$\bar{x}$ irreducible in $F[x,y]/(y^2-x^3+x)$ I want to show that $\bar{x}$ is irreducible in $F[x,y]/(y^2-x^3+x)$ with $F$ a field of characteristic not equal to 2. I was thinking of construct homomorphism to $F[t]$ sending $x$ and $y$ to polynomials of $t$, but no choice I found was optimal. Any help is much appreciated.
 A: The trick here is to use a norm function to reduce factorization questions about your ring (which is not a UFD) to one which is a UFD. (You may have already seen examples of this with, say, $\mathbb{Z}[\sqrt{-5}]$ and $\mathbb{Z}$.)
Let $R$ be a commutative ring, and let $d \in R$. Consider the ring $R[\sqrt{d}] := R[Y]/\langle Y^{2} - d\rangle$; we will use $\sqrt{d}$ to denote the class $\overline{Y}$. Note that $R[\sqrt{d}]$ is a free $R$-module of rank $2$ with basis $1, \sqrt{d}$. One may accordingly define a norm map $N \colon R[\sqrt{d}] \to R$ which sends $a+b\sqrt{d}$ to $a^{2}-b^{2}d$ for $a, b \in R$. It is easy, if tedious, to check that $N$ is multiplicative, i.e. $N(uv) = N(u)N(v)$ for any $u, v \in R[\sqrt{d}]$.
Let $R = F[X], d = X(X-1)(X+1)$, so that your ring of interest is $A := R[\sqrt{d}]$. (To avoid possible confusion, $\sqrt{d}$ is playing the role of the residue class $\overline{y}$ in your notation.) Suppose $X = uv$ for some $u, v \in A$. Then $N(X) = X^{2} = N(u)N(v)$. Up to scaling by units of $F$ (which you may check does not affect the ensuing analysis), we have $N(u)$ must be one of $1, X, X^{2}$. If $N(u) = 1$, then $u$ is a unit; this is an easy exercise. Likewise, if $N(u) = X^{2}$, then $N(v) = 1$, so $v$ is a unit. It thus remains to show that we cannot have $N(u) = N(v) = X$.
Suppose there exist $f(X), g(X) \in R$ such that for $u = f(X)+g(X)\sqrt{d}$,
$$N(u) = f(X)^{2}-g(X)^{2}(X^{3}-X) = X.$$
Since $X$ divides the RHS and divides $g(X)^{2}(X^{3}-X)$, it must also divide $f(X)^{2}$, hence $f(X)$. If $f(X) = Xh(X)$, we obtain
$$Xh(X)^{2}-g(X)^{2}(X^{2}-1) = 1$$
after cancelling all redundant factors of $X$. If this were the case, then the leading terms of $Xh(X)^{2}$ and $g(X)^{2}(X^{2}-1)$ must cancel. But we can see that if $n := \deg(h), m := \deg(g)$, then $Xh(X)^{2}$ has degree $2n+1$, and $g(X)^{2}(X^{2}-1)$ has degree $2m + 2$; hence, the leading terms cannot cancel. This is our desired contradiction.
