1. Find all complex numbers z so that \frac{z^3+8}{z^3-8}=i. Write your solutions in the standard form, and use square roots like \frac{\sqrt 3}{2}.

  • \frac{z^3+8}{z^3-8}=i \Rightarrow z^3=\frac{-8-8i}{1-i}=\frac{8\sqrt{2}e^{\frac{5\pi}{4}i}}{\sqrt{2}e^{-\frac{\pi}{4}i}}=8e^{\frac{3\pi}{2}i} \Rightarrow z=\{2e^{\frac{\pi}{2}i+\frac{2k\pi}{3}i}|k=0,1,2\}=\{2i,-\sqrt{3}-i,\sqrt{3}-i\}.

2. Let \mathbb Q[i][\sqrt 2]=\{\alpha+\beta\sqrt 2:\alpha,\beta\in\mathbb Q[i]\} and \mathbb Q[i,\sqrt 2]=\{a+bi+c\sqrt 2+di\sqrt 2:a,b,c,d\in\mathbb Q\}. Show that:

(1) \mathbb Q[i][\sqrt2]=\mathbb Q[i,\sqrt2].

  • (x \in \mathbb{Q}[i][\sqrt{2}] \Leftrightarrow x=\alpha+\beta\sqrt{2},\alpha,\beta\in\mathbb{Q}[i] \xLeftrightarrow{\alpha=a+bi,\beta=c+di} x=a+bi+c\sqrt{2}+di\sqrt{2},a,b,c,d \in \mathbb{Q} \Leftrightarrow x \in \mathbb{Q}[i,\sqrt{2}])\Rightarrow\mathbb{Q}[i][\sqrt{2}]=\mathbb{Q}[i,\sqrt{2}].

(2) \mathbb Q[i,\sqrt2] is a field.

  • \because \mathbb{F} is a field\Rightarrow \mathbb{F}[x] is a field.
    \therefore \mathbb{Q} is a field.\Rightarrow\mathbb{Q}[i] is a field.\Rightarrow\mathbb{Q}[i][\sqrt{2}] is a field.\xRightarrow{\mathbb{Q}[i][\sqrt{2}]=\mathbb{Q}[i,\sqrt{2}]}\mathbb{Q}[i,\sqrt{2}] is a field.

3. Show that x^2+x+1|x^{3m}+x^{3k+1}+x^{3n+2} in \mathbb Q[x] for any m,n,k\in\N.

  • x^{3m}+x^{3k+1}+x^{3n+2}=\sum_{i=0}^{m-1}(x^{3(i+1)}-x^{3i})+1+\sum_{j=0}^{k-1}(x^{3(j+1)+1}-x^{3j+1})+x+\sum_{t=0}^{n-1}(x^{3(t+1)+2}-x^{3t+2})+x^2=(x^3-1)\sum_{i=0}^{m-1}x^{3i}+(x^3-1)\sum_{j=0}^{k-1}x^{3j+1}+(x^3-1)\sum_{t=0}^{n-1}x^{3t+2}+(1+x+x^2)=(x^2+x+1)[(x-1)(\sum_{i=0}^{m-1}x^{3i}+\sum_{j=0}^{k-1}x^{3j+1}+\sum_{t=0}^{n-1}x^{3t+2})＋1]\Rightarrow x^2+x+1|x^{3m}+x^{3k+1}+x^{3n+2}.

4. Let f(x)\in\mathbb Q[x] with \deg(f(x))=2n+1 where n\in\N. If x^n+1|f(x)+1 and x^n-1|f(x)-1, find all f(x).

  • g(x):=f(x)-x^n\therefore (x_0^n+1=0\Rightarrow g(x_0)=f(x_0)+1=u(x_0)(x_0^n+1)=0)\wedge (x_0^n-1=0\Rightarrow g(x_0)=f(x_0)-1=v(x_0)(x_0^n-1)=0).
    \therefore x^n+1|g(x)\wedge x^n-1|g(x)\xRightarrow{(x^n+1,x^n-1)=1}(x^n+1)(x^n-1)|g(x).
    \because \deg(g(x))=\deg(f(x))=2n+1\therefore g(x)=a(x-t)(x^n-1)(x^n+1),a\neq0,t\in\mathbb{Q}\therefore f(x)=a(x-t)(x^n+1)(x^n-1)+x^n,a\neq0,t\in\mathbb{Q}.

5. Let f(x) \in \R[x] such that f(a) \ge 0 for all a \in \R. Show that there are g(x),h(x) \in \R[x] such that f(x) = g(x)^2 + h(x)^2. (This was proved by David Hilbert (1862-1943) in 1893. A generalization is called Hilbert’s seventeenth problem which was solved by Emil Artin (1898-1962) in 1927.)

