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.
Best Answer
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.