Why $1+(-1)=0$ in the Hamiltonian Quaternion but not in the group ring $RQ_8$? 
*

*An example from Dummit & foote says that $1 + (-1) = 0$ in the Hamiltonian Quaternion, but nonzero in $RQ_8$ ($R$ is the real line, $Q_8$ is the Quaternion group).  
In the Hamiltonian Quaternion, $1 + (-1)$ are actually written as $$1 + 0i + 0j + 0k + (-1) + 0i + 0j + 0k = (1-1) + 0i + 0j + 0k,$$
and I do not understand why this equals zero.  Do we know what $1-1$ amounts to in the Quaternion group? 

*Also is a ring $R$ really a subring of $R[x]$.
Strictly speaking, a constant term in a polynomial ring is actually written as $(r,0,0,0,0,0,\ldots)$ which is not an element in $R$. So I do not really think $R$ is even an subset of $R[x]$.
$R$ is isomorphic to a subring of $R[x]$. But why can we conclude that $R$ is a subring of $R[x]$?
 A: You should really think of $\mathbb{H}$ (the Hamiltonians) as an algebra structure on $\mathbb{R}^4$ in the same way that $\mathbb{C}$ is an algebra structure on $\mathbb{R}^2$. In both cases, addition is pointwise (inherited from $\mathbb{R}$). In $\mathbb{H}$, multiplication given by the identification $1=(1,0,0,0)$, $i=(0,1,0,0)$, $j=(0,0,1,0)$ and $k=(0,0,0,1)$. So, for example,
$$(0,1,0,0)(0,1,0,0)=i^2=-1=(-1,0,0,0)$$
and
$$(0,1,0,0)(0,0,1,0)=ij=k=(0,0,0,1)$$
etc..
From this point of view, it is clear why $$1-1=(1,0,0,0)-(1,0,0,0)=(1-1,0,0,0)=(0,0,0,0)=0.$$
Now, for your second question, you are right that $R$ is ``isomorphic'' to a subring of $R[x]$, but it is common in mathematics to identify [groups, rings, etc.] with their image under an injective morphism (sometimes even when you shouldn't). In this particular case, there is nothing to be gained by distinguishing $R$ from the set of constant polynomials in $R[x]$.
A: *

*Elements of $RQ_8$ are formal linear combinations of elements of $Q_8$. Think of the elements of $Q_8$ as a basis for a vector space. Then 1 and $-1$ are linearly independent. In particular, $1 + (-1) \ne 0$.

*You're right, however $R$ is always canonically isomorphic to a subring of $R[x]$. So we identify the two.
