factorial ring basis $\mathbb{Z}/2\mathbb{Z}$ 
*

*i) $\mathbb{Z}/2\mathbb{Z}$ has no $\mathbb{Z}$ basis.   

*ii) $\mathbb{Z}/2\mathbb{Z}$ has no $\mathbb{Z/2\mathbb{Z}}$ basis.   

*iii) Suppose $R= \mathbb{Z}[t]$. Then $2R+tR$ has no R-basis. 
i) Let $R= \mathbb{Z}$ an $L_{module}=\mathbb{F}_{2}=\mathbb{Z}/2\mathbb{Z}$. Then $L\subset \mathbb{Z} 1 (1 \in \mathbb{F}_{2})$. And $L=\mathbb{Z}1$. But $\mathbb{Z}1$ is not linearly independent, because $2\in \mathbb{Z}, 2 \ne 0$, but $21 = 0$
So $\mathbb{Z}/2\mathbb{Z}$ does not have a $\mathbb{Z}$ basis. 
ii)  Let $R= \mathbb{Z}/2\mathbb{Z}$ an $L_{module}=\mathbb{F}_{2}=\mathbb{Z}/2\mathbb{Z}$
Then $L= R = \mathbb{Z}/2\mathbb{Z}$. So it is a basis of itself, and thats why the assumption is not correct. 
iii) It is given that $R= \mathbb{Z}[t]$ an $I=2R+Rt$. (I believe) that it is a principal ideal domain. It is a R module, with generators 2,t an N=2. It can not have a basis because a basis (w) (n=1) would mean $I=Rw$, but that is not possible (I believe). Also (2,t) can not be a basis since $2,t$ are dependent : $-t, 2 \in R $ not all 0 , $(-t)2+2t=0$.
Everywhere where I believe something, I am not able to prove it. Can somebody help me prove my beliefs? Thanks. 
 A: Aside: Why introduce new letters and new symbols to clutter up things? What's the point of introducing $L$ if the only thing we are going to do with $L$ is say that it is equal to $\mathbb{F}_2$, and why introduce $\mathbb{F}_2$ if it's just $\mathbb{Z}/2\mathbb{Z}$?
Also, "$L_{module}$" does not really clarify things; it makes them worse. It's clunky notation, doesn't seem to mean anything (what does it mean when you write "module" under a structure? To specify that we are considering a certain abelian group $M$ as an $R$ module, we usually write $M_R$ or ${}_RM$, not $M_{module}$, and we certainly don't write $R_{module}$).

(i) The reason $\mathbb{Z}/2\mathbb{Z}$ has no $\mathbb{Z}$-basis is that every element of $\mathbb{Z}/2\mathbb{Z}$ satisfies $2x = 0$; this means that no nonempty set is linearly independent, and since $\mathbb{Z}/2\mathbb{Z}\neq\{0\}$, then the empty set does not generate the module. So no set can be both linearly independent and span.
(ii) On the other hand, $\mathbb{Z}/2\mathbb{Z}$ does have a $\mathbb{Z}/2\mathbb{Z}$ basis. The singleton $\{\overline{1}\}$ is a basis (we are dealing with vector spaces here, and vector spaces always have bases).
(iii) $\mathbb{Z}[t]$ is not a principal ideal domain: if $\mathbb{Z}[t]$ were a principal ideal domain, then $(2,t)$ would be generated by a $\gcd$ of $2$ and $t$. But $2$ and $t$ have no common divisors other than $1$ and $-1$ ($\mathbb{Z}[t]$ is a UFD). Yet $(2,t)\neq (1)$. 
The reason $(2,t)$ does not have a basis: as noted above, any generating set must have more than one element (the ideal is principal). But in $\mathbb{Z}[t]$, any two elements are linearly dependent over $\mathbb{Z}[t]$: given $p(t)$ and $q(t)$, not both zero, we have $q(t)p(t) + (-p(t))q(t) = 0$. If they are both zero, then $1p(t)+1q(t)=0$. So no set with more than one element can be independent, and no set with one element or fewer can generate $(2,t)$, so $(2,t)$ does not have a basis.
A: Ummmm...given any commutative ring with identity $R$, there is a basis for $R$ as an $R$-module, namely $\{1\}$.  In fact, an $R$-module $M$ has a basis (of elements in $R$) precisely when $M$ is a free $R$-module.  Since $\mathbb{Z}/2\mathbb{Z}$ is finite and any free $\mathbb{Z}$ module (except $0$) is infinite, it is clear that $\mathbb{Z}/2\mathbb{Z}$ is not a free $\mathbb{Z}$ module and hence has no basis in $\mathbb{Z}$.
Of course, $R$ has a basis as an $R$-module, namely $\{1\}$.  Therefore ii) is incorrect.
Also, $\mathbb{Z}[t]$ (where, presumably, $t$ is an indeterminate) is not a principal ideal domain, since $I=(2,t)=2R+tR$ is not principal (if you're not sure why this is the case, it would be worth writing down a proof).
