# Why do we usually quotient a polynomial ring by a monic irreducible polynomial?

I'm studying some ring theory, and I'm wondering for a polynomial ring $$F[x]$$ over a field $$F$$, if we quotient by some element $$f(x) \in F[x]$$ to get $$F[x]/(f(x))$$, why do we usually want $$f(x)$$ to be monic irreducible? How about arbitrary polynomials $$g(x)$$?

• If $f(x)$ is irreducible, the quotient will be a field. Sep 2 '19 at 5:02
• Note: if $F$ is a field then $(af(x))=(f(x))$ for all $a\ne0$ Sep 2 '19 at 5:03
• We can form such quotient rings with any $g(x)$. The resulting rings are useful for many a purpose. They are fields only when $f(x)$ is a scalar multiple of a monic irreducible one. May be you have only seen such examples in your studies so far. Sep 2 '19 at 6:24
• Thanks for the comments, I understand better why we usually make those assumptions now! Sep 2 '19 at 6:25

Quotient by irreducible to get an integral domain because if it is not irreducible say $$F[x]/g(x)f(x)$$ then $$f(x),g(x)$$ are non zeros element in the quotient ring such that $$f(x)g(x)=0$$