Isomorphism between two finite fields We have $k_1:= \mathbb F_7(\alpha)$ and $k_2 := \mathbb F_7(\beta)$ where $\alpha^2 = 3$ and $\beta^2 = -1$ in $\mathbb F_7$. I have to show that these two are isomorphic. 
Let $\phi:k_1 \rightarrow k_2$ be a homomorphism which preserves $1 \in k_1$. Then 
$$\phi(\alpha^2)= \phi(3) = 3 = \phi(\alpha)^2$$
where 
$$\phi(\alpha) = x + y\beta\;,\;\;x,y \in \mathbb F_7$$
Thus 
$$(x+y\beta)^2 = \phi(\alpha)^2  = x^2 + 2xy\beta +y ^2 \beta^2 = x^2 +2xy\beta -y^2=3$$
So $x$ or $y$ must be $0$. But $y$ can't be zero because $3$ has no root in $\mathbb F_7$. So $x = 0$ such that 
$$-y^2 = 3 \rightarrow y \in \{-2, 2\}$$
Is the function $\phi$ with $\phi(x) = x\;$ for $\;x \in \mathbb F_7\;$ and $\;\phi(\alpha) = 2\beta\;$ then an isomorphism ?
 A: Well, you have a candidate; why don't you check whether $\phi((a+b\alpha)(c+d\alpha)) = \phi(a+b\alpha)\phi(c+d\alpha)$, for all $a,b,c,d \in \mathbb{F}_7$? For sums it's trivial, as you already define it as a linear map over the base field.
A: Hint: $ $ mod $\,7\!:\  \alpha^2 \equiv 3 \equiv -4 = 4\beta^2\! = (2\beta)^2,\,$ so check  $\,\alpha \to 2\beta\,$ preserves sums and products.
A: An idea: since
$$k_1=\Bbb F_7[x]/\langle\;x^2-3\;\rangle\;,\;\;k_2=\Bbb F_7[x]/\langle\;x^2+1\;\rangle$$
Now, in both cases there exists a primitive element, i.e.: an element that generates the multiplicative group $\,k_i^*\;$ , so if you find both elements in both fields you shall be done...
For example: let $\,\alpha\in\overline{\Bbb F_7}\;$ be a root of $\,x^2-3\in\Bbb F_7[x]\,$ , then in $\,k_1\,$ we have
$$\alpha^2=3\implies k_1:=\{p(\alpha)\;;\;p(x)\in\Bbb F_7[x]\;,\;\;\deg p\le 1\}$$
For example, we have that $\,\alpha +5\;,\;3\alpha\;,\;2\alpha -5\,$ , etc. are element in $\,k_1\,$ . 
Since $\,|k_1^*|=48\;$ ,every non-zero element in the field has multiplicative order a divisor of $\,48\,$ , so we can try the following: pick (nd remember: we work modulo $\,7\,$ all the time!)
$$\begin{align*}r:&=\alpha+1\\
 r^2&=\alpha^2+2\alpha+1=3+2\alpha+1=2\alpha+4\\
r^4&=4\alpha^2+2\alpha+2=2\alpha\\
r^6&=(2\alpha)^3=\alpha(3)=3\alpha\\
r^{12}&=(3\alpha)^2=2\cdot 3=-1\end{align*}$$
and we've found that $\;r\;$ is not a generator of $\,k_1^*\,$ (why? check what happens with $\,r^{24}\,$ ...) . Try other elements until you find an generator and then try to do the same with $\,k_2\,$ and then you'll have an easy choice for your isomorphism...
