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.
2. Let \( R \) denote the ring \( \mathbb{C}[t, t^{-1}] \) of Laurent polynomials over \( \mathbb{C} \).
(i) Give a rigorous proof that \( R \) is isomorphic to \( \mathbb{C}[x, y]/(xy - 1) \).
(ii) Show that every ideal in \( R \) is a principal ideal.
(iii) Give an example (with proof) of a non-zero prime ideal in \( R \).
Proof (i). Firstly, we want to define a map \(\varphi: \mathbb{C}[x, y]\to \mathbb{C}[t, t^{-1}]\) such that \[ \varphi(f(x, y)) = f(t, t^{-1}). \] Hence, we want to show tht \(\varphi\) is a ring homomorphism. Firstly, we can know that \(\varphi\) is well-define since if \(g(x, y) = h(x, y)\). then we will have \(g(t, t^{-1}) = h(t, t^{-1})\). Then, we can know that \(\varphi\) is a ring homomorphism since \[ \begin{align} \varphi(g(x, y) + h(x, y)) &= g(t, t^{-1}) + h(t, t^{-1}) = \varphi(g(x, y)) + \varphi(h(x, y)), \\ \varphi(g(x, y)h(x, y)) &= g(t, t^{-1})h(t, t^{-1}) = \varphi(g(x, y))\varphi(h(x, y)). \end{align} \] For any \(f(t, t^{-1})\in \mathbb{C}[t, t^{-1}]\), we can know that \[ f(t, t^{-1}) = \sum_{i = -n}^{m} a_i t^i,\text{ where }n, m\in \mathbb{N}. \] In general, lets assume that \(n\neq 0\). Then, we can get that \[ \begin{align} f(t, t^{-1}) &= \sum_{i = -n}^{m} a_i t^i = a_{-n}t^{-n} + \dots + a_{-1}t^{-1} + a_0 + a_1t + \dots + a_m t^m, \\ &= a_{-n}(t^{-1})^n + a_{-n+1}(t^{-1})^{n-1} + \dots + a_{-1}t^{-1} + a_0 + a_1t + \dots + a_m t^m, \\ \end{align} \] Then, we can see that when \[ f(x, y)= a_{-n}(y)^n + a_{-n+1}(y)^{n-1} + \dots + a_{-1}y + a_0 + a_1x + \dots + a_m x^m, \\ \] we have \[ \varphi(f(x, y)) = f(t, t^{-1}). \] It shows that \(\varphi\) is surjective. Now, we want to show that \(\ker(\varphi) = (xy - 1)\). Suppose that \(f(x,y)\in (xy - 1)\), then \(f(x, y) = g(x, y)(xy - 1)\) for some \(g(x, y)\in \mathbb{C}[x, y]\). Then, we can know that \[ \varphi(f(x, y)) = f(t, t^{-1}) = g(t, t^{-1})(t\cdot t^{-1} - 1) = g(t, t^{-1})(1 - 1) = 0. \] Thus, we showed that \((xy - 1)\subset \ker(\varphi)\). Suppose that \(f(x, y)\neq 0\) and \(f(x, y)\in \ker(\varphi)\).
3. Consider a field \( K \), and an irreducible polynomial \( f(x) \in K[x] \). Prove that if \( L \) is a field extension of \( K \) such that \([L : K] \) is relatively prime to the degree of \( f(x) \), then \( f(x) \) remains irreducible in \( L[x] \).
