$\ {-}4\equiv 1\pmod{\!5}\ $ so $\ x^2+1 = (x-2)(x+2)\ $ in $\Bbb Z_5[x]$

Question

I am a bit confused with the idea of a polynomials in a polynomial ring. In a worked example, I am given the ring $\mathbb{Z}_5[x]$ and we consider the polynomial $x^2+1$. Now it says that we can factor it into $x^2+1=(x+2)(x-2)$ and thus it has $\pm 2$ as roots.
I know that we are working on $\mathbb{Z}_5=\mathbb{Z}/5\mathbb{Z}$, but what exactly does that mean? It seems impossible to me that $x^2+1$ is $(x+2)(x-2)=x^2-4$. Hence, how exactly do we treat polynomials in such a ring?

Answer 1

In $\mathbb{Z}/5$, $-4=1$, since $1+4=5$ i.e, $1+4=0$ mod $5$.

Answer 2

The field of coefficients is ${\Bbb Z}_5$. So we have $-4\equiv 1\mod 5$.

Answer 3

What seems impossible to you needn't be impossible.

First of all, do you know arithmetic in $\mathbb{Z}_5$? There, the elements $-4$ and $1$ are the same, since $1-(-4)=1+4=5=0$ (every calculation done in $\mathbb{Z}_5$). Due to this, the polynomials are in fact equal.

You cannot simply assume that what holds for the polynomials you know, those in $\mathbb{R}$, automatically holds for polynomials in an arbitrary ring $R$. For example there are different polynomials, whose polynomial functions (which are not the same as the polynomials!) are the same, which cannot happen in the infinite field $\mathbb{R}$.

There is plenty of great material to read for an introduction to basic ring theory, so you might want to start there.