$f(x) \in R[X]$ irreducible $\Rightarrow (f(x))$ ideal? If $R$ is a field, and $f(x) \in R[X]$ is an irreducible polynomial in $R[X]$, does then follow that $(f(x))$ is a (principal) ideal in $R[X]$?
[Edit]
My understanding (which may be wrong) of (f(x)) is the subset that is generated by taking all (finite) linear combinations of f(x), and I want to know if this is an ideal of $R[X]$.

Your question boils down to basic logic.
Given any element of a ring we can consider the ideal generated by it. Irreducibility is irrelevant.
So your question is the following: If I consider the ideal generated by (the one polynomial) $f(x)\in R[x]$ then is it generated by one element?
The answer is yes by the reflexive property of equality...
