How to find square root of 2 and homomorphisms in the quotient ring $L=\frac{\mathbb{F_{5}[x]}}{(x^{2}+2x-1)}$ I am trying to work through the following problem.
Suppose we are considering the field $$L=\frac{\mathbb{F_{5}[x]}}{(x^{2}+2x-1)}$$ (where $F_{5}$ is $\mathbb{Z}/5\mathbb{Z}$)
We also note it is indeed a field as $f(x)=x^2+2x-1$ is irreducible in F.
Then I am wanting to find the square root of two in L, and denote is as z,
then show that
$\beta : \mathbb{F[x]} \to L$ defined by $x \to z$ is a ring homomorphism, and find the kernel of $\beta$ and finally to show it induced an isomorphism $$\frac{\mathbb{F[x]}}{(x^2-2)} \cong L$$
My thoughts;
To me , I thought that L is a field with $25$ elements, and each is of the form $ax+b$ where $a, b\in \mathbb{F_{5}}$
So to me , finding the root of two is like asking which of these elements squared gives us 2.
I also know we can write $x^{2}=1+3x$ in L.
So I just tried squaring some of these elements, and I wasn't getting anywhere, so I thought there must be a much better way to do it.
I thought maybe I could write $(ax+b)(ax+b)=a^{2}x^{2}+2abx+b^{2}=a^{2}(1+3x)+2abx+b^{2}=(3a^{2}+2ab)x+(a^{2}+b^{2})=2$
and then solve $3a^{2}+2ab=0$ and $a^{2}+b^{2}=0$ as a system of equations, as pointed out below, after I fixed the typo this leads me to 
$z=(x+1)$ or $z=(x-1)=(x+4)$
Now I am interested in the next part, I will choose $z=(x+1)$ to be the z we refer to now.
I know for it to be a homomorphism it must satisfy ;
$\beta(1)=1$
$\beta(ab)=\beta(a)\beta(b)$
$\beta(a+b)=\beta(a)+\beta(b)$
But here I am already confused, so the identity in F is just $1$ and we are saying $1 \to (x+1)$ so if it is a homomorphism it must be that $(x+1)$ is the identity of L. Is this true? Is not the identity of $L$ $1+I=1+(x^2+2x-1)$
So now I have been told that actually it just sends that x,
so $0x+1 \to 1 $ under the map and the other requirements can be verified.
But how can I find $Ker(\beta)$? I know it is $=\{a \in F : \beta(a)=0_{L}\}$
 but isnt $0_{L}=I$ , so the kernel consists of anything that gets sent to I? Is this even the correct approach? If anyone at all can comment please do so.
But now I am just confused, for how is this a map if we only know what it does to x? what about $x^{2}$ for example

You have a field $F=\mathbb{Z}_5[x]/\langle p(x)\rangle$, where $p(x)=x^2+2x-1$ (Note that this polynomial has no roots in $\mathbb{Z}_5$ so $F$ is a field). Indeed in $F$ you have $x^2=3x+1$. Your approach  to calculate the $\sqrt{2}$ is correct, but you have little mistake: $(ax+b)^2=a^2x^2+2abx+b^2 = a^2(1+3x)+2abx+b^2= x(3a^2+2ab)+a^2+b^2=2$. Solving the system of equations gives $a=\pm 1$ and $b= \pm 1$. And indeed $(\pm x \pm 1)^2=2$ in $F$.
For the second part: You have chosen $z=x+1$. So now you have map $\beta: \mathbb{F}_5[x]\rightarrow L$, $x \mapsto x+1$. To see this is a ring homomorphism note that for $a,b \in \mathbb{F}_5[x]$ you have $\beta(a+b)=\beta(r(x)+q(x))=r(x+1)+q(x+1)=\beta(r(x))+\beta(q(x))=\beta(a)+\beta(b)$ (Here I just denoted $a=r(x) \text{ and } b=q(x)$). (I think you can check the other conditions yourself) So indeed we have ring homomorphism which has kernel Ker$(\beta)=\{p(x) \in \mathbb{F}_5[x] \, | \, \beta(p(x))=0_L\}$. So which elements of $\mathbb{F}_5[x]$ will map to $0_L$? Indeed the answer is $\langle x^2-2 \rangle$. Now apply the isomorphism theorem..
