Ring of Witt vectors over an integral domain Let $p$ be a prime and $A$ is a commutative ring. The ring of Witt vectors $W(A)$ is defined via $X_0^{p^n}+...+p^nX_n$ (not the general one for arbitrary $n$).
I have read somewhere ring of Witt vectors over an integral domain of prime characteristic is an integral domain. Why this is true? And why do we need the prime characteristic condition?

Alternatively:
Note the functoriality of Witt vectors: A ring homomorphism $f:R \rightarrow S$ induces a morphism $W(f): W(R)\rightarrow W(S)$ by the rule $W(f)((r_0, r_1, r_2, \dots ))=(f(r_0), f(r_1), f(r_2), \dots)$. In particular, observe that if $f$ is injective, then so is $W(f)$.
Now a very classical fact about Witt vectors is: If $k$ is a perfect field of characteristic $p>0$, then $W(k)$ is a DVR (This is e.g. somewhere in Serre's Local Fields I believe).
So now, if $R$ is a domain of charactestic $p$, we can embed it into its fraction field $k_0$ and in turn embed $k_0$ into its algebraic closure $k$. This way, we get an injective morphism $R \hookrightarrow k$, which induces an injection $W(R)\hookrightarrow W(k)$. In particular, $W(R)$ is a subring of the domain $W(k),$ hence a domain itself.
