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 $g(x)$?

  • $\begingroup$ If $f(x)$ is irreducible, the quotient will be a field. $\endgroup$ Sep 2 '19 at 5:02
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    $\begingroup$ Note: if $F$ is a field then $(af(x))=(f(x))$ for all $a\ne0$ $\endgroup$ Sep 2 '19 at 5:03
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    $\begingroup$ 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. $\endgroup$ Sep 2 '19 at 6:24
  • $\begingroup$ Thanks for the comments, I understand better why we usually make those assumptions now! $\endgroup$
    – eatfood
    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 $f(x)g(x)=0$


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