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.
Problem 2. Let \(V\) be a finite dimensional vector space over the complex numbers and \(T\) a linear operator on \(V\) such that the subspaces of \(V\) generated by all the eigenvectors is one dimensional. Prove that the minimal polynomial of \(T\) is \((x − \lambda)^n\) where \(n\) is the dimension of \(V\) .
Proof. Given that the subspaces of \(V\) generated by all the eigenvectors is one dimensional, we can know that the basis of the subspace is \(\{v\}\), where \(v\) is the eigenvector of \(T\). Then we have \(T(v) = \lambda v\), where \(\lambda\) is the correspondent eigenvalue of \(v\). Hence, if \(v'\) is an eigenvector of \(T\). Then, we can know that \(v' = cv\) for some \(c\in\mathbb{C}\). Hence, we have \[ T\cdot v' = T\cdot cv = cT\cdot v = c\lambda v = \lambda cv = \lambda v'. \] Thus, we show that there is only one eigenvalue of \(T\), which is \(\lambda\). Then, we can know that the characteristic polynomial of \(T\) is \((x - \lambda)^n\), where \(n\) is the dimension of \(V\). In that case, the minimal polynomial of \(T\) is \((x - \lambda)^m\), where \(m\leq n\) (to be continued...)
5. Consider the ring \(R = \mathbb{Q}[x, y, z]/I\), where \(I = (x^2y - z^5)\).
(i) Prove that \(R\) is not a field. Then decide whether \(R\) is an integral domain, and justify your answer.
(ii) Given \(f\in \mathbb{Q}[x, y, z]\), let \(\overline{f}\) denote the element \(f + I\) of \(R\). Is the ideal \(J = (\bar x, \bar y)\) a prime ideal of \(R\)? Prove your answer.
(iii) Show that \(\overline{z^5} \in J^2\), but that \(\overline{z^4}\notin J^2\).
Proof (i). We know that \(\mathbb{Q}[x, y, z]/I\) is a field if and only if \(I\) is a maximal ideal. We know that \((x^2y - z^5)\subset (x^2y - z^5, z)\) since \(x^2y - z^5 = 1\cdot (x^2y - z^5) + 0\cdot z\). Now, I want to show that \((x^2y - z^5, z)\) is a proper ideal of \(\mathbb{Q}[x, y, z]\). Firstly, for any \(f\neq 0\in \mathbb{Q}[x, y, z]\), we can know that \(f\cdot (x^2y - z^5)\) does not contain a term of constant. Similarly, we can know that \(g\cdot z\) is a polynomial without constant term for any \(g\neq 0\in \mathbb{Q}[x, y, z]\). Hence, we can know that if \(f\cdot (x^2y - z^5) + g\cdot z\neq 0\), then \(f\cdot (x^2y - z^5) + g\cdot z\notin \mathbb{Q}\). In other words, if \(i\in (x^2y - z^5, z)\), then \(i\) is not a unit. Thus, we have \((x^2y - z^5, z)\neq \mathbb{Q}[x, y, z]\), which implies that \((x^2y - z^5, z)\) is a proper ideal. Then, we can conclude that \((x^2y - z^5)\) is not a maximal ideal. Therefore, we can know that \(R\) is not a field. \(\blacksquare\)
Now, we want to show that \(R\) is an integral domain.
