# 15/16: Group 3

## Contents

## \(f(x)=x^{4}-16x^{2}+4\)

We were asked to study this polynomial given the roots $\alpha=\sqrt{3}+\sqrt{5}$ $\in$ \({\mathbb C}\). We will looks at its reducibility/irreducibility over different fields. We will also look at field extensions of $f(x)$ looking at its properties.

## Reducibility

### Is $f(x)$ reducible in \({\mathbb Q}\)?

To begin with, we find out whether \(f(x)\) is irreducible over \({\mathbb Q}\).

We can use the Rational Root Theorem which states that:

If \(f(x)=a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}+a_{n}x^{n} \in{\mathbb Q}\) has integer coefficients, \(f(x)\) has a root \(\frac{k}{l} \in{\mathbb Q}\) with \(gcd(k,l)=1\) and \(k,l\in{\mathbb Z}\) only if \(k \mid a_{0}\) and \(l \mid a_{n}\).

With \(f(x)=x^{4}-16x^{2}+4\), \(a_{0}=4\) and \(a_{n}=1\) so we need \(\frac{k}{l}\) where \(k \mid 4\) and \(l \mid 1\).

Thus, we can say the possible roots are:

\[4, \hspace{2mm}2, \hspace{2mm}1, \hspace{2mm}-4, \hspace{2mm}-2, \hspace{2mm}-1\]

As \(x^{4}\geq 0\) and \(x^{2}\geq 0\) for all values of \(x\) , we are only required to test the positive possible roots.

Next we plug the positive roots into \(f(x)\): \[f(1)=11 \hspace{5mm}f(2)=-44 \hspace{5mm}f(4)=4\]

We can observe that there are no roots of \(f(x)\) in \({\mathbb Q}\), thus a linear factor of \(f(x)\) in \({\mathbb Q}\) does not exist.

Our next step is to find a quadratic factor of \(f(x)\), which will be in the form\[(x^{2}+ax+b)(x^{2}+cx+d)\]

We can now expand this\[x^{4}+(a+c)x^{3}+(b+ac+d)x^{2}+(bc+ad)x+bd\]

Comparing this to \(f(x)\), we can devise the following equations\[1)bd=4 \hspace{10mm}2)bc+ad=0 \hspace{10mm}3)b+ac+d=-16 \hspace{10mm}4)a+c=0\]

If we refer to equation \(4)\)

\(a+c=0\)

we can rearrange to obtain\[a+c=0 \Rightarrow c=-a\]

We can then plug this into \(2)\)

\(bc+ad=0\) \( \Rightarrow b(-a)+ad=0\) \( \Rightarrow a(-b+d)=0\) \( \Rightarrow a(d-b)=0\)

Now we can say that\[a=0 \hspace{10mm}or \hspace{10mm}d-b=0\]

First we consider \(a=0\)

and we plug this into \(3)\)

\(b+(0)c+d=-16 \Rightarrow b+d=-16 \Rightarrow b=-16-d\)

Then we can plug this into \(1)\)

\(bd=4 \Rightarrow d(-16-d)=4 \Rightarrow d^{2}+16d+4=0\)

Then using the Quadratic Formula, we can solve to find \(d\)

\(d=-8+2\sqrt{15}\)\(\notin{\mathbb Q}\) or \(d=-8-2\sqrt{15}\)\(\notin{\mathbb Q}\)

Therefore \(a\neq0\)

We now consider \(d-b=0\)

\( \Rightarrow d=b \) and we plug this into \(1)\)

\( \Rightarrow d^{2}=4 \Rightarrow d=2\hspace{5mm}or \hspace{5mm} d=-2\)

Considering $b=d=-2$, we can plug this into \(3)\) which goes to\[ -2+ac-2=-16 \Rightarrow ac-4=-16 \Rightarrow ac=-12 \]

We know $c=-a \Rightarrow -a^{2}=-12 \Rightarrow a^{2}=12 \Rightarrow a=2\sqrt{3} \Rightarrow c=2\sqrt{3}$

Therefore $a \hspace{1mm}and \hspace{1mm} c \hspace{1mm}$\(\notin{\mathbb Q}\) so we cannot use this set of coefficients.

