$f(x) \in R[X]$ 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 $R[X]$.
 A: Ok, take any arbitrary ring $S$, a family of elements element $F ⊂ S$ and examine the set of finite linear combinations of elements of $F$ in $S$
$$S_F = \{Σ_{f ∈ F'}^n a_f f;~\text{$F' ⊂ F$ is finite and for all $f ∈ F'$, $a_f ∈ S$}\}.$$
Prove that:


*

*$S_F = \bigcap \{I ⊂ S;~\text{$I$ is an ideal of $S$ with $F ⊂ I$}\}$.

*Arbitrary intersections of ideals are ideals.


The right set in (1.) should be taken as the definition of the ideal generated by $F$. And to be clear, it is the intersection of all ideals in $S$ containing $F$. By (2.) $S_F$ is an ideal.
Your claim follows for $F = \{f\}$.

Edit: Actually, for your case just prove that $S_{\{f\}} = \{af;~a ∈ S\}$.

So this has nothing to with $R$ being a field or $f$ being irreducible.
A: 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...