Proof (i). We know that \(\mathbb{Q}[x, y, z]/I\) is an integral domain if \(I\) is a prime ideal. Since \(\mathbb{Q}[x, y, z] = \mathbb{Q}[x, z][y]\), we can know that \(x^2y - z^5\) is a degree \(1\) polynomial of \(\mathbb{Q}[x, z][y]\). Hence, \(x^2y - z^5\) is irreducible, which implies that \(x^2y - z^5\) is prime. Thus, we can know that \(I = (x^2y - z^5)\) is a prime ideal. Therefore, we have \(R\) is an integral domain. \(\blacksquare\)
Proof (ii). We want to show that \(J = (\bar x, \bar y)\) is a prime ideal of \(R\). If \(\bar r\in (\bar x, \bar y)\), we can know that \(\bar r = \bar x\bar f + \bar y\bar g\) for some \(\bar f, \bar g\in R\). Then, we we can know that \(\bar r\) does not contain a term of \(a_nz^n\) for any \(n\in \mathbb{N}\) and \(a_n\in\mathbb{Q}\) (including the case that \(n = 0\). In the other direction, if \(\bar r\) is an element of \(R\) without a term of \(a_nz^n\) for any \(n\in \mathbb{N}\) and \(a_n\in\mathbb{Q}\), then we can know that \(\bar r = \bar x\bar f + \bar y\bar g\) for some \(\bar f, \bar g\in R\). Hence, if \(\bar f, \bar g\in R\) do not contain a term of \(a_nz^n\) for any \(n\in \mathbb{N}\) and \(a_n\in\mathbb{Q}\) (i.e. \(\bar f\notin J, \bar g\notin J\)), then we can know that \(\bar f\cdot \bar g\) does not contain a term of \(a_nz^n\) for any \(n\in \mathbb{N}\) and \(a_n\in\mathbb{Q}\) (i.e. \bar f\bar g\notin J). Therefore, we can know that \(J\) is a prime ideal of \(R\). \(\blacksquare\)
Proof (iii). Firstly, we want to show that \(J^2 = ({\bar x}^2, \bar x\bar y, {\bar y}^2)\). Let \(\bar p, \bar q\in J\). Then, we can know that \(\bar p = \bar x\bar f + \bar y\bar g\) and \(\bar q = \bar x\bar h + \bar y\bar k\) for some \(\bar f, \bar g, \bar h, \bar k\in R\). Then, we can know that \(\bar p\bar q = \bar x\bar x\bar f\bar h + \bar x\bar y\bar f\bar k + \bar x\bar y\bar g\bar h + \bar y\bar y\bar g\bar k\). Hence, we can know that \(\bar p\bar q\in ({\bar x}^2, \bar x\bar y, {\bar y}^2)\), which implies that \(J^2\subseteq ({\bar x}^2, \bar x\bar y, {\bar y}^2)\). In the other direction, if \(\bar r\in ({\bar x}^2, \bar x\bar y, {\bar y}^2)\), then we can know that \(\bar r = \bar x\bar x\bar f + \bar x\bar y\bar g + \bar y\bar y\bar h\) for some \(\bar f, \bar g, \bar h\in R\).
4. Recall that if \( G \) is a group and \( x, y \in G \), \( x \) and \( y \) are conjugate if \( y = gxg^{-1} \) for some \( g \in G \), and that conjugacy is an equivalence relation on \( G \). Let \( [x] \) denote the conjugacy class of an element \( x \in G \); i.e., \( [x] = \{gxg^{-1} \mid g \in G\} \).
(i) Prove that for \( x \in G \), \( [x] \) has exactly one element if and only if \( x \in Z(G) \), where \( Z(G) \) denotes the center of \( G \).
(ii) More generally, prove that the number of elements in \( [x] \) equals \( |G : C_G(x)| \), where \( C_G(x) \) is the centralizer of \( x \) in \( G \), i.e., \( C_G(x) = \{g \in G \mid gxg^{-1} = x\} \). Hint: When is \( gxg^{-1} \) equal to \( hxh^{-1} \)?
(iii) Suppose that \( G \) is a finite group of odd order, and suppose that \( N \) is a normal subgroup of \( G \) of order 3. Prove that \( N \leq Z(G) \). Hint: What does the normality of \( N \) say about the conjugacy classes of elements \( x \in N \)? How can one use this to partition \( N \)?
