KU 2015 (January) ALGEBRA QUALIFYING EXAM

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.

Proof. Let \(\varphi: G\to G\) such that \(g\mapsto g^2\) for all \(g\in G\). Firstly, we want to show that \(\varphi\) is well-defined. Suppose that \(g = h\) for some \(g, h\in G\), then we have \(g^2 = h^2\). Then, we want to show that \(\varphi\) is a group homomorphism. For any \(g, h\in G\), we have \(\varphi(gh) = (gh)^2 = g^2h^2 = \varphi(g)\varphi(h)\) since \(G\) is abelian. Now, we want to show that \(\varphi\) is injective. Suppose that \(\varphi(g) = \varphi(h)\) for some \(g, h\in G\), then we have \(g^2 = h^2\). Hence, we can have \(g^2h^{-2} = (gh^{-1})^2 = e\). Given that \(G\) is group of odd order. Hence, we can know that there does not exist any element of even order in \(G\). If \(gh^{-1}\) is an non identity element of \(G\), then we can say that \(gh^{-1}\) has order of \(2\), which is a contradiction. Hence, we have \(gh^{-1} = e\), which implies that \(g = h\). Thus, we showed that \(\varphi\) is injective. Given that \(\varphi: G\to G\) and \(G\) has finite order, according to \(\textbf{Pigeon Hole Principle}\), we can know that \(\varphi\) is surjective. Thus, we can have for any \(x\in G\), there exists a unique \(y\in G\) such that \(y^2 = x\). \(\blacksquare\)

Proof. Given that \(435 = 3\cdot 5\cdot 29\), we can know that there exists a sylow 3-subgroup \(G\), sylow 5-subgroup \(H\), and sylow 29-subgroup \(K\). And the sylow 3-subgroup has order of \(3\), which is a prime. Hence, we can know that the sylow 3-subgroup is cyclic and isomorphic to \( \mathbb{Z}/3\mathbb{Z} \). For the similar reason, we can have the sylow 5-subgroup is cyclic and isomorphic to \( \mathbb{Z}/5\mathbb{Z} \) and the sylow 29-subgroup is cyclic and isomorphic to \( \mathbb{Z}/29\mathbb{Z} \). Suppose that \(g\neq e\in G\cap H\), then we can know that \(g\) has order either \(3\) or \(5\), but cannot be both. Thus, we can know that \(G\cap H = \{e\}\). For the similar reason, we can have \(H\cap K = \{e\}\) and \(G\cap K = \{e\}\). Now, according to the \(\textbf{Third Sylow's Theorem}\), we can know that the number of sylow 3-subgroup, sylow 5-subgroup, and sylow 29-subgroup are \(3a + 1\), \(5b + 1\), \(29c + 1\) for some \(a, b, c\in \mathbb{Z}\). Moreover, we have \(3a + 1 \mid 435\), \(5b + 1 \mid 435\), and \(29c + 1 \mid 435\). Hence, we can conclude that \(a = b = c = 0\). And the number of the sylow 3-subgroup, sylow 5-subgroup, and sylow 29-subgroup are \(1\), \(1\), and \(1\), respectively. For any \(g\) in our original group, we can know that \(gGg^{-1}\) is a group of order \(3\) and there is only one subgroup of order \(3\), which is \(G\). Hence, we can know that \(gGg^{-1} = G\), which implies that \(G\) is normal in our original group. Similarly, we can know that \(H\) and \(K\) are normal in our original group. The inner direct product of two normal subgroups are normal. Hence, we can know that \(G H\) is normal in our original group. Besides that, we can have \[ \begin{align} |GH| = \frac{|G||H|}{|G\cap H|} = 15, |GHK| = \frac{|GH||K|}{|GH\cap K|} = 435. \end{align} \] Thus, we know that our original group is \(GHK\). Since we know that both \(G\) and \(H\) are normal and \(G\cap H = \{e\}\), we can know that \(GH = G\times H\). Since \(GH\) and \(K\) are normal, and \(GH\cap K = \{e\}\), we can know that \(GHK = GH\times K\). Thus, we can know that \(GHK = G\times H\times K = \mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/29\mathbb{Z}\). Since \(\mathbb{Z}/3\mathbb{Z}\times \mathbb{Z}/5\mathbb{Z}\times \mathbb{Z}/29\mathbb{Z}\) is abelian, we can know that our original group is abelian. \(\blacksquare\)

