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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.

xxxxxx
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  • Your attempt for $\implies$ doesn't makes sense. Recall that the following definitions: If $R$ is a ring, $u \in R$ is a unit if there exists $v \in R$ such that $uv=vu=1$. Also, if $I$ is an ideal of $R$, for every $a$ and $b$ in $R$ we have that $a+I = b+I !\iff! a-b \in I$ (not $a=b$). – azif00 Feb 07 '21 at 03:55

3 Answers3

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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.

Aaron
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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 $x$ & $x\!-\!1\iff f(0)\neq 0\neq f(1)$

Bill Dubuque
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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 $x^2+I=x+I$.

egreg
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