Galois $\text{GF}(5)$, does $α^4=2$ make any sense? 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.
 A: 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 $K^\times=K\setminus\{0\}$ has order $\lvert K^\times \rvert=624$. So $a^{624}=1$ for any $a\in K^\times$, in particular 
$$a^{625}=a\cdot a^{624}=a\cdot 1=a,\quad\text{for any  }a\in K^\times$$
Remark: Obviously, this equality holds for $a=0$ as well, so that every $a\in K$ is a root of $x^{625}-x\in\mathrm{GF}(5)[x]$. 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 $x^{\lvert L\rvert}-x$ 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 $a\in \mathrm{GF}(5)$ 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): 


*

*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 $a,b,c,d\in \mathrm{GF}(5)$, getting a contradiction.

*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 $x^n-1\in k[x]$, 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 
$$(\zeta\alpha)^4=\zeta^4\alpha^4=\alpha^4=2,$$
so $\zeta\alpha$ is a root of $f$ as well, for any fourth root of unity $\zeta$. Now $x^4-1\in\mathrm{GF}(5)[x]$ 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 $\alpha,\zeta\alpha,\eta\alpha,\tau\alpha$ 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 $a\in \mathrm{GF}(5)^\times$ - that means, the $4$-th roots of unity are nothing but $1,2,3,4\in\mathrm{GF}(5)$, and the roots of $f$ in $\mathrm{GF}(5)(\alpha)$ are thus 
$$\alpha,2\alpha,3\alpha,4\alpha.$$
A: So here's what i ended up doing.
a) $\alpha^4=2$
$\alpha^{625}=(\alpha^4)^{156}\alpha=(2)^{156}\alpha=(2\times2)^{78}\alpha=(4)^{78}\alpha=(4\times4)^{39}\alpha=(1)^{39}\alpha=\alpha$
b)$\alpha^5=\alpha\times\alpha^4=2\alpha$,  $\alpha^{25}=\alpha^5\alpha^5=2\alpha\times2\alpha=4\alpha^2$, $\alpha^{125}=\alpha^{25}\alpha^{5}=2\alpha4\alpha^2=8\alpha^3=3\alpha^3$
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)$x^4-2=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$
So the roots are $+-\alpha,+-i\alpha$ where $\alpha=\sqrt[4]{2}$.
Is this all okay or am I doing anything wrong ?
