Group 3

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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 \({\mathbb Q}\).

References

http://en.wikipedia.org/wiki/Finite_field

http://en.wikipedia.org/wiki/Root_test

http://en.wikipedia.org/wiki/Polynomial_long_division

http://en.wikipedia.org/wiki/Minimal_polynomial_(field_theory)

http://en.wikipedia.org/wiki/Eisenstein's_criterion