Proof (i). Firstly, we can know that \(x\in [x]\) since \(exe^{-1} = exe = x\). Thus, if \([x]\) has exactly one element, then we can know that \(gxg^{-1} = x\) for all \(g\in G\). Hence, we have \(x = gxg^{-1}\) for all \(g\in G\), which implies that \(x\in Z(G)\). If \(x\in Z(G)\), then we have that \(gxg^{-1} = x\) for all \(g\in G\), which implies that \(\{x\} = [x]\). Thus, we can know that \([x]\) has exactly one element. \(\blacksquare\)
Proof (ii). Since the number of elements in \([x]\) is the number of elements in the orbits. We define the map \(\varphi: [x]\to G/C_G(x))\) by \(\varphi(gxg^{-1}) = gC_G(x)\). Firstly, we can know that the map is surjective since for any \(gC_G(x)\), we have \(gxg^{-1}\in [x]\) such that \[ \varphi(gxg^{-1}) = gC_G(x). \] Now, suppose that \(gC_G(x) = hC_G(x)\). Then, we have \(h^{-1}g\in C_G(x)\), which implies that \(h^{-1}gxg^{-1}h = x\). Hence, we have \[ \begin{align} h^{-1}gxg^{-1}h &= x\\ h(h^{-1}gxg^{-1}h)h^{-1} &= hxh^{-1}\\ gxg^{-1} &= hxh^{-1}. \end{align} \] Thus, we can know that the map is injective. Hence, we can know that the number of elements in \([x]\) equals \(|G/C_G(x)| = |G:C_G(x)|\). \(\blacksquare\)
Proof (iii).We define a map \(\varphi: G\to \text{Aut}(N)\) by \(\varphi(g) = gng^{-1}\). The reason that \(\varphi(g)\) is a group automorphism of \(N\) is \(N\) is normal in \(G\). We can see that \(\varphi\) is well-defined since for any \(g = h\in G\), we have \(gng^{-1} = hnh^{-1}\). Then, we can know that \(\varphi\) is a group homomorphism since for any \(g, h\in G\), we have \[ \varphi(g)\varphi(h) = \varphi(g)\circ \varphi(h) = g h n h^{-1}g^{-1} = (gh)n(gh)^{-1} = \varphi(gh). \] Hence, we can see that the image of \(\varphi\), \(\varphi(G)\), is a subgroup of \(\text{Aut}(N)\). Given that \(N\) has order of \(3\), we have \(N\cong \mathbb{Z}/3\mathbb{Z}\), which implies that \(N\) is cyclic. Hence, any non-trivial element is a generator of \(N\). Thus, we have \(N = \{e, x, x^2\}\), where \(x, x^2\) are the generators of \(N\). Once we know where \(x\) is mapped to, we can know where \(x^2\) is mapped to since \(\varphi\) is a group homomorphism. There are only two options, either \(x\to x\) or \(x\to x^2\). Hence, we can see that \(|\text{Aut}(N)| = 2\), which implies that \(\mathbb{Z}/2\mathbb{Z}\cong \text{Aut}(N)\). Since \(\varphi(G)\) is a subgroup of \(\text{Aut}(N)\), we have \(|\varphi(G)| = 1\) or \(2\). According to the \(\textbf{First Isomorphism Theorem}\), we have \(G/\ker(\varphi) \cong \varphi(G)\)\). Since \(|\varphi(G)| = |G/\ker(\varphi)| = [G:\ker(\varphi)]\), we have \([G:\ker(\varphi)] = 1\) or \(2\). Given that \([G:\ker(\varphi)]\mid |G|\) and \(|G|\) is odd, we have \([G:\ker(\varphi)] = 1\). Thus, we can see that \(\varphi(G) = \{e\}\), which implies that \(\ker(\varphi) = G\). And \(\ker(\varphi) = \{g\in G\mid \varphi(g) = e\}\), which implies that \(gng^{-1} = n\) for all \(n\in N\). Therefore, we can get that \(N\leq Z(G)\). \(\blacksquare\)
6. Consider the polynomial \( p(x) = x^3 - 2 \in \mathbb{Q}[x] \). Find \( \alpha, \beta \in \mathbb{C} \) for which \( F = \mathbb{Q}(\alpha, \beta) \) is a splitting field for \( p(x) \), and prove that this is the case. Then find, with justification, the degree of the field extension \( \mathbb{Q} \subseteq F \).
