Intuition about turning a polynomial ring into a field Let's consider the polynomial ring $R=F[x]$ over a field $F$. Then by taking the quotient by a principal ideal $I=(f(x))$ generated by an irreducible polynomial $f(x)$, we obtain a field $R'=F[x]/(f(x))$.
It's easy to see that $R'$ is indeed a field. Since the ideals of $R$ which contain $I$ are in bijective correspondence with the ideals of $R'$, we can conclude that $R'$ has only two ideals and is therefore a field (as $I$ is maximal in $R$ since $f(x)$ is irreducible).
I wanted to ask, is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field? I would ideally like some way of demonstrating that the existence of a nonzero polynomial equivalent to zero in $R'$ somehow allows us to describe an algorithm to calculate multiplicative inverses...
 A: This is more an answer to your first question is there an intuitive way of understanding why taking the quotient by some ideal makes $F[x]$ into a field?
The main idea, is that for a commutative ring $R$ with unity, and an ideal $J$ of $R$, the following are equivalent :


*

*$J$ is a maximal ideal.

*$R/J$ is a field.


This is a general result on commutative rings with unity. Proving that 1. implies 2. is easy. Take $x \notin J$. Then as $J$ is maximal, $(x)+J=R$. Which means that it exists $(a,j) \in R \times I$ such that $ax+j=1$ or $\bar a \bar x = \bar 1$ in $R/J$ proving that $R/J$ is a field.
Now take the ring $R=F[x]$. It is commutative with unity. But it has more specificities. It is an Euclidean domain as on the polynomials over a field, the polynomial division is an Euclidean division.
And in a Euclidean domain, an irreducible polynomial generates a maximal ideal. Applying the result I mentioned in introduction, you get that $F[x]/(f(x))$ is a field.
Not sure that it is intuitive. However if you can look in details at the proof of the introduction result on maximal ideals, it can be a way to get a kind of intuition on your question.