Consider $b=d=2$ we can plug this into \(3)\)

\(2+ac+2=-16 \Rightarrow ac+4=-16 \Rightarrow ac=-20\)

\(c=-a \Rightarrow a^{2}=20 \Rightarrow a=2\sqrt{5} \Rightarrow c=-2\sqrt{5}\)

Therefore $a \hspace{1mm}and \hspace{1mm} c \hspace{1mm}$\(\notin{\mathbb Q}\).

We have now considered all possibilities of \(a,b,c \hspace{1mm} and \hspace{1mm} d \), therefore there are no quadratic factors of $f(x)$ in \({\mathbb Q}\). We already know there are no linear factors so \(f(x)\) is irreducible over \({\mathbb Q}\).

__Minimal Polynomials__

We know that $f(x)$ is irreducible in \({\mathbb Q}\). It is also monic as the leading coefficient is 1. $f(x)$ has root $\alpha=\sqrt{5}+\sqrt{3}$. $\alpha$ is $\in$ \({\mathbb Q[\sqrt{3},\sqrt{5}]}\). Therefore $f(x)$ is the minimal polynomial of $\alpha$ which is in \({\mathbb Q[\sqrt{3},\sqrt{5}]}\) over \({\mathbb Q}\).

### Is $f(x)$ reducible in \({\mathbb R}\)?

We find out next whether $f(x)$ is irreducible over \({\mathbb R}\).

Above, we have showed that $f(x)$ has quadratic factors, which will be of the form\[(x^{2}+ax+b)(x^{2}+cx+d)\]

with 2 possible sets of coefficients\[a=2\sqrt{3} \hspace{10mm} b=-2 \hspace{10mm} c=-2\sqrt{3} \hspace{10mm} d=-2 \] \(\hspace{100mm}\) \(a=2\sqrt{5} \hspace{10mm} b=2 \hspace{10mm} c=-2\sqrt{5} \hspace{10mm} d=-2 \)

We know \(\alpha=\sqrt{3}+\sqrt{5}\) is in \({\mathbb R}\) is a root of $f(x)$. Therefore $(x-(\sqrt{3}+\sqrt{5}))$ is a linear factor of $f(x)$. So we know that one of our quadratic factors can be simplified further.

Using long division, we calculated that x^{2}-2\sqrt{5}+2=(x-(\sqrt{3}+\sqrt{5}))(x+(\sqrt{3}-\sqrt{5})).

From this we know that $(\sqrt{5}-\sqrt{3})$ is also a root of $f(x)$ in \({\mathbb R}\).

We can see that these two roots are of similar form, so we can guess 2 other roots of $f(x)$ to be $(\sqrt{3}-\sqrt{5})$ and $(-\sqrt{3}-\sqrt{5})$ We now test this claim by expanding $(x-(\sqrt{3}-\sqrt{5}))(x-(-\sqrt{3}-\sqrt{5}))$. When we expand this we get $x^{2}+2\sqrt{5}x+2$. This confirms that all claims are correct and $f(x)$ can now be rewritten as

$f(x)=(x^{2}-2\sqrt{5}x+2)(x^{2}+2\sqrt{5}+2)=(x-(\sqrt{3}+\sqrt{5}))(x-(\sqrt{5}-\sqrt{3}))(x-(-\sqrt{3}-\sqrt{5}))(x-(\sqrt{3}-\sqrt{5}))$

This cannot be reduced further as all the factors are of degree 1. All roots of $f(x)$ are in \({\mathbb R}\), therefore $f(x)$ is reducible in \({\mathbb R}\).

### Is $f(x)$ reducible in \({\mathbb C}\)?

Repeating the process as in \({\mathbb R}\) all roots found are also in \({\mathbb C}\), therefore $f(x)$ is reducible over \({\mathbb C}\) such that:

$f(x)=(x^{2}-2\sqrt{5}x+2)(x^{2}+2\sqrt{5}+2)=(x-(\sqrt{3}+\sqrt{5}))(x-(\sqrt{5}-\sqrt{3}))(x-(-\sqrt{3}-\sqrt{5}))(x-(\sqrt{3}-\sqrt{5}))$

Stating the Fundamental Theorem of Algebra:

Given a polynomial $P(z)$ of degree $n$. This polynomial has $n$ values $z_{i}$ such that $p(z_{i})=0$. Such values are polynomial roots.

$f(x)$ is of $degree 4$ therefore we know it has 4 roots all of which are in \({\mathbb C}\) so $f(x)$ cannot be reduced further in \({\mathbb C}\).