Proof. Suppose that \( [L : K] = m \) and \(f(x)\) is an irreducible polynomial over \(K\) where its degree is \(n\) and \(\gcd(n, m) = 1\). Suppose that \(\alpha\) is a root of \(f(x)\) in \(L\). Then, we can know that \(f(x)\) is a minimal polynomial of \(\alpha\). And \(K(\alpha) = K[x]/(f(x))\) is a field extension of \(K\), which has degree of \(n\) (i.e. \( [K(\alpha) : K] = n \)) since the minimal polynomial of \(\alpha\) is \(f(x)\), which has degree of \(n\). Since \(L\) is an extension of \(K\), if \(f(x)\) is irreducible in \(L[x]\), then we have \( [L(\alpha) : L] = n \). If \(f(x)\) is reducible in \(L[x]\), then we have \( [L(\alpha) : L] \lt n \). In general we have \( [L(\alpha) : L] \leq n \). According to \(\textbf{Tower Theorem}\), we have \[ [L(\alpha) : K]=[L(\alpha) : L][L : K] \leq n \cdot m. \] Now, for the other direction, we want to show that \([L(\alpha) : K]\geq mn\). Firstly, we have \([L(\alpha) : K]=[L(\alpha) : L][L : K]\), which implies that \(m=[L : K] \mid [L(\alpha) : K]\). Secondly, since \(L\) is an extension of \(K\), we can know that \(L(\alpha)\) is an extension of \(K(\alpha)\). Hence, we have \( [L(\alpha) : K]=[L(\alpha) : K(\alpha)][K(\alpha) : K]\), which implies that \(n=[K(\alpha) : K] \mid [L(\alpha) : K]\). Since \(\gcd(m, n)\cdot \text{lcm}(m, n)=mn\) and \(\gcd(m, n)=1\), we can know that \(\text{lcm(m, n)=mn}\). Since \(n\mid [L(\alpha) : K]\) and \(m \mid [L(\alpha) : K]\), we can know that \([L(\alpha) : K]\) is the common multiple of \(m\) and \(n\) and \(mn \mid [L(\alpha) : K]\), which implies that \(mn \leq [L(\alpha) : K]\). Thus, we showed that \[ [L(\alpha) : K]=mn. \] Since \([L(\alpha) : K]=[L(\alpha) : L][L : K]=[L(\alpha) : L]\cdot m=mn\), we can know that \([L(\alpha) : L]=n\). In that case, we can know that the minimal polynomial of \(\alpha\) over \(L\) has degree of \(n\), which is \(f(x)\). Since \(f(x)\) is the minimal polynomial of \(\alpha\) over \(L\), we can know that \(f(x)\) is irreducible in \(L[x]\). \(\blacksquare\)
Proof. We will prove it with mathematical induction on \( n \). When \(n = 1\), we know that there exists \( f \in V^* \) such that \( f(v_1) \neq 0 \) given that \( v_1 \neq 0 \). It is because if all \( f \in V^* \) satisfies \( f(v_1) = 0 \), then we have \( v_1 = 0 \), which is a contradiction. Now, we assume that the statement is true for \( n \) where \( n \geq 1 \) and we want to show that the statement is true for \( n + 1 \). Let \(f\in V^*\) such that \(f(v_i) \neq 0\) for all \(1 \leq i \leq n\). If \(f(v_{n+1})\neq 0\), then we are done. If \(f(v_{n+1}) = 0\), then we need to build a new \(h\in V^*\) such that \(h(v_i) \neq 0\) for all \(1 \leq i \leq n+1\). Firstly, we know that there exists \(g\in V^*\) such that \(g(v_{n+1}) \neq 0\) since \(v_{n+1} \neq 0\) from the base case. Now, we denote \(c_i\in F\) such that \(c_ig(v_i) = f(v_i)\). If \(g(v_i)\neq 0\) for \(1\leq i\leq n\), then we can know that \(c_i\) is unique. If \(g(v_i) = 0\) for some \(i\), then there exists no \(c_i\in F\) such that \(c_ig(v_i) = f(v_i)\) since \(f(v_i) \neq 0\). Then, we define \(C = \{c_{n_1}, \ldots, c_{n_k}\}\) where \(1 \leq n_1 \lt \ldots \lt n_k \leq n\). Hence, we can know that \(C\) is a finite set. Besides that, we can know there exists infinitely many \(c\in F\) such that \(cg(v_{n+1})\neq 0\). Hence, pick non-zero \(a\in F\) such that \(a\neq c_i\) for all \(c_i\in C\) and define \(h = f - a\cdot g\in V^*\). In that case, we can know that \(h(v_i) \neq 0\) for all \(1 \leq i \leq n+1\). \(\blacksquare\)
6. Let \( V \) be a finite dimensional vector space over the field \( F \) and suppose \( T : V \rightarrow V \) is a linear operator on \( V \).
Proof (i). Since \(\mu_T(x) = x^2 - x = x(x - 1)\), then we can know that \[ V = \ker(T) \oplus \ker(T - I), \] where \(\ker(T)\) and \(\ker(T - I)\) are \(T\)-invariant subspaces. Thus, we can see that \(\ker(T-I)\) is a \(T\)-invariant complement of \(\ker(T)\). \(\blacksquare\)
Solution (ii). Let \(T\) be a linear operator such that \(T = 0\).