# Group 3

### Our Polynomial

$f(x)=x^{4}-20x^{2}+16$

We know that $\alpha=\sqrt{3}+\sqrt{7}$ is a root in ${\mathbb C}$ and as $\alpha\in{\mathbb R}$ that $\alpha$ is a root in ${\mathbb R}$

### Factorise $f(x)$

#### Over ${\mathbb Q}$

Firstly we looked to see if our polynomial is irreducible over ${\mathbb Q}$ using "Eisenstein's criterion". For this to work we want a prime $p$ that divides $16$ and $20$ but does not divide $1$ and also $p^{2}$ does not divide $16$. The only prime number that divides $16$ and $20$ is $2$. However, $2^{2} = 4$ divides $16$ so we cannot apply Eisenstein's to show the function is irreducible over ${\mathbb Q}$.

For $f(x)$ to be reducible, we need to find a polynomial $g(x)$ and a polynomial $h(x)$ such that $f(x)=g(x)h(x)$ and $deg(g(x))$and $deg(h(x))\neq 0$. When know that the $deg( f(x))=4$ and that $deg (f(x))=deg (g(x)) + deg (h(x))$. So we can assume without loss of generality that for $f(x)$ to be reducible in ${\mathbb Q}$ then the $deg(g(x))=1$ or $2$ and $deg(h(x))= 3$ or $2$. If $deg(g(x))=1$, then we know that $f(x)$ has a root in ${\mathbb Q}$.

So to disprove this, we need to show that $f(x)$ has no linear roots in ${\mathbb Q}$, to do this we will use the "root test" (also known as the rational root theorem). In this example $k|16$ and $l|1$ so the possibilities for roots are $k/l=\pm1, \pm2, \pm4, \pm8, \pm16$

So we then have to evaluate $f(x)$ for each of the possible values for $k/l$.

$k/l$ $f(k/l)$
$\pm1$ -3
$\pm2$ -48
$\pm4$ -48
$\pm8$ 2832
$\pm16$ 60432

As none of these values are equal to $0$ there is no linear factor.

We have shown that there is no linear factor when factorising over ${\mathbb Q}$. We now want to see whether our polynomial $f(x)=x^{4}-20x^{2}+16$ factorises into two polynomials in ${\mathbb Q}$ both of degree $2$.

$x^{4}-20x^{2}+16=(x^{2}+ax+b)(x^{2}+cx+d)$ with $a,b,c,d\in {\mathbb Q}$.

Multiplying out we get $x^{4}+(a+c)x^{3}+(b+ac+d)x^{2}+(cb+ad)x+bd$ this leaves us with four conditions to satisfy. These are:

• $1)$ $a+c=0$
• $2)$ $b+ac+d=-20$
• $3)$ $bc+ad=0$
• $4)$ $bd=16$

These four conditions lead to two cases:

• Case $1.$ $c=0 \Rightarrow a=0$
• Case $2.$ $c\neq 0 \Rightarrow b-d=0 \Rightarrow b=d$

Case $1:$

If $a=0$

$a=-c=0$

$b+d=-20$

$b=-20-d$

$\Rightarrow (-20-d)d=16$

$-20d-d^{2}=16$

$d^{2}+20d+16=0$

$\Rightarrow$ $d = \frac {-20 \pm \sqrt{400 - 64}}{2}$ $=\frac{-20 \pm 4\sqrt{21}}{2}$ $=-10\pm 2\sqrt{21}$

This leads to a contradiction as there are no solutions to this equation that are elements in ${\mathbb Q}$.

Case $2:$

If $d-b=0$

$\Rightarrow b=d$

$-a^{2}+2b=-20$

$b^{2}=16 \Rightarrow b=\pm4$

$a^{2}=2b+20$

If $b=-4$ then $a^{2}=12$ $\Rightarrow$ $a=\pm \sqrt{12}$

If $b=4$ then $a^{2}=28$ $\Rightarrow$ $a=\pm \sqrt{28}$

This again leads to a contradiction as neither values of $a$ are elements in ${\mathbb Q}$.

So $f(x)$ can't be reduced in ${\mathbb Q}$. This is due to the fact that it does not have any linear factors and does not factorise into $2$ polynomials $f(x)$ and $g(x)$ both of degree $2$.

#### Over ${\mathbb R}$ and ${\mathbb C}$

After "polynomial division" ie, dividing $f(x)$ by $x-\sqrt{3}-\sqrt{7}$ we get $x^{3}+(\sqrt{3}+\sqrt{7})x^{2}+(-10+2\sqrt{21})x+4\sqrt{7}-4\sqrt{3}$ and so $f(x)=x^{4}-20x^{2}+16$ is reducible in ${\mathbb R}$ and ${\mathbb C}$. This shows that a linear factor exists in ${\mathbb R}$, but we want to see if we can reduce it further. After showing that our root is a linear factor, naturally we looked to see if $x +\sqrt{3} +\sqrt{7}$ was also a root.

