Csdrhrt

I just want to see if this is a misprint or am I making some really silly mistake.
the question says let $α$ be a root of $f(x) = x^4 − 2$ over $text{GF}(5)$.

Obviously what one should do in such cases is take it s.t. $α^4=2$.

However my difficulty arises through the fact that this doesn't seem to be true for any element of $text{GF}(5):={0,1,2,3,4}$.

what am I doing wrong, because honestly I doubt its a misprint, and suspect it's much more likely that because I'm very new to galois fields that I'm doing something incorrectly.

I think there are some important things to remark concerning what you wrote above:

To a). Your calculation is absolutely correct, but the fact that the question as you posted it in your comment speaks about explaining an "obvious" fact in "a few sentences", I would suspect that it aimed more for something like the following (I'm not entirely sure though, as the ordering of the given subtasks seems to disagree with my interpretation):

Suppose we already know that $f(x)=x^4-2$ is irreducible over $mathrm{GF}(5)$ (part d)). This would tell us that the degree of $alpha$ over $mathrm{GF}(5)$ is $4$ (part c)), so $K:=mathrm{GF}(5)(alpha)$ is a field with $5^4=625$ Elements. Its multiplicative group $K^times=Ksetminus{0}$ has order $lvert K^times rvert=624$. So $a^{624}=1$ for any $ain K^times$, in particular
$$a^{625}=acdot a^{624}=acdot 1=a,quadtext{for any }ain K^times$$

Remark: Obviously, this equality holds for $a=0$ as well, so that every $ain K$ is a root of $x^{625}-xinmathrm{GF}(5)[x]$. Considering that a polynomial of degree $n$ over a field can have only a maximum of $n$ roots (in some algebraic closure), one thus gets that every finite field $L$ of characteristic $p$ is a splitting field of $x^{lvert Lrvert}-x$ over $mathrm{GF}(p)$.

To c). Here I am not sure what the first line of your solution is used for? An later on, I think it would have been more precise to explicitly give some $mathrm{GF}(5)$-Basis of $mathrm{GF}(5)(alpha)$, you might for example prove that
${1,alpha,alpha^2,alpha^3}$ is such a basis (your argument already shows it is generating, so you only had to use the irreducibility of $f$ to prove linear independence), hence
$$[mathrm{GF}(5)(alpha):mathrm{GF}(5)]=dim_{mathrm{GF}(5)}(mathrm{GF}(5)(alpha))=4.$$

To d). How exactly did you end up with only checking there is no $ain mathrm{GF}(5)$ with $a^2=2$ to ensure that $f$ has no irreducible factor of degree $2$? You were definitely right about checking for roots to exclude factors of degree $1$, but the point of excluding factors of degree $2$ could need some supplementation, I think. Basically, there are two natural ways to do this (unfortunately both are a bit lengthy):

1. As $f$ is monic, you may without loss of generality assume both degree $2$ factors monic. So you could multiply out the assumption
   $$f(x)=(x^2+ax+b)(x^2+cx+d)$$
   compare the coefficients and try to solve for $a,b,c,din mathrm{GF}(5)$, getting a contradiction.




2. As $mathrm{GF}(5)$ consists of only five elements, there are just $5^2=25$ monic polynomial of degree $2$ in $mathrm{GF}(5)[x]$. By testing for roots in $mathrm{GF}(5)$ you might find and list all irreducible ones, and then, for any of these, test if it divides $f$.

To e). If I'm not mistaking what you wrote, this is an important misconception, I think! You could of course label $alpha$ as $sqrt[4]{2}$, which makes perfect sense, but the use of $i$ suggests you are thinking about complex numbers. However, the question happens in characteristic $5$! The procedure is similar, though. You already know one root of $f$ in $K=mathrm{GF}(5)(alpha)$, which is $alpha$ (by definition). Now, what are the other roots? This is where the roots of unity come in. By definition, a $n$-th root of unity over a field $k$ (in some fixed algebraic closure of $k$) is nothing but a root of the polynomial $x^n-1in k[x]$, that is, the roots of unity are the elements $zeta $ with $zeta ^n=1$. Now, if $alpha$ is a root of $f$, that is $alpha^4=2$, and $zeta$ is a fourth root of unity over $mathrm{GF}(5)$, then
$$(zetaalpha)^4=zeta^4alpha^4=alpha^4=2,$$
so $zetaalpha$ is a root of $f$ as well, for any fourth root of unity $zeta$. Now $x^4-1inmathrm{GF}(5)[x]$ is separable (as the characteristic $5$ does not divide $4$), so there are four different 4th roots of unity over $mathrm{GF}(5)$, call them $1,zeta,eta,tau$ (actually, the group of roots of unity is cyclic as a finite subgroup of a multiplicative group of a field, so we might as well write them as $1,zeta,zeta^2,zeta^3$), and the four elements $alpha,zetaalpha,etaalpha,taualpha$ are all roots of $f$. As $f$ is of degree $4$, these are of course all roots. Now, what are the fourth roots of unity over $mathrm{GF}(5)$? As the multiplicative group $mathrm{GF}(5)^times$ has order $4$, we immediately have $a^4=1$ for all $ain mathrm{GF}(5)^times$ - that means, the $4$-th roots of unity are nothing but $1,2,3,4inmathrm{GF}(5)$, and the roots of $f$ in $mathrm{GF}(5)(alpha)$ are thus
$$alpha,2alpha,3alpha,4alpha.$$

So here's what i ended up doing.

a) $alpha^4=2$

$alpha^{625}=(alpha^4)^{156}alpha=(2)^{156}alpha=(2times2)^{78}alpha=(4)^{78}alpha=(4times4)^{39}alpha=(1)^{39}alpha=alpha$

b)$alpha^5=alphatimesalpha^4=2alpha$, $alpha^{25}=alpha^5alpha^5=2alphatimes2alpha=4alpha^2$, $alpha^{125}=alpha^{25}alpha^{5}=2alpha4alpha^2=8alpha^3=3alpha^3$

c)$x^4=2$ over GF(5)

$x^4-2$ is irreducible over GF(5)

so $GF(5)(alpha)cong GF(5)[x]/<x^4-2>$

and $GF(5)(alpha):={a+bx+cx^2+dx^3|a,b,c,d in Bbb Q }$

so $alpha$ has degree 4 over GF(5).

d) there is no $a in GF(5)$ s.t. $ a^2=2$ so it cant be factored into quadratics and further no elements of GF(5) are solutions to $x^4-2=0$ so it is irreducible

e)$x^4-2=(x-sqrt[4]{2})(x+sqrt[4]{2})(x-isqrt[4]{2})(x+isqrt[4]{2})$

So the roots are $+-alpha,+-ialpha$ where $alpha=sqrt[4]{2}$.

Is this all okay or am I doing anything wrong ?