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?
On
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.
Upgrading comments to an answer:
Let $p$ be a prime, $A$ a (commutative, unital) ring and $W(A)$ its ring of ($p$-typical) Witt vectors. References are [Bou] Bourbaki, Algebre Commutative ch. IX §1, and [Haz] Hazewinkel, Witt vectors. Part 1, arXiv:0804.3888v1.
In particular let $\Phi_n = \sum_{i=0}^n p^i X_i^{p^{n-i}}$ be the $n$-th Witt polynomial. For clarity, I will use $\times$ for the multiplication in $W(A)$, while multiplication in $A$ is just written with a dot $\cdot$.
Fact 1: If $W(A)$ is an integral domain, then so is $A$.
Proof: For $a,b \in A$, one has $(a,0,0,....) \times (b,0,0,...) = (a \cdot b, 0,0,...)$ ([Bou] no. 6 Prop. 4; [Haz] 5.21.). It follows that non-trivial zero divisors in $A$ give rise