Which polynomial gcd is "correct"? [unit normalization of GCDs] Find the gcd of $x^4-2x^3-x+3$ and $x^2-1.$ 
Note: I am using $a=bq+r$
First approach:
$$
\begin{align*}
x^4-2x^3-x+3&=(x^2-1)(x^2-2x+1)+(4x+4)\\
x^2-1&=(4x+4)(2x-2)+0
\end{align*}
$$
Therefore, $\text{gcd}(x^4-2x^3-x+3,\:x^2-1)=x+1.$
Second approach:
$$
\begin{align*}
x^4+5x^3+6x+3&=(x^2+6)(x^2-5x-6)+(4x+4)\\
x^2+6&=(4x+4)(2x+5)+0
\end{align*}
$$
Therefore, $\text{gcd}(x^4+5x^3+6x+3,\:x^2+6)=x+1.$
In the first approach I used the given polynomials while in the second approach I first used $\text{mod}\:7$ to change the negatives to positives and then proceeded with the calculation.
For both approaches, the remainders shown are in $\text{mod}\:7$ i.e. 
in the first approach the remainder is actually $-3x+4$ and in the second its $-24x+39.$ 
Which approach is the correct way of solving such problems? If it matters, why? Also, suppose I used the first approach, would the final answer be $-3x+4$, $4x+4$, or $x+1$?
 A: When calculating in quotient rings (or, equivalently, with ring congruences), we enjoy the freedom to choose any representative of an equivalence class, e.g. $\bmod 7\!:\ 8^n\equiv 1^n\equiv 1$ is simpler by choosing the rep $1$ vs. $8$ of $\,8+7\Bbb Z.\,$ So in your gcd calculation we can choose the coefficient reps as we please. Usually smallest magnitude reps yield simpler arithmetic, here $\,0,\pm1,\pm2,\pm3\,$ for $\,\Bbb Z/7.$
In general domains, gcds and lcms are defined only up to associates, i.e. up to unit (invertible) factors. For integers the units are $\pm1\,$ and we normalize a gcd $\,g\neq 0$ by choosing the positive choice from $\pm g.\,$  For polynomials over a field the units are all coef's $\,c\neq 0.\,$ Hence the associates of $\,g\neq 0\,$ are its constant multiples $\,cg\neq 0.\,$ The standard normalization convention here is to choose the rep that is monic (lead coef $= 1),\,$ i.e. if $\,0 \neq g\,$ has lead coef $\,a\,$ then we unit normalize it to the monic $\,a^{-1} g,\,$ e.g. your gcd $\,g = -3x\!-\!3\,$ times $\,(-3)^{-1}$ yields $\,x+1\,$ as its unit normalized standard rep.
Worth remark is that we can compute your gcd more simply. Notice $\,x^2-1 = (x\!-\!1)(x\!+\!1)\,$ is a product of nonassociate primes therefore $\,\gcd(f,(x\!-\!1)(x\!+\!1)) = \gcd(f,x\!-\!1)\gcd(f,x\!+\!1).\,$ Furthermore  $\, g := \gcd(f,x\!-\!a) = x\!-\!a\,$ if $\,f(a)=0\,$ else $\,g = 1,\,$ by the Factor Theorem. Hence the gcd in your example is $\,x\!+\!1\,$ because $\,f(-1)\equiv 0\,$ but $f(1)\not\equiv 0\pmod{\!7}$.
