Finding zero divisors in a polynomial quotient ring

Question

Is $x^2+x+I$ a zero divisor in $\mathbb{Z}_7[x]/I$, where $I=(x^3+5x^2+2x+5)$?

I know that $\gcd(x^3+5x^2+2x+5,x^2+x)=x+1$, and that it means that $x^2+x$ is indeed a zero divisor.

What I struggle with is finding $g(x)$ such that $$(f(x)+\langle x^3+5x^2+2x+5\rangle)(g(x)+\langle x^3+5x^2+2x+5\rangle)=\langle x^3+5x^2+2x+5\rangle$$ in the quotient ring.

I know how to find an inverse, is this any similar? How is it done?

Answer 1

The long division of $x^3+5x^2+2x+5$ by the GCD $x+1$ yields $x^2+4x+5$, so we have (in the given ring) $$(x^2+x)(x^2+4x+5)=x(x+1)(x^2+4x+5)=x(x^3+5x^2+2x+5)=x\cdot0=0$$ Thus if $f(x)=x^2+x$, $g(x)$ may be $x^2+4x+5$.

Answer 2

Hint:

Note that $6$ is a root of given polynomial. That is, in $\Bbb Z_7$, $$x^3+5x^2+2x+5=(x-6)(x^2+4x+5)=(x+1)(x^2+4x+5)$$ So take $f(x)=x+1$ and $g(x)=x^2+4x+5$ and let $I=\langle x^3+5x^2+2x+5 \rangle$. Then $f+I \neq I$ and $g+I \neq I$ but $$(f+I)(g+I)=fg+I=I$$

Answer 3

$ f=x(x\!+\!1)$ is a $0$-divisor mod $g\!\iff\! d:=\gcd(f,g)\neq 1,g$ $\!\iff\! g(0)=0\,$ xor $g(-1)=0\,$

Then we have $\!\bmod g\!:\ (\color{#c00}{g/d})f \equiv (f/d)g\equiv 0,\,$ both $\,g/d,f\not\equiv 0.\,$ Thus you seek $\,\color{#c00}{g/d}$

This is true in the OP: $\,g(0)= 5\neq 0,\,\ g(-1)=7=0\,$ in $\Bbb Z_{\large 7},\ $ hence $\,d = x\!+\!1$