Proof (i). Given that \(I\) is a proper ideal, we can know that \(1\notin I\). Hence, we have \(1 + I\in R/I\), which implies that the multiplicative identity exists in \(R/I\). And it is not hard to see that \(0 + I = I\in R/I\), which implies that the additive identity exists in \(R/I\).

Solution. Let \( \alpha = \sqrt{2} + \sqrt[3]{2} \). Then we have \[ \begin{align} \alpha - \sqrt{2} &= \sqrt[3]{2} \\ (\alpha - \sqrt{2})^3 &= 2 \\ \alpha^3 - 3\alpha^2\sqrt{2} + 6\alpha - 2\sqrt{2} &= 2 \\ -3\alpha^2\sqrt{2} - 2\sqrt{2} &= 2 - \alpha^3 - 6\alpha \\ \sqrt{2}(-3\alpha^2 - 2) &= 2 - \alpha^3 - 6\alpha \\ \sqrt{2} &= \frac{2 - \alpha^3 - 6\alpha}{-3\alpha^2 - 2} \\ &= \frac{\alpha^3 + 6\alpha - 2}{3\alpha^2 + 2}. \end{align} \] Hence, we can know that \( \sqrt{2} \) is in the field \( \mathbb{Q}(\alpha) \). And we have \(\mathbb{Q}(\sqrt{2}) \subset \mathbb{Q}(\alpha)\). Since \(\mathbb{Q}(\sqrt{2})\subset \mathbb{Q}(\sqrt{2}, \sqrt{3})\). Given that \(\sqrt{2} + \sqrt[3]{2}\) is in the field \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2})\), we can know that \(\mathbb{Q}(\sqrt{2} + \sqrt[3]{2}) \subset \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})\). Since \(\sqrt{2}\in \mathbb{Q}(\sqrt{2} + \sqrt[3]{2})\), we can know that \(\sqrt[3]{2} = (\sqrt{2} + \sqrt[3]{2}) - \sqrt{2}\) is in the field \(\mathbb{Q}(\sqrt{2} + \sqrt[3]{2})\). Thus, we have \(\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}) \subset \mathbb{Q}(\sqrt{2} + \sqrt[3]{2})\). Therefore, we can know that \(\mathbb{Q}(\sqrt{2} + \sqrt[3]{2}) = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})\). We know \([\mathbb{Q}(\sqrt{2}): \mathbb{Q}] = 2\) since \(x^2 - 2\) is an irreducible polynomial over \(\mathbb{Q}\) and \(\sqrt{2}\) is a root of \(x^2 - 2\) (i.e \(x^2 - 2\) is the minimal polynomial of \(\sqrt{2}\) over \(\mathbb{Q}\)). And \(\sqrt[3]{2}\notin \mathbb{Q}(\sqrt{2})\) since there is no \(a, b\in \mathbb{Q}\) such that \(a + b\sqrt{2} = \sqrt[3]{2}\). Let \(f(x)= x^3 - 2\) and we want to show that \(f(x)\) is irreducible over \(\mathbb{Q}(\sqrt{2})\) by contradiction. Suppose that \(f(x)\) is reducible over \(\mathbb{Q}(\sqrt{2})\), then we can have \(f(x) = (x + a)(x^2 + bx + c)\) for some \(a, b, c\in \mathbb{Q}(\sqrt{2})\). Hence, \(f(x)\) has a root in \( \mathbb{Q}(\sqrt{2}) \). We know that \(\sqrt[3]{2}\) is a root of \(f(x)\) but \(\sqrt[3]{2}\notin \mathbb{Q}(\sqrt{2})\). And \(f(x)\) has two other roots in \(\mathbb{C}\), which is a contradiction. Hence, we have \(f(x)\) is irreducible over \(\mathbb{Q}(\sqrt{2})\). Thus, we can know that \[ \mathbb{Q}(\sqrt{2})[x]/(x^3 - 2) \cong \mathbb{Q}(\sqrt{2})[\sqrt[3]{2}]. \] Since \(\sqrt[3]{2}\) is algebraic over \(\mathbb{Q}(\sqrt{2})\), we have \( \mathbb{Q}(\sqrt{2})[\sqrt[3]{2}] = \mathbb{Q}(\sqrt{2})(\sqrt[3]{2}) = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2}) = \mathbb{Q}(\sqrt{2} + \sqrt[3]{2})\). Given that \(x^3 - 2\) is the minimal polynomial of \(\sqrt[3]{2}\) over \(\mathbb{Q}(\sqrt{2})\), we can conclude that \([\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}(\sqrt{2})] = 3\). Then, we have \[ \begin{align} [\mathbb{Q}(\sqrt{2} + \sqrt[3]{2}): \mathbb{Q}] &= [\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}] \\ &=[\mathbb{Q}(\sqrt{2}, \sqrt[3]{2}): \mathbb{Q}(\sqrt{2})][\mathbb{Q}(\sqrt{2}): \mathbb{Q}] \\ &= 3\cdot 2 = 6. \end{align} \] Now, we know that the minimal polynomial of \(\sqrt{2} + \sqrt[3]{2}\) over \(\mathbb{Q}\) is degree \(6\). We already had \[ \sqrt{2} = \frac{\alpha^3 + 6\alpha - 2}{3\alpha^2 + 2}. \] Then, we square both sides and simplify it, \[ \begin{align} 2 &= \frac{(\alpha^3 + 6\alpha - 2)^2}{(3\alpha^2 + 2)^2} \\ 2(3\alpha^2 + 2)^2 &= (\alpha^3 + 6\alpha - 2)^2 \\ 2(9\alpha^4 + 12\alpha^2 + 4) &= \alpha^6 + 12\alpha^4 - 4\alpha^3 + 36\alpha^2 - 24\alpha + 4 \\ 18\alpha^4 + 24\alpha^2 + 8 &= \alpha^6 + 12\alpha^4 - 4\alpha^3 + 36\alpha^2 - 24\alpha + 4 \\ \alpha^6 - 6\alpha^4 - 4\alpha^3 + 12\alpha^2 - 24\alpha - 4 &= 0. \end{align} \] Since \(x^6 - 6x^4 - 4x^3 + 12x^2 - 24x - 4\) had degree of \(6\) and \(\sqrt{2} + \sqrt[3]{2}\) is a root of \(x^6 - 6x^4 - 4x^3 + 12x^2 - 24x - 4\), we can know that \(x^6 - 6x^4 - 4x^3 + 12x^2 - 24x - 4\) is the minimal polynomial of \(\sqrt{2} + \sqrt[3]{2}\) over \(\mathbb{Q}\).

Solution (i). The dual space \( V^* \) is the set of all linear transformations from \( V \) to \( F \).

