0
$\begingroup$

There is a theorem in algebra which says: "let $F$ be a field and assume $f(x) \in F[x]$ is irreducible, then $\frac{F[x]}{(f(x))}$ is a field".

I have the proof of this theorem, but I can't get the meaning of $\frac{F[x]}{(f(x))}$. I don't know what does it really mean, so, although the proof of this theorem seems very easy, I can't get it.

I would really appreciate it if someone helped me understand the meaning of this fraction!

  • 2
    It is a [quotient ring](https://en.wikipedia.org/wiki/Quotient_ring).2017-01-13
  • 0
    I don't know if this could be helpful: http://math.stackexchange.com/questions/1253187, http://math.stackexchange.com/questions/1145015, http://math.stackexchange.com/questions/18261882017-01-13

1 Answers 1

1

First you have to get the meaning of the quotient ring $R/I$ of a ring $R$ by an ideal $I$. Then one can take $R$ as the polynomial ring $F[x]$ over a field $F$ and $I$ the ideal generated by a polynomial $f\in F[x]$. Then $R/I=F[x]/(f(x))$. The Theorem you mentioned has been proved in this duplicate. The next task is to study some simple examples, e.g., $$ \mathbb{R}[x]/(x^2+1),\; \mathbb{Z}[x]/(x^2+3,p) $$ see here, and here.