## Irreducibility/reducibility in \({\mathbb Z_p}[x]\)

To begin with, we can manually check as many primes as we can:

In \({\mathbb Z_2}[x],x^{4}-16x^{2}+4= x^{4}=x.x.x.x\) thus reducible as none of the factors are of $deg 4$ or $deg 0$

In \({\mathbb Z_3}[x]\)

$x^{4}-16x^{2}+4=x^{4}+x^{2}+1=(x^{2}+2)(x^{2}-1)$

We derived this by checking the roots of the equation. This shows that 1 and 2 are roots so $x+2$ and $x+1$ are linear factors $f(x)$ respectively.By expanding $(x+1)(x+2)$ and computing long division of $f(x)$, we obtain the answer shown above. Therefore,$f(x)$ is reducible in \({\mathbb Z_3}[x]\).

In \({\mathbb Z_5}[x]\)

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

When testing for roots of $f(x)$ in \({\mathbb Z_5}[x]\) we found that no linear factors existed. We checked for quadratic factors of the form $f(x)=(x^{2}+ax+b)(x^{2}+cx+d)$. We then solved for $a,b,c$ and $d$ by solving the following equations:

$a+c=0 \hspace{5mm}ac+b+d=4 \hspace{5mm}ad+bc=0 \hspace{5mm}bd=4$

and we obtain the following

$a=0 \hspace{10mm}b=2 \hspace{10mm}c=0 \hspace{10mm}d=2$

And this gets us the result shown above and therefore $f(x)$ is reducible in \({\mathbb Z_5}[x]\)

In \({\mathbb Z_7}[x]\),

$f(x)=x^{4}-16x^{2}+4=x^{4}+5x^{2}+4=(x^{2}+1)(x^{2}+4)$

Again, $f(x)$ does not have any roots in \({\mathbb Z_7}[x]\) and therefore no linear factors. This then indicates that if $f(x)$ is reducible, it must have quadratic factors. Similarly to \({\mathbb Z_5}[x]\), we test for quadratic factors by solving for $a,b,c$ and $d$. This gets us the above result and therefore $f(x)$ is reducible in \({\mathbb Z_7}[x]\).

In \({\mathbb Z_{11}}[x]\),

$f(x)=x^{4}-16x^{2}+4=x^{4}+6x^{2}+4=(x+1)(x+7)(x+9)(x+10)$

We obtain this result by checking for roots and we find that $1,2 4$ and $10$ are roots. These give us the linear factors of $(x+10), (x+9), (x+7), (x+10)$ respectively. So, $f(x)$ is reducible in \({\mathbb Z_{11}}[x]\). With the irreducible factors $(x+1)(x+7)(x+9)(x+10)$.

In \({\mathbb Z_{13}}[x]\),

$f(x)=x^{4}-16x^{2}+4=x^{4}+10x^{2}+4$

We checked for roots of $f(x)$ in \({\mathbb Z_{13}}[x]\) this is resulted in no roots, hence no linear factors. We checked for quadratic factors however $a,b,c,d \notin{\mathbb Z_{13}}$, therfore $f(x)$ is irreducible in \({\mathbb Z_{13}}[x]\).

Note $f(x)=x^{4}+10x^{2}+4=(x^{2}+5)^{2}+5$ however 5 has a $deg=0$ which confirms that $f(x)$ is irreducible.

## Field Extensions

To construct a field $E$ we consider the ideal $I=f(x)$. This is the principal ideal of \(\mathbb Q[x]\) generalised by \(f(x)\in\mathbb Q[x]\) As we have shown, $f(x)$ is irreducible over \(\mathbb Q\), $I$ is the maximal ideal .

Considering the following theorem:

Let R be a commutative ring the ideal $\subseteq$ R is a maximal if and only if \(\frac{R}{M}\) is a field. Applying this, we can say that \(\frac{\mathbb Q[x]}{(f(x))}\) is a field. By The First Isomorphism Theorem \(E=\frac{\mathbb Q[x]}{(f(x))}\) is a field extension of \(\mathbb Q\).

If we consider \(\theta =x+I \in E\).

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

Therefore:

$f(\theta)=\theta^{4}-16\theta^{2}+4=16\theta^{2}-4-16\theta^{2}+4=0$

So, $\theta$ is a root of $E$ in $f(x)$.

