Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$. Let $F$ be a field and let $f(x) \in F[x]$. Prove that $f(x) + \langle x^2 − x \rangle$ is a unit in the
quotient ring $F[x]/ \langle x^2 − x \rangle$ if and only if $f(0)f(1) \neq 0$.
$\textbf{My Attempt:}$
prove "$\implies$":
Assume $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring $F[x]/ \langle x^2 − x \rangle$.
Then, $(f(x) + \langle x^2 − x \rangle) \cdot (f(x) + \langle x^2 − x \rangle) = f(x)f(x) + \langle x^2 − x \rangle = 1 + \langle x^2 − x \rangle$.
Then, $f(x)f(x) = 1$ $\implies$ $f(x) \neq 0$.
Then, neither $f(0)$ nor $f(1)$ can be equal to $0$. So does $f(0)f(1)$. So, $f(0)f(1) \neq 0$.
prove "$\Longleftarrow$":
Assume $f(0)f(1) \neq 0$. Then, $f(0) \neq 0$ and $f(1) \neq 0$.
Which I am getting stucked on how to related $f(0) \neq 0$ and $f(1) \neq 0$ to proving that $f(x) + \langle x^2 − x \rangle$ is a unit in the quotient ring.
Also, I don't know if the prove of "$\implies$" is correct as well.
 A: We first unravel the definition.  Let us write $[f(x)]$ for the image of $f(x)$ in the quotient ring.  Then $[f(x)]$ is a unit if and only if there exists a $g(x)$ such that $[f(x)][g(x)]=[f(x)g(x)]=[1]$.  So $[f(x)]$ is a unit if and only if there exist $g(x),h(x)$ such that $f(x)g(x)=1+h(x)(x)(x-1)$.
In one direction, evaluating this expression at $x=0$ and $x=1$ shows that $f(1)g(1)=1$ and $f(0)g(0)=1$, so $f$ is non-zero at both $0$ and $1$.  In the other direct, assume that $f(0)=a, f(1)=b$, $a,b\neq 0$.  Then let $g(x)$ be any polynomial such that $g(0)=1/a, g(1)=1/b$.  Then $f(x)g(x)-1$ evaluates to $0$ at both $0$ and $1$, so it is divisible by bother $x$ an $(x-1)$.  Since those are relatively prime polynomials, $f(x)g(x)-1$ is divisibly be $x^2-x$.  Thus, $[f(x)][g(x)]=[1]$, so $f$ is a unit.
A: Hint: as for integers, $f$ is coprime to $\,x(x\!-\!1)\color{#c00}\iff f$ is coprime to $x$ & $x\!-\!1,\,$ cf. here, so
$$\begin{align}
&(1)=(f)\ \,{\rm in}\,\ F[x]/(x(x\!-\!1))\\[.2em]
\iff\ & (1)=(f,\,x(x\!-\!1))\,\ {\rm in}\,\ F[x]\\[.2em]
\color{#c00}\iff\ & (1)=(f,x)=(f(0),x),\ (1) = (f,x\!-\!1)=(f(1),x\!-\!1)\\[.2em]
\iff\ & f(0)\neq 0\neq f(1),\ {\rm by\ eval\ at}\ x=0,1,\ \&\,\ K\ \rm field\\[.2em]
\iff\ & f(0)f(1)\neq 0,\ {\rm by}\ K\ \rm domain
\end{align}\qquad$$
Remark $ $ employing CRT this multiplicativity of coprimality can be expressed more structurally, viz. the CRT factorization $\,F[x]/(x(x\!-\!1))\cong F[x]/x\times F[x]/(x\!-\!1)\,$  descends to unit groups, hence $f$ is a unit mod $\,x(x\!-\!1)\iff f$ is a unit mod $x$ & $x\!-\!1\iff f(0)\neq 0\neq f(1)$
A: Let me analyze your attempt, rather than offer a solution.
It's easier if you write $I=\langle x-x^2\rangle$, so to reduce the clutter.
“$\Rightarrow$” Saying that $f(x)+I$ is a unit means there exists $g(x)+I$ such that $(f(x)+I)(g(x)+I)=1+I$. You're using $f$ again, but it's wrong.
The condition now reads $f(x)g(x)+I=1+I$ and therefore $f(x)g(x)-1\in I$, so
$$
f(x)g(x)-1=(x-x^2)h(x)
$$
for some polynomial $h$. Evaluate at $0$ and $1$ and finish up.
“$\Leftarrow$” The key here is to write $f(x)=(x-x^2)q(x)+ax+b$ using polynomial division. Evaluating at $0$ yields $b=f(0)$ and evaluating at $1$ yields $a+b=f(1)$.
Now this is much simpler: you just need to find the inverse of $ax+b+I$, using the fact that $b\ne0$ and $a+b\ne0$. Why? Because $f(x)+I=ax+b+I$.
Hint: the inverse has the form $cx+d+I$, for some $c,d\in F$. Moreover $x^2+I=x+I$.
