There are simliar questions here: normalization of the nodal cubic, Normalization of a variety, but I think my purpose of question is slightly different.
I am trying to show that $\operatorname{Spec} k[t] \to \operatorname{Spec} k[x,y]/(y^2-x^3-x^2)$ given by $(x,y)\mapsto (t^2-1,t^3-t)$ is a normalization.
By basic theory it is enough to show that $k[t]$ is isomorphic to the integral closure of $k[x,y]/(y^2-x^3-x^2)$.
I am try to follow the following steps:
Show that $k[t]$ and $k[x,y]/(y^2-x^3-x^2)$ have the same fractional fields.
Show that $k[t]$ is integrally closed.
Show that $k[t]$ is contained in the integral closure of $k[x,y]/(y^2-x^3-x^2)$.
For 1, I cannot see why the map $k[x,y]/(y^2-x^3-x^2)\to k[t]$ given above is injective. It is well-defined. If this is injective, then it induces an inclusion of fractional fields, and this inclusion must be in fact surjective because its image contains $t=y/x$.
I've shown 2: $k[t]$ is a UFD, and 3: $t=y/x$ satisfies the equation $t^2-x-1$. 2 and 3 show that $k[t]$ is the integral closure of $k[x,y]/(y^2-x^3-x^2)$.
So what only left is showing injectivity of $k[x,y]/(y^2-x^3-x^2)\to k[t]$. Why does $f(t^2-1,t^3-t)=0$ imply $f(x,y) \in (y^2-x^3-x^2)$?
Write $$f(x,y)=(y^2-x^3-x^2)g(x,y)+a(x)y+b(x).$$ We have $a(t^2-1)(t^3-t)+b(t^2-1)=0$. What can you say about the degrees of $a(t^2-1)(t^3-t)$, respectively $b(t^2-1)=0$?