Therefore (x-$\theta$) is a linear factor of $f(x)$ and then we can factorise $f(x)$ using long division.

$f(x)=(x-\theta)(x^{3}+(\theta)x^{2}+(\theta^{2}+10)x+(\theta^{3}+10\theta))$

### Finite field

So for $m\in E$ $f(m)$

$f(m)=m^{4}-16m^{2}+4-(\theta^{4}-16\theta{2}+4)$=$(m^{2}-\theta^{2})(m^{2}+\theta^{2})-16(m^{2}-\theta^{2})=(m-\theta)(m+\theta)(m^{2}+\theta^{2}-16)$

We know that $f(x)$ is irreducible in \({\mathbb Z}_{13}\), therefore:

\[\left\vert{E}\right\vert = \left\vert{\mathbb Z_{13}}\right\vert^{[E:\mathbb Z_{13}]}=13^{4}\]

That is, $E$ a field of $28561$ elements.

## $\alpha=\sqrt{3} + \sqrt{5}$

### Is $\alpha$ constructible?

To consider if $\alpha$ is constructible, we must first familarise ourselves with the following Proposition.

* Proposition*
All the integers can be constructed.

We will not prove this proposition here. However, the proof can be found in the link provided in our sources named 'Constructible Numbers' under Proposition 3.2.1.

*First we consider if $\sqrt{3}$ is constructible.*

We do this by considering a right-angled triangle ABC such that

$AC=2$ where $AC$ is the hypotenuse

$AB=1$

$BC$ is unknown

From this and basic trigonometry we can derive the following equation:

\[ (AB)^{2} + (BC)^{2} = (AC)^{2}\]

Subbing in the values, we get: \[1^{2} + (BC)^{2} = 2^{2} \Rightarrow 1 + (BC)^{2} = 4\]

From this we can see that $(BC)^{2} = 3$ and hence $BC=\sqrt{3}$ . Therefore $\sqrt {3}$ is constructible.

*Now we consider if $\sqrt{5}$ is constructible.*

We do this in a similar way to $\sqrt{3}$ by considering a right-angled triangle ABC such that,this time,

$AC=3$ where $AC$ is the hypotenuse

$AB=2$

$BC$ is unknown

From this and basic trigonometry we can derive the following equation:

\[ (AB)^{2} + (BC)^{2} = (AC)^{2}\]

Subbing in the values, we get: \[2^{2} + (BC)^{2} = 3^{2} \Rightarrow 4 + (BC)^{2} = 9\]

From this we can see that $(BC)^{2} = 5$ and hence $BC=\sqrt{5}$ . Therefore $\sqrt {5}$ is constructible.

To determine if $\alpha$ is constructible we look at the following proposition.

* Proposition*
Once $a$ and $b$ have been constructed, then $a+b$ can be constructed.

The proof for this proposition can be found under Proposition 3.2.5 in our Constructible Numbers link below.

Applying this to our $\alpha$, if we let $a=\sqrt{3}$ and $b=\sqrt{5}$ then $a+b=\sqrt{3}+\sqrt{5}$ is constructible. Hence $\alpha$ is constructible.

### Is $\alpha$ algebraic?

*Definition*

Let $K \supseteq F$ be a field extension and $\alpha \in K$. Then $\alpha$ is algebraic over F if there exists a non-zero polynomial $g(x)$ with coefficients in $F[x]$ st $g(x)=0$.

We know $f(\alpha)=0$. Therefore $\alpha$ is algebraic over $F$ st $K \supseteq F$ and $\alpha \in K$.

From this we can deduce that $\alpha$ is algebraic over the following:

over ${\mathbb Z}$ where ${\mathbb R} \supseteq {\mathbb Z}$ and $\alpha \in {\mathbb R}$

over ${\mathbb Q}$ where ${\mathbb R} \supseteq {\mathbb Q}$ and $\alpha \in {\mathbb R}$

This can be considered for a number of different cases.

## Sources

http://mathworld.wolfram.com/FundamentalTheoremofAlgebra.html - Fundamental Theorem of Algebra

https://www.artofproblemsolving.com/wiki/index.php/LaTeX:Symbols - LaTex: Symbols

http://www.math.utah.edu/~bertram/courses/4030/Constructible.pdf - Constructible Numbers