After another "polynomial division" dividing by $x +\sqrt{3} +\sqrt{7}$. This left us with a quadratic $x^{2} - 10 +2\sqrt{21}$. We then used the quadratic formula to find the remaining two roots of the equation:

$\frac {-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac {-0\pm\sqrt{0^{2}-4(-10+2\sqrt{21})}}{2}$
$= \pm\sqrt{10-2\sqrt{21}}$

Hence, we have shown that our polynomial $f(x)$ is reducible over ${\mathbb R}$ and therefore reducible over ${\mathbb C }$. We can express $f(x)$ in the following form:

$f(x) = (x-\sqrt{3}-\sqrt{7})(x+\sqrt{3}+\sqrt{7})(x+\sqrt{10-2\sqrt{21}})(x-\sqrt{10-2\sqrt{21}})$

The above shows the factorisation of $f(x)$ as a product of irreducible polynomials over ${\mathbb R}$. This is also the same factorisation over ${\mathbb C }$.

### Reducible over ${\mathbb Z}_{p}$

We are interested in whether our function $f(x)=x^{4}-20x^{2}+16$ is irreducible over ${\mathbb Z}_{p}$ and reducible over ${\mathbb Z}_{q}$ where $p,q$ are prime integers. In order to find out we use the root test starting for $p=2$ :

$x$ $f(x)$ $f(x)\in{\mathbb Z}_{2}$
0 16 0
1 -3 1

As we can see from the table above the function has a root ($f(0)=0$) in ${\mathbb Z}_{2}$ . Therefore ${\mathbb Z}_{q}$ is reducible for the prime $q=2$. Next we do the same for $p=3$:

$x$ $f(x)$ $f(x)\in{\mathbb Z}_{3}$
0 16 1
1 -3 0
2 -48 0

This time we can see that the function has a root in ${\mathbb Z}_{3}$ . However, we are looking for a prime that the function is irreducible over. Next $p=5$:

$x$ $f(x)$ $f(x)\in{\mathbb Z}_{5}$
0 16 1
1 -3 2
2 -48 2
3 -83 2
4 -48 2

On this occasion we see that $f(x)$ does not have a root in ${\mathbb Z}_{5}$ . Thus the primes we were looking for are $p=5$ and $q=2$ or $q=3$ for $f(x)$ irreducible over ${\mathbb Z}_{p}$ and reducible over ${\mathbb Z}_{q}$.

### Using $f(x)$ constructing a field $E$ of the form ${\mathbb Q}[x]/I$ for some ideal $I$

We know that $f(x)$ is irreducible over ${\mathbb Q}[x]$ and hence we know that $(f(x))$ is a maximal ideal of ${\mathbb Q}[x]$.

Looking at our field $E$ which equals:

${\mathbb Q}[x]/(x^{4}-20x^{2}+16)$

Now if we let $\theta=x+I$ where $\theta\in {\mathbb Q}[x]/I$. Then $f(\theta)=0 \Rightarrow \theta^{4}-20\theta^{2}+16=0$ so $\theta^{4}=20\theta^{2}-16$.

We can write any polynomial of $\theta$ in the form

$a+b\theta+c\theta^{2}+d\theta^{3}$

i.e. with the basis {${1,\theta,\theta^{2},\theta^{3}}$} and hence the $dim_{\mathbb Q}(E)=4$

We then wanted to see how calculations work within this field $E$. We already know $x^{4}+I=20x^{2}-16+I$ and we know wanted to calculate $\theta^{5}$, these are the following calculations required to calculate $\theta^{5}:$

$\theta^{5}=\theta^{4}\theta$
$=(x^{4}+I)(x+I)$
$=(20x^{2}-16+I)(x+I)$
$=20x^{3}-16x+I$

### Contructing a field with finite elements using $f(x)$

To construct a field with size $5^{4}=625$, we start with our polynomial $f(x)=x^{4}-20x^{2}+16$ which we have already shown is irreducible in ${\mathbb Z}_{5}$ then a field with $625$ elements is given by:

${\mathbb Z}_{5}[x]/( x^{4}-20x^{2}+16)$

Its elements have the form $ax^{3}+bx^{2}+cx+d$ where $a,b,c$ and $d\in {\mathbb Z}_{5}$ and the multiplication is defined by $x^{4}-20x^{2}+16=0$, or by rearranging $x^{4}=4$.

### Algebraic Elements of $f(x)$

$\alpha=\sqrt{3}+\sqrt{7}$
$\alpha^{2}=(\sqrt{3}+\sqrt{7})^{2}=10+2\sqrt{21}$
$\alpha^{2}-10=2\sqrt{21}$
$(\alpha^{2}-10)^{2}=(2\sqrt{21})^{2}$
$\alpha^{4}-20\alpha^{2}+100=84$
$\alpha^{4}-20\alpha^{2}+16=0$

$\Rightarrow f(\alpha)=0$

$f(x)\in{\mathbb Q}[x]\Rightarrow\alpha$ is an algebraic number

$\Rightarrow {\mathbb R}\supseteq{\mathbb Q}$ is a field extension and $\alpha\in{\mathbb R}$ is algebraic over ${\mathbb Q}$

$\Rightarrow$ Our field is ${\mathbb Q}$.

Hence for our field ${\mathbb Q}$ and polynomial $f(x)$ we have that this polynomial is the "minimal polynomial" of $\alpha$ over ${\mathbb Q}$.