Tutorial 1 : Algebraic Extensions and Ruler and Compass Constructions

1. Let

be an extension of degree

(a) For any $a \in K,$ prove that the map $\mu_a: K \rightarrow K$ defined by $\mu_a(x)=ax$ for all $x \in K,$ is a linear transformation of the

-vector space

Show that

is isomorphic to a subfield of the ring $F^{n \times n}$ of $n \times n$ matrices with entries in

(b) Prove that

is a root of the characteristic polynomial of $\mu_a.$ Use this procedure to find monic polynomials satisfied by $\sqrt[3]{2}$ and $1+\sqrt[3]{2}+\sqrt[3]{4}.$

(a) It is clear that the map $\mu_a: K \rightarrow K$ defined by $\mu_a(x)=ax$ for all $x \in K,$ is a linear transformation of the -vector space Let be the ring of linear maps from to Define $\phi :K\rightarrow L(K,K)$ by $\phi (a)=\mu _a.$ Then $\mu_{ab}(x)=abx=a(\mu _b(x))=\mu _a\mu _b(x)$ and $\mu _{a+b}=(a+b)x=ax+bx=(\mu _a+\mu _b)(x)$ . Hence $\phi$ is a ring homomorphism and $Ker\phi =\{a\mid \mu _a=0\}.$ But $\mu _a(x)=ax=0 \: \forall x\in K\Rightarrow a=0$ (take ). Hence $\phi$ is injective. Thus $K\simeq$ a subfield of $F^{n \times n}$ since $L(K,K)\simeq F^{n \times n}.$
(b) Let $f(x) \in F[x]$ be the characteristic polynomial of $\mu_a.$ By the Cayley-Hamilton Theorem, $f(\mu_a)=0.$ This shows that is a root of Let $\beta =\sqrt[3]{2}.$ Then a $\mathbb{Q}$ -basis of $\mathbb{Q}(\beta)$ is given by $B=\{1,\beta ,\beta ^2\}.$ Consider the $\mathbb{Q}$ -isomorphic $\mu_\beta:\mathbb{Q}(\beta )\rightarrow\mathbb{Q}(\beta ).$ Thus $\mu _\beta (1)=\beta ,\mu _\beta(\beta )=\beta ^2,\mu_\beta(\beta^2)=\beta ^3=2.$ Hence the matrix of $\mu_\beta$ with respect to the basis is:

$\displaystyle M=\begin{bmatrix} 0 & 0 & 2\\ 1 & 0 & 0\\ 0 & 1 & 0 \end{bmatrix} \;\;$ and $\displaystyle \; M(\lambda )=\begin{vmatrix}-\lambda && 0 && 2\\ 0 && -\lambda && 0\\ 0 && 1 && -\lambda \end{vmatrix} = -\lambda ^3+2.$

(c) Let us find $\;$ irr $\;(\alpha ,\mathbb{Q}$ ), where $\alpha =1+\beta +\beta ^2.$ Since $\displaystyle \mu_\alpha (1)=1+\beta+\beta ^2 \;\; ,\mu _\alpha (\beta )=\beta +\beta ^2+2 \;\; ,\mu _\alpha(\beta^2)=\beta^2+2+2\beta.$

Therefore

$\displaystyle M(\mu_\alpha )=\begin{bmatrix}1 && 2 && 2\\ 1 && 1 && 2\\ 1 &... ...1-\lambda && 2 && 2\\ 1 && 1-\lambda && 2 \\ 1 && 1&& 1-\lambda\end{vmatrix}.$

$\displaystyle C(\lambda)$

$\displaystyle =$

$\displaystyle -\lambda^3+3\lambda^2+3\lambda+1$

2. Prove that

is not a sum of squares in the field $\mathbb{Q}(\beta)$ where $\beta = \sqrt[3]{2}\; e^{2 \pi i/3}.$

Let $\phi:\mathbb{Q}(\beta \zeta _3)\rightarrow\mathbb{Q}(\beta)$ be the $\mathbb{Q}$ -automorphism given by

$\phi(\beta\zeta_3)=\beta.$

If $-1=\Sigma x_i^2$ in $\mathbb{Q}(\beta\zeta_3)$ ,then $\phi(-1)=-1=\Sigma \phi(x_i)^2$ in $\mathbb{Q}(\beta)\subseteq\mathbb{R}.$ This is a contradiction.

