The quotient in $ \mathbb{Z} _{5} [x] $ by a irreducible polynomial.

Question

Let $ f(x) = x^2+x+2$ in $ \mathbb{Z} _{5} [x] $. Find the quotient $ \mathbb{Z} _{5} [x] / (f(x)) $

I know that $f$ is irreducible, but I don't see how that helps me. I would say that the quotient is the set of all polynomials in $ \mathbb{Z} _{5} [x] $ with a degree less than 2.

In general, when I look at the quotient of $F[X] $ by a ideal generated by a polynomial $(f(x)) $, where $F$ is a field, the quotient is the set of all remainders when you divide any polynomial from $F[X] $ by $f(x) $?

Answer 1

If we're being technical or studying the fundamentals, the quotient is a set of residue classes, and the remainders are representatives of these classes. With this in mind, your guess about polynomials in $\Bbb Z_5[x]$ with degree less than $2$ would be correct.

Answer 2

Let me add this answer to the pure description given by Fimpellizieri:

As $f$ is irreducible it generates a maximal ideal (since $\mathbb{F}_5[x]$ is a pid) and therefore the quotient $\mathbb{F}_5[x]/(f)$ is a field. It is (up to isomorphism) the unique field with $25$ elements. What actually happens when you do this is that you add a root of the polynomial to the field $\mathbb{F}_5$ and then turn that into a field again (or actually you take the intersection of all field extensions such that the base field and the root are contained). What you constructed is thus the smallest field extension of $\mathbb{F}_5$ such that your polynomial has a root (actually both then since it was a degree $2$ polynomial).