  • Given f(x)=k\prod_{i=1}^m(x-a_i)^{l_i}\cdot\prod_{j=1}^m(x^2+p_jx+q_j),k\neq0,a_i are pairwise distinct, we can find that f(x)\geq0 \forall x\Rightarrow2|l_i,k>0,q_j-\frac{p_j^2}{4}>0\Rightarrow f(x)=k\prod_{i=1}^m(x-a_i)^{l_i}\cdot\prod_{j=1}^m[(x+\frac{p_j}{2})^2+(\sqrt{q_k-\frac{p_j^2}{4}})^2].
    Noticed that (c^2+d^2)=(ac+bd)^2+(ad-bc)^2, and a,b,c,d\in\mathbb{R}[x]\Rightarrow ac+bd,ad-bc\in\mathbb{R}[x], so by applying the formula on \prod_{j=1}^m[(x+\frac{p_j}{2})^2+(\sqrt{q_k-\frac{p_j^2}{4}})^2], we can finally get f(x)=k\prod_{i=1}^m(x-a_i)^{l_i}\cdot(u^2(x)+v^2(x))=[\sqrt k\prod_{i=1}^m(x-a_i)^{\frac{l_i}{2}}u(x)]^2+[\sqrt k\prod_{i=1}^m(x-a_i)^{\frac{l_i}{2}}v(x)]^2, which exactly is what we want to prove.

6. Let n\in\N and a_1,a_2\cdots,a_n\in\Z be pairwise distinct. Show that f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)-1 cannot be written as a product of two positive degree polynomials in \Z[x]. Consequently, f(x) is irreducible over \mathbb Q. (This was proved by Issai Schur (1875-1941) in 1908.)

  • Suppose that f(x)=g_1(x)g_2(x),\deg g_i(x)<n,g_i(x)\in \mathbb{Z}[x],i=1,2.
    \therefore f(a_j)=-1=g_1(a_j)g_2(a_j),j=1,2,\dots,n.
    \because g_1(a_j),g_2(a_j)\in\mathbb{Z} \therefore g_1(a_j)+g_2(a_j)=0.
    As a_1,a_2,\dots,a_n are pairwise distinct roots of g_1(x)+g_2(x), and \deg(g_1(x)+g_2(x))<n, therefore g_1(x)+g_2(x)=0,f(x)=-g_1^2(x).
    As the leading coefficient of f(x) is 1, which >0, proves that the supposition should be false.
    Therefore, f(x) cannot be written as a product of two positive degree polynomials in \mathbb{Z}[x].
    Consequently, f(x) is irreducible over \mathbb{Q}.

7. Let f(x) = x^6 +x^5 −8x^4−10x^3 −19x^2 +x+2. It is known that \sqrt2+\sqrt3 is a zero of f(x). Find all zeros of f(x) in \mathbb C.

  • Notice that f(\sqrt{2}+\sqrt{3})=f(\sqrt{2}-\sqrt3)=f(-\sqrt2+\sqrt3)=f(-\sqrt2-\sqrt3)=0.
    Apparently, f(-\frac{1}{2}-\frac{\sqrt7}{2}i)=f(-\frac{1}{2}+\frac{\sqrt7}{2}i)=0.

8. Let p be a prime and A\in M_n(\mathbb Q) satisfying A^p = I_n. Show that for any \lambda_0 \in \mathbb C and any integer 0 < k <p, if \lambda_0I_n−A is a singular matrix, then \lambda_0^kI_n − A is also singular.

  • A^p=I_n\Rightarrow f(\lambda)=x^p-1 is an annihilating polynomial of A.
    Notice that f(\lambda)=(x-1)\sum_{i=0}^{p-1}x^i, and \sum_{i=0}^{p-1}x^i is irreducible over \mathbb{Q}.
    So it can be divided into three cases:
    1. m(\lambda)=x-1\Rightarrow\lambda_0=1\Rightarrow\lambda_0^k=1\Rightarrow\lambda_0^k I_n-A is singular.
    2. m(\lambda)=\sum_{i=0}^{p-1}x^i\Rightarrow\lambda_0\in{e^{2\pi i\frac{t}{p}}|t\in\mathbb{Z}_p^\times}\Rightarrow\lambda_0^k\in{e^{2\pi i\frac{tk}{p}}|t\in\mathbb{Z}_p^\times}={e^{2\pi i\frac{t}{p}}|t\in\mathbb{Z}_p^\times}\Rightarrow\lambda_0^k I_n-A is singular.
    3. m(\lambda)=x^p-1\Rightarrow\lambda_0\in{e^{2\pi i\frac{t}{p}}|t\in\mathbb{Z}_p}\Rightarrow\lambda_0^k\in{e^{2\pi i\frac{tk}{p}}|t\in\mathbb{Z}_p}={e^{2\pi i\frac{t}{p}}|t\in\mathbb{Z}_p}\Rightarrow\lambda_0^k I_n-A is singular.
    All in all, for all \lambda_0\in\mathbb{C},k\in\mathbb{Z}_p^\times, if \lambda_0I_n-A is a singular matrix, then \lambda_0^kI_n-A is also singular.