Proof (ii). Firstly, we define \(L_v\) to be a map from \(V^*\) to \(F\) such that \(L_v(f) = f(v)\) where \(v\in V\) for all \(f \in V^*\). Then, we define a map \(\Phi : V \to V^{**}\) such that \(\Phi(v) = L_v\) for all \(v \in V\). Firstly, we show that \(\Phi\) is linear. Let \(v_1, v_2 \in V\) and \(c \in F\). If \(f\in V'\), then \[ \begin{align} \Phi(v_1 + v_2)(f) &= f(v_1 + v_2) = f(v_1) + f(v_2) = L_{v_1}(f) + L_{v_2}(f) = \Phi(v_1)(f) + \Phi(v_2)(f), \\ \Phi(cv_1)(f) &= f(cv_1) = cf(v_1) = cL_{v_1}(f) = c\Phi(v_1)(f). \end{align} ] After that, we want to know that \(\Phi\) is injective. Let \(v \in V\) such that \(\Phi(v) = 0\). Since \(\Phi(v) = L_v(f) = f(v) = 0\) for all \(f \in V^*\), we can know that \(v = 0\). Then, we can get that \(\ker(\Phi) = \{0\}\) and \(\Phi\) is injective and invertible. Since \(\dim(V) = \dim(V^*)\) and \(\dim(V^*) = \dim(V^{**}) \) and \(\Phi\) is invertible, we can know that \(\Phi\) is an isomorphism. Therefore, we have \(V \cong V^{**}\). \(\blacksquare\)

Solutions (i). Suppose that \(v\in \ker(A)\cap W\). Then we have \(A\cdot v = v = 0\), which implies that \(W\cap \ker(\phi_A) = 0\). Besides that, with \( \dim \ker(\phi_A) + \dim W = n \), we can know that \( \ker(\phi_A)\oplus W = \mathbb{R}^n \). Since we know that \(\ker(\phi_A)\oplus \text{Range}(\phi_A) = \mathbb{R}^n\), we can know that \(\text{Range}(\phi_A) = W\). Hence, for any \(v\in \mathbb{R}^n\), we can write \(v = v_1 + v_2\), where \(v_1\in \ker(\phi_A)\) and \(v_2\in W\). Then, we have \(A\cdot v = Av_1 + Av_2 = Av_2 = v_2\). And \(A(Av) = A(Av_1 + Av_2) = A(Av_2) = Av_2 = v_2\). Thus, we can get \[ A^2\cdot v - A\cdot v= (A^2 - A)\cdot v = 0, \text{ for all } v\in \mathbb{R}^n. \] Hence, we can know that \(A^2 - A = 0\). Then, we can conclude that the minimal polynomial \(\mu_A \mid x^2 - x\). Suppose that \(\mu_A\) is a polynomial of degree \(1\). Then we can know that \(\mu_A = x\) or \(\mu_A = x - 1\). If \(\mu_A\) is a polynomial of degree \(2\), then we can know that \(\mu_A = x(x - 1)\).

Solution (ii). If \(\mu_A = x\), then we have \(A = 0\). Hence, the Jordan canonical form of \(A\) is \(0\). If \(\mu_A = x - 1\), then we have \(A = I\). Hence, the Jordan canonical form of \(A\) is \(I\). If \(\mu_A = x(x - 1)\), then we want to prove that the Jordan Canonical Form of \(A\) is a diagonal matrix with \(0\) and \(1\) on the diagonal entries. Since \(W = \{v\in \mathbb{R}^n \mid Av = v\}\), we can know that \(W = \ker(A - I)\). Given that \(\text{dim}\ker(A) + \text{dim}W = n\), we have \(\text{dim}\ker(A - 0\cdot I) + \text{dim}\ker(A - I) = n\). Thus, we can know that \(A\) is diagonalizable. Hence, the Jordan Canonical Form of \(A\) is a diagonal matrix with \(0\) and \(1\) on the diagonal entries.

Proof (iii). Suppose that \(A\) is a symmetric matrix. Then we have \(A = A^T = A^*\) since \(A\) is a real matrix. Let \(w\in W\) and \(v\in \ker(\phi_A)\). Then we have \[ \langle w, v\rangle = \langle Aw, v\rangle = \langle w, A^*v\rangle = \langle w, Av\rangle = \langle w, 0\rangle = 0. \] Hence, we can know that \(W\) is orthogonal to \(\ker(\phi_A)\). \[\tag*{\(\blacksquare\)}\]

References

• Qualifying Exams Kansas State Mathematics Ph.D. Program (Algebra Archive)