3. Let

be a field and

be an indeterminate. Find the irreducible polynomial of

over

where

Clearly

is a root of $T^3-(T+1)y\;\in k[y][T].$ Since

is transcendental over

is an irreducible element of

Therefore the polynomial

is irreducible by Eisenstein criterion. Hence $\;$ irr $\;(x, k(y)=g(T).$

4. Find an algebraic extension

of $\mathbb{Q}(x)$ such that the polynomial $f(y)=y^2 - x^3/(x^2+1) \in \mathbb{Q}(x)[y]$ has a root in

Let $p(x)/q(x)\in \mathbb{Q}(x)$ be a root of

Then we have

and hence

But then deg of L.H.S. is even and deg of R.H.S. is odd.Thus

is irreducible over $\mathbb{Q}(x).$ Hence

is a maximal ideal in $\mathbb{Q}(y)[x].$ Hence $\mathbb{Q}(x)[y]/(f(y))$ is an algebraic extension of $\mathbb{Q}(x)$ containing a root

5. The construction of a regular

-gon amounts to the construction of the real number $\alpha=\cos(2\pi/7).$ Show that $\alpha$ is a root of

Hence conclude that a regular

-gon is not constructible by ruler and compass.

Consider the diagram of fields:

$\displaystyle \xymatrix{ & \mathbb{Q}(\zeta_7) \ar@{-}[d]^2 & \\ & \mathbb{Q}( \zeta_7+\frac{1}{\zeta_7}=2\cos2\pi/7) \ar@{-}[d]^3 & \\ & \mathbb{Q}&\\ }$

Since $\;$ irr $\;(\zeta_7,\mathbb{Q}) = \Phi_7(x)=x^6+x^5+x^4+x^3+x^2+x+1,$ we have $[\mathbb{Q}(\zeta_7):\mathbb{Q}(\cos2\pi/7)] = 2.$ Hence $\deg \;$ irr $\;(\cos2\pi/7,\mathbb{Q}) = 3.$ As

$\displaystyle \frac{1}{x^3}\phi_7(x)$	$\displaystyle =$	$\displaystyle x^3+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3},$
	$\displaystyle =$	$\displaystyle (x+\frac{1}{x})^3-2(x+\frac{1}{x})+(x+\frac{1}{x})^2-1$

Let $y = x+\frac{1}{x}.$ The $\zeta_7+\frac{1}{\zeta_7} = 2\cos2\pi/7 = 2\alpha$ satisfies the equation

. Hence $[\mathbb{Q}(\cos2\pi/7):\mathbb{Q}] = 3 \neq 2^n\;\forall\; n\in\mathbb{N}$ . Thus $\cos2\pi/7$ is not constructible by ruler and compass.

6. Show that an angle of

degrees, $n \in \mathbb{N},$ is constructible if and only if $3 \mid n.$

First we show that $1^\circ$ is not constructible. Suppose it is,then a regular

-gon is constructible.But $3^2 \mid 360$ .Hence $1^\circ$ is not constructible. Since

-gon is constructible. Thus $3^\circ$ is constructible. Suppose that $n^\circ$ is constructible and $3 \nmid n,$ then

for some $x,y \in \mathbb{Z}$ . Hence $1^\circ$ is constructible.This is a contradiction. Hence $3 \mid n$ . Conversely,suppose $3 \mid n$ , then

and since $3^\circ$ is constructible, $n^\circ$ is constructible.