Homework 1

Homework 1.pdf

1. (30 pts) Decide if the following statements are true or false and justify your answer:

   1. Let \(F\) be a field and \(a \in F\). If there exists a natural number \(n\) such that \(a^n = 0\) (where \(a^n = a \cdot a \cdot \dots \cdot a\), \(n\) times), then \(a = 0\).

      True

      Assume we have \(a^n = 0\), then if \(a=0\), we are done

      If \(a\neq 0\), then \(\exists a^{-1}\in\mathbb{F}\) (multiplicative inverse), then \(a^{-1}\cdot a^{n} = a^{-1}\cdot0\Rightarrow a^{n-1}=0\)​

      Iteratively, we will have \(a^{n-\left(n-1\right)}=0\Rightarrow a=0\)

      Thus \(a=0\)​ 2. Let \(F\) be a field and \(a \in F\). If there exists a natural number \(n\) such that \(n \cdot a = 0\) (where \(n \cdot a = a + a + \dots + a\), \(n\) times), then \(a = 0\).

      False

      Consider \(\mathbb{F}=\mathbb{Z}_3\), then then we know \(n\cdot a=3\cdot 1=0\), but here \(a=1\neq 0\)​ 3. Let \(F\) be a field. If there exists \(a \in F\), \(a \neq 0\) and a natural number \(n\) such that \(n \cdot a = 0\), then \(n \cdot x = 0\) for every \(x \in F\).

      True

      Since \(a \in F\), \(a \neq 0\), then \(\exists a^{-1}\in F\) (multiplicative inverse) and \(1=a\cdot a^{-1}\)​

      Consider \(n\cdot x=n\cdot x\cdot 1=n\cdot x\cdot (a\cdot a^{-1})\overset{\text{commutativity}}{=} n\cdot a\cdot x\cdot a^{-1}\)​

      And we know \(n \cdot a = 0\), thus \(n\cdot x=0\cdot x\cdot a^{-1}\), thus \(n\cdot x=0\)​

2. (20 pts) Consider \(G_n = \{ w \in \mathbb{C} : w\) is an \(n\)-th root of unity\(\}\). Reduce the following expressions:

   1. \(w^{22}+ (w^{2} + w)^{2} + \bar{w}^{106}- w^{31}+ w^{138}\) for any \(w \in G_7\).

      We know \(w \in G_7\), then by definition we have \(w^7=1\) and then \(\overline{w^7}=\bar 1\Rightarrow \bar w ^7=1\)

      Thus \(w^{22}+(w^{2}+w)^{2}+\bar{w}^{106}-w^{31}+w^{138}=w\cdot w^{3\times7}+w^{4}+w^{2} +2w^{3}+\bar{w}\cdot\bar{w}^{15\times7}-w^{3}\cdot w^{4\times7}+w^{5}\cdot w^{19\times7}\)

      \(=w+w^{4}+w^{2}+2w^{3}+\bar{w}-w^{3}+w^{5}=w+\bar{w}+w^{2}+w^{3}+w^{4}+w^{5}\)​

      Also we know \(w\cdot \bar w=|w|^2=1\), then \(\bar{w}=w^{-1}\)​

      \(=\) \(w+w^{-1}+w^{2}+w^{3}+w^{4}+w^{5}\) and we know \(w^{-1}=w^{-1}\cdot w^{7}=w^{6}\)​

      \(=w+w^{2}+w^{3}+w^{4}+w^{5}+w^{6}\)​

      Since \(w^{7}=1\Rightarrow w^{7}-1=0\Rightarrow\left(w-1\right)\left(w^{6}+\ldots+1\right )=0\), then \(w=1\) or \(w^6+...+1=0\)​

      Finally, if \(w=1\), the result is \(6\). If \(w\neq 1\), the result is \(-1\)​ 2. \(w^{-51}- \bar w^{74}+ w^{33}\cdot w^{-355}+ \overline{w^{-127}}\) for any \(w \in G_5\).

      We know \(w \in G_{5}\), then by definition we have \(w^{5}=1\) and then \(\overline{w^5}=\bar{1}\Rightarrow\bar{w}^{5}=1\)​

      Then \(w^{-51}-\bar{w}^{74}+w^{33}\cdot w^{-355}+\overline{w^{-127}}=w^{5\cdot11}\cdot w^{-51}-\bar{w}^{14\cdot5+4}+w^{6\cdot5+3}\cdot w^{-350-5}+\overline{w^{-125}}\cdot \overline{w^{-2}}\)​

      \(=w^{4}-\bar{w}^{4}+w^{3}\cdot w^{-5}+\bar{w}^{-2}=w^{4}-w+w^{3}+\bar{w}^{3}=w^{4} -w+w^{3}+w^{-3}=w^{4}-w+w^{3}+w^{2}\)​

