorphism. Suppose that \(g, h\in G\), then we have \[ \varphi(h)\cdot \varphi(g) = \varphi(g)\circ\varphi(h) = g^{-1}\circ h^{-1} = (hg)^{-1} = (gh)^{-1} = \varphi(gh). \] Hence, we can know that \(\varphi\) is a group homomorphism. Suppose that \(\varphi(g) = \varphi(h)\), then we have \(g^{-1} = h^{-1}\), which implies that \(g = h\). It shows that \(\varphi\) is injective. Given that \(G\) is a finite group, we can know that \(\varphi\) is surjective by the \(\textbf{Pigeonhole Principle}\). It means that \(\varphi\) is an automorphism of \(G\). For any \(g\in G\), we have \(\varphi^{2}(g) = g\), which implies that \(\varphi\) has order of \(2\). Hence, we can know that \(2\mid |\mathbb{Z}_p| = p\). If \(p\neq 2\), we can know that it is impossible that \(2\mid p\) since \(p\) is a prime number. Hence, \(p\) has to be \(2\) when \(G\neq \mathbb{Z}_2\). Let \(G = \mathbb{Z}_3\). For any \(\varphi\in \text{Aut}(G)\), we have \(\varphi(0) = 0\), \(\varphi(1) = 1\) and \(\varphi(2) = 2\), which implies that \(\varphi\) is the identity automorphism. Besides that \(\psi\in \text{Aut}(G)\) such that \(\psi(0) = 0\), \(\psi(1) = 2\) and \(\psi(2) = 1\), which implies that \(\psi\) has order of \(2\). And those are the only automorphisms of \(\mathbb{Z}_3\), which means \(\text{Aut}(\mathbb{Z}_3) = \mathbb{Z}_2\). When \(G = \mathbb{Z}_2\), let \(\varphi\in \text{Aut}(G)\) such that \(\varphi(0) = 0\) and \(\varphi(1) = 1\). Then we have \(\varphi\) is the identity automorphism, and it is the only automorphism of \(\mathbb{Z}_2\). Same for \(G = \mathbb{Z}_1\). Thus, we can know that the only possible \(p\) is \(2\).
4. Let \( A \) be an \( n \times n \) matrix over a field \( F \). Prove \( \det(A) \) equals \( (-1)^n \) times the constant term of the characteristic polynomial of \( A \).
Proof. Since the characteristic polynomial of \( A \) is \( \chi_A(x) = \det(xI - A) \), which is a polynomial of degree \( n \) with variable \( x \). And we know that \(\det(-A) = (-1)^n\det(A)\). Hence, we plug in \(x = 0\), we have \[ (-1)^n\det(A) = \det(-A) = \det(0I - A). \] This implies that \( \det(A) \) equals \( (-1)^n \) times the constant term of the characteristic polynomial of \( A \). \(\blacksquare\)
5. Let \( V \) denote the complex vector space of complex polynomials having degree at most five, endowed with the usual inner product. In other words, for \( f(x), g(x) \in V \), \(\langle f(x), g(x) \rangle := \int_0^1 f(x) \overline{g(x)} \, dx \).
(a) Find an orthonormal basis for the space \( W \) spanned by \( x + x^2, x^2 + x^3, x^3, x^3 + x^4 \).
(b) Find (with proof) a self adjoint operator \( T : W \to W \) that is not the identity map.
Solution (a). We will first find an orthogonal basis for \( W \). Let \( v_1 = x + x^2 \), \( v_2 = x^2 + x^3 \), \( v_3 = x^3 \), and \( v_4 = x^3 + x^4 \) by \(\textbf{Gram-Schmidt orthogonalization}\). We normalize \(v_1\) firstly and have \[ u_1 = \dfrac{v_1}{\|v_1\| } = \dfrac{x + x^2}{\sqrt{\langle x + x^2, x + x^2 \rangle}} = \dfrac{x + x^2}{\sqrt{\int_0^1 (x + x^2)(x + x^2) \, dx}} = \dfrac{x + x^2}{\sqrt{\dfrac{31}{30}} = \sqrt{\dfrac{30}{31}}(x + x^2)}. \] Then we have \[ \begin{align} u_2 &= v_2 - \langle v_2, u_1 \rangle u_1 \\ &= x^2 + x^3 - \sqrt{\dfrac{30}{31}} \int_0^1 (x^2 + x^3)(x + x^2) \, dx (x + x^2)(\sqrt{\dfrac{30}{31}}) \\ &= x^2 + x^3 - \dfrac{30}{31} \int_0^1 (x^2+x^3)(x + x^2) \, dx (x + x^2) \\ &= -\frac{49}{62}x + x^3 - \dfrac{13}{62} x^2. \\ u_3 &= v_3 - \langle v_3, u_1 \rangle u_1 - \langle v_3, u_2 \rangle u_2 \\ \end{align} ]