Solution (6). Firstly, we want to find the all the roots of \(p(x)\). Suppose that \(a\) is the root of \(p(x)\). Then we have \(p(a) = a^3 - 2 = 0\) and \(a^3 = 2\). Hence, we can know that \(a^3\) is a cube of \(2\). Thus, we can know that \(a = \sqrt[3]{2}\), which is a real number. Now, if \(a^3 = 2\cos(2\pi) + 2\sin(2\pi)i = 2(\cos(2\pi) + sin(2\pi)i) = 2e^{2\pi i}\), then we have \(a = \sqrt[3]{2}e^{2\pi i/3}\). Hence, \(a = \sqrt[3]{2}(\cos(2\pi/3) + \sin(2\pi/3)i) = \sqrt[3]{2}(-1/2 + \sqrt{3}/2i)\). Similarly, we can have \(a^3 = 2\cos(-2\pi) + 2\sin(-2\pi)i = 2e^{-2\pi i}\), which implies that \(a = \sqrt[3]{2}e^{-2\pi i/3}\). Hence, we have \(a = \sqrt[3]{2}(\cos(-2\pi/3) + \sin(-2\pi/3)i) = \sqrt[3]{2}(-1/2 - (\sqrt{3}/2)i)\). Then, we have the three distinct roots of \(p(x)\), which are \[ a_1 = \sqrt[3]{2},\quad a_2 = \sqrt[3]{2}(-1/2 + (\sqrt{3}/2)i),\quad a_3 = \sqrt[3]{2}(-1/2 - (\sqrt{3}/2)i). \] Thus, we can know that the splitting field of \(p(x)\) is \(F = \mathbb{Q}(a_1, a_2, a_3)\). Since \(a_2 - a_3 = \sqrt[3]{2}\cdot\sqrt{3}i\), we have \[ \dfrac{\sqrt[3]{2}\cdot\sqrt{3}i}{\sqrt[3]{2}} = \sqrt{3}i. \] Hence, \(\mathbb{Q}(\sqrt[3]{2}, \sqrt{3}i)\subset \mathbb{Q}(a_1, a_2, a_3)\). It is not hard to see that \(\mathbb{Q}(a_1, a_2, a_3)\subset \mathbb{Q}(\sqrt[3]{2}, \sqrt{3}i)\). Since \(\sqrt{3}i\) is a root of \(x^2 + 3 = 0\). Since there is no rational roots of \(x^2 + 3 = 0\), we can know that \(x^2 + 3\) is irreducible over \(\mathbb{Q}\). Thus, we can know that \[ \mathbb{Q}(\sqrt{3}i\) \cong \mathbb{Q}[x]/(x^2 + 3). \] Hence, \(\mathbb{Q}(\sqrt{3}i\)\) has degree \(2\) since the minimal polynomial of \(\sqrt{3}i\) has degree \(2\). Then, we want to show that \(x^3 - 2\) is irreducible over \(\mathbb{Q}(\sqrt{3}i)\) by contradiction. If \(x^3 - 2\) is a reducible polynomial over \(\mathbb{Q}(\sqrt{3}i)\), then we have a root of \(x^3 - 2\) in \(\mathbb{Q}(\sqrt{3}i)\). Since \(\mathbb{Q}(\sqrt{3}i\) = \{a + b\sqrt{3}i\mid a, b\in \mathbb{Q}\), we can know that the root of \(x^3 - 2\) is \(a + b\sqrt{3}i\) for some \(a, b\in \mathbb{Q}\). Then suppose that \(a + b\sqrt{3}i) is a root of \(p(x)\), \[ \begin{align} (a + b\sqrt{3}i\)^3 &= 2, \\ (a^2 + 2ab\sqrt{3}i - 3b^2)(a + b\sqrt{3}i) = 2, \\ (a + b\sqrt{3}i\)^3&= a^3 - 9 a b^2 + 3 i \sqrt{3} b (a^2 - b^2) = 2 \end{align} \]\Hence, we can know that \(3b(a^2 - b^2)\) =0\), which implies that \(b = 0\) or \(a-b = 0\) or \(a + b = 0\). Given that \(b\neq 0\), or \(a + b\sqrt{3}i = a\) is a root of \(p(x)\), which is a contradiction. Suppose that \(a = b\), then we can have \(a^3 - 9a^3 = -8a^3 = 2\), which implies that \(a^3 = -1/4\), which is a contradiction. Suppose that \(a = -b\), then we can have \(a^3 - 9a^3 = -10a^3 = 2\), which implies that \(a^3 = -1/4\), which is a contradiction. Hence, there exists no roots in \(p(x)\) and \(p(x)\) is irreducible. Then \(\mathbb{Q}(\sqrt{3}i, \sqrt[3]{2})\cong \mathbb{Q}(\sqrt{3}i)[x]/(x^3 - 2)\). Therefore, we have \[ [\mathbb{Q}(\sqrt{3}i, \sqrt[3]{2}): \mathbb{Q}(\sqrt{3}i)][\mathbb{Q}(\sqrt{3}i): \mathbb{Q}] = 3\cdot 2 = 6. \]