GCD of $f=X^3 +9X^2 +10X +3$ and $g= X^2 -X -2$ in $\mathbb{Z}/5\mathbb{Z}$.

Question

I have to calculate the gcd of $f=X^3 +9X^2 +10X +3$ and $g= X^2 -X -2$ in $\mathbb{Q}[X]$ and $\mathbb{Z}/5\mathbb{Z}$.

In $\mathbb{Q}[X]$ I got that $X+1$ is a gcd and therefore $r(X+1)$ since $\mathbb{Q}$ is a field.

But I dont know how to do polynomial division in $\mathbb{Z}/5\mathbb{Z}$. Can somebody please help me?

Thank you!

Answer 1

You got the GCD in $\mathbb Z[X]$, i.e. $X+1$. The GCD in $\mathbb Z / 5 \mathbb Z[X]$ is obtained by reducing each coefficient of the GCD in $\mathbb Z$ in $\mathbb Z / 5 \mathbb Z$.

Which means that the GCD in $\mathbb Z / 5 \mathbb Z[X]$ is equal to $\bar{1}X + \bar{1} = X + \bar{1}$.

Answer 2

HINT

It's easy to see that $x^2-x-2 = (x+1)(x-2)$, but in $\mathbb{Z}/5\mathbb{Z}$ you have $5 \equiv 0$, so $$ x^3+9x^2+10x+3 \equiv x^3-x^2+3 = (x+1)\times \ldots $$