      Since \(w^{5}=1\Rightarrow w^{5}-1=0\Rightarrow\left(w-1\right)\left(w^{4}+\ldots+1\right )=0\), then \(w=1\) or \(w^{4}+...+1=0\)

      If \(w=1\), then the result is \(2\). If \(w\neq 1\), then \(=w^{4}+w^{3}+w^{2}-w=-w-1-w=-1-2w\)​ 3. \(\sum_{k=8}^{19}w^{2k+1}\) for any \(w \in G_{12}\)​

      We know \(w \in G_{12}\), then by definition we have \(w^{12}=1\)​

      Then \(\sum_{k=8}^{19}w^{2k+1}=w^{17}+w^{19}+\cdots+w^{39}=w^{5}+w^{7}+w^{9}+w^{11}+w+w ^{3}+\ldots+w^{11}+w+w^{3}\)​

      \(=2(w+w^{3}+w^{5}+w^{7}+w^{9}+w^{11})=2w(1+w^{2}+w^{4}+w^{6}+w^{8}+w^{10})\)

      Since \(w^{2}\in G_6\), then \((w^{2})^{6}=1\Rightarrow\left(w^{2}-1\right)\left(\left(w^{2}\right)^{5}+\cdots+ 1\right)=0\)​

      If \(w=1\), then the result is \(12\).

      If \(w=-1\), then the result is \(-12\)​

      If \(w\neq \pm 1\), then the result is \(0\)​

3. (20 pts) Find a polynomial \(p(x) \in \mathbb{R}[x]\) of minimal degree satisfying the following properties:

   • The solutions of \(z^{2}+ z = \bar z\) are roots of \(p(x)\).
   • The sum of all the roots of \(p(x)\) is equal to zero.
   • One of the real roots of \(p(x)\) has multiplicity \(m = 2\).

   Let \(z=a+bi\), then \(z^{2}+z=\bar{z}\Rightarrow a^{2}-b^{2}+2abi+a+bi=a-bi\Rightarrow a^{2}-b^{2}+2ab i+2bi=0\Rightarrow a^{2}-b^{2}=0,2b\left(a+1\right)=0\)

   Thus \(a=\pm b\) and \(b=0\) or \(a=-1\), then \(z=0,-1+i,-1-i\)

   Thus \(x(x+1-i)(x+1+i)=x(x^{2}+(1-i)(1+i)x+(1-i)(1+i))=x(x^{2}+2x+2)=x^{3}+2x^{2}+2x=p(x)\)

   But here the sum of roots are \(-2\), thus we need another root \(x=2\), then we need to multiply \((x-2)\) in it

   And one of the real roots of \(p(x)\) has multiplicity \(m = 2\), thus we can multiply \(x\) in it

   Finally, \(p\left(x\right)=x^{2}\left(x-2\right)\left(x^{2}+2x+2\right)\)

4. Let \(p(x) = 6x^{5}- 25x^{4}+ 39x^{3}- 2x^{2}- 6x\). Write \(p(x)\) as a product of irreducible polynomials over the following fields:

   1. \(\mathbb{R}\).

      We know roots of \(p(x)=6x^{5}-25x^{4}+39x^{3}-2x^{2}-6x=x\cdot q\left(x\right)\) in \(\mathbb{Q}\), then use theorem in lecture

      \(\left\{\frac{a}{b}:\gcd(a,b)=1,q\left(\frac{a}{b}\right)=0\right\}\subseteq\left \{\frac{a}{b}:a\mid6,b\mid6\right\}\)​\(=\left\{\frac{a}{b}:\gcd(a,b)=1,a=\pm1,\pm2,\pm3,\pm6,b=\pm1,\pm2,\pm3,\pm6\right \}\)

      \(=\left\{\pm1,\pm2,\pm3,\pm6,\pm\frac{1}{2},\pm\frac{3}{2},\pm\frac{1}{3},\pm\frac{2}{3} ,\pm\frac{1}{6}\right\}\)

      \(q(\frac{1}{2})=0,q(-\frac{1}{3})=0\), then \(p(x)=6x^{5}-25x^{4}+39x^{3}-2x^{2}-6x=6x\cdot\left(x-\frac{1}{2}\right)\left(x+\frac{1}{3} \right)\cdot h\left(x\right)\)​

      Here \(h(x)=x^{2}-4x+6\) has no root in \(\R\)

      Thus \(p(x)=6x\cdot\left(x-\frac{1}{2}\right)\left(x+\frac{1}{3}\right)\left(x^{2}-4x+6 \right)\) in \(\R\)​ 2. \(\mathbb{C}\).

      \(h(x)=x^{2}-4x+6\) has roots: \(x=2\pm i\sqrt{2}\), then \(x^{2}-4x+6=(x-(2+i\sqrt{2}))(x-(2-i\sqrt{2}))\)​

      Thus \(p(x)=6x(x-\frac{1}{2})(x+\frac{1}{3})(x-(2+i\sqrt{2}))(x-(2 -i\sqrt{2}))\)​