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Here is the question I am trying to understand the solution of letter $(c)$ in it:

Let $x^3 - 2x + 1$ be an element of the polynomial ring $E = \mathbb Z[x]$ and use the bar notation to denote the passage to the quotient ring $\mathbb Z[x]/(x^3 - 2x + 1).$ Let $p(x) = 2x^7 - 7x^5 + 4x^3 - 9x + 1$ and let $q(x) = (x - 1)^4.$\

$(a)$ Express each of the following elements of $\overline E$ in the form $\overline{f(x)}$ for some polynomial $f(x)$ of degree $\leq 2: \overline{p(x)}, \overline{q(x)}, \overline{p(x) + q(x)}$ and $\overline{p(x)q(x)}.$

$(b)$ Prove that $\overline{E}$ is not an integral domain.

$(c)$ Prove that $\overline{x}$ is a unit in $\overline{E}$.

Here is the solution I found online here https://linearalgebras.com/solution-abstract-algebra-exercise-7-4-17.html:

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My question is:

Why the inverse of $\bar{x}$ should be of degree at most 2, could someone clarify this to me please?

Emptymind
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    Because every element of your quotiented ring can be represented by a polynomial of degree $\leq 2$. It is not specific of $\bar{x}$. This can be achieved through euclidean division – julio_es_sui_glace Sep 09 '23 at 21:30
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    $x^3-2x+1 = 0\Rightarrow -x(x^2+2) = 1$ so $x^{-1} = -(x^2+2),$ by uniqueness of inverses. Your prior question's answers showed every polynomial is congruent to its remainder $!\bmod x^3-2x+1,,$ which has degree $\le 2.,$ This is a dupe and will soon be closed – Bill Dubuque Sep 09 '23 at 21:35
  • There is no need compare coef's as in the (very poor) solution. Find a better source to learn algebra. – Bill Dubuque Sep 09 '23 at 21:41
  • @julio_es_sui_glace how did you know that every element in $\bar{E}$ ......? is there a proof for this? – Emptymind Sep 09 '23 at 21:57
  • @BillDubuque I do not think that the previous question showed that every polynomial is ...... – Emptymind Sep 09 '23 at 21:58
  • What did you not understand in the explanation in the dupe to the prior? In any case that play no role here as I mention above, since $,-(x^2+2)$ already has degree $\le 2,,$ i.e. it is already in normal form. – Bill Dubuque Sep 09 '23 at 22:03
  • @BillDubuque The solution "pulls the inverse out of thin air." The balance of that part of the answer is intended to motivate that selection. – Robert Shore Sep 09 '23 at 22:57
  • @Robert It's not ad-hoc: to invert $,x$ mod $f,$ simply divide $f,$ by $x$ to get $,f = x g + c,,\ c=f(0),,$ so $\bmod f!:\ xg \equiv -c\Rightarrow x(-g/c)\equiv 1\Rightarrow x^{-1}\equiv -g/c \equiv (1-\hat f)/x,\ \ \hat f = f/f(0).,$ This is a special (one step) case of using the extended Euclidean algorithm to compute the inverse, as explained here. Using undetermined coef's works but is brute force and unenlightening (and it is redundant to do so after they already found the inverse (presumably) the way I described). – Bill Dubuque Sep 09 '23 at 23:31
  • Said more simply: $,f = x:!(\overbrace{f_n x^{n-1}+\cdots + f_2 x + f_1}^{\textstyle\color{#c00} g}) + f_0,\Rightarrow, x^{-1}\equiv -\color{#c00}g/f_0\pmod{!f}\ \ $ Presumably their strange notation means $,\bar x (\overline{-x^2+2})=\bar 1,,$ i.e. $,\bar x(- \bar g) = \bar c,,$ a coset notation form of my above proof (in congruence language). $\ \ $ – Bill Dubuque Sep 10 '23 at 00:07

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