The definition for an ideal to be principal is:
Let $R$ be a commutative ring with a unit element. An ideal $I$ of $R$ is $\textbf{principal}$ if there exist $a\in R$ such that $I=\{ar\mid r\in R\}$. In this case, $a$ is said to be $\textbf{generate}$ $I$.
There is something that I am still confused about from the wording of the definition. Since $R$ is a commutative ring with a unit element, that implies that $R$ contains the identity element. If an ideal $I$ is a principal ideal, in non mathematical notational worded description, it means that all the element of the ideal $I$ can be generated by some single element $a\in R$. Nowhere does the definition state that $a$ has to be an element of $I$. Can we not have one of the following two scenarios $(1)$ where both $a$ and $r$ are in $R\setminus I$ or $(2)$ $a\in I$, $r\in R\setminus I$, but $ar \in I$.
I posted several times in regards to the concept of principal ideal because when I look over the proof of why the ideal of the set of polynomials with constant term being even, or similarly the constant terms being divisible by three, the usual proof is to show that why the ideals $I=\langle 2, x\rangle$ or $J=\langle3, x\rangle$ of the commutative ring $\mathbb{Z}[x]$, are not principal ideals. For either of these proofs, why can I not use the identity element in $\mathbb{Z}[x]$ as a generator to serve as the single element generator to obtain all the elements in either $I$ or $J$, even thought the identity element is neither in $I$ or $J$.
Thank you in advance.
Best Answer
If the ring is unital (i.e. has a unit) then $a=1\cdot a\in I$. Hence the principal ideal $\langle a\rangle$ always contains its generator $a$ in such a setting. You have to be more careful if you ring is missing a unit, though. Regarding to your two cases: $(1)$ will never happen in a unital ring and $(2)$ always holds by definition.
Let's take a closer look at the ideal $I=\langle2,x\rangle<\mathbb Z[x]$. For $I$ to be principal we have to find some $f(x)\in\mathbb Z[x]$ such that $I\mathbf{=}\langle f(x)\rangle$. However, if you take $f(x)=1$ then $\langle f(x)\rangle=\mathbb Z[x]\supsetneq I$ missing the crucial equality. Slightly more general: any ideal containing a unit (a multiplicatively invertible element) will be the whole ring. See here for instance for the details.