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I have the following question :

Let $R=\mathbb{Z}_7[x]$, $f(x)=x^2-6\in R$, and $I=\langle f(x) \rangle$ maximal ideal. Is $(x^2-1+I)$ invertible in $\frac{R}{I}$ to $(4+I)$?

What I did : $$(x^2-1+I)(4+I)=(4x^2-4+I)$$

and in order to be invertible we need to check if $(4x^2-4+I)=(1+I)$ Since $(1+I)$ is the trivial coset.

For some reason in the answer they reached that $(4x^2-4+I)=(-1+I)$ so the answer is no. that also stated that $3+I$ is the invertible since $(3x^2-3+I)=(1+I)$

Could anyone explain why $(4x^2-4+I)=(-1+I)\in \mathbb{Z}_7[x]$ and also why $(3x^2-3+I)=(1+I)\in \mathbb{Z}_7[x]$.

Thank you!

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    @DanielRobert-Nicoud You'r right sorry, edited. – 2017-02-01
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    $3$ is not a square in $\mathbb{F}_7$, so $x^2-6$ is irreducible, and $K = \mathbb{F}_7[x]/(x^2-6)$ is a **finite** integral domain, so it also a field, and its non-zero elements are all inversible, in particular the one you mentioned $x^2-1= x^2-1-(x^2-6) = 5$ in $K$ – 2017-02-01

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Yes. The equation $(4x^2-4+I)=(-1+I)\in \mathbb{Z}_7[x]$ means this: there exists a polynomial $f(x)$ in $\mathbb{Z}_7[x]$ such that $$ 4x^2-4 = -1 +f(x)\cdot (x^2-6) $$ in $\mathbb{Z}_7[x]$. This is equivalent to require that $4x^2-3\in \langle x^2-6 \rangle$. Now take as $f(x)$ the constant polynomial 4. You get (in $\mathbb{Z}_7[x]$): $4(x^2-6)=4x^2-24\equiv 4x^2-3.$ Done.

You can do similarly for the other question you posed, method of proof is the same.

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    Do you mean $4x^2-4\in $ right? – 2017-02-01
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    @JaVaPG yes, of course. The procedure I mentioned explains why the two cosets are equal in the quotient ring (which by the way in this case is a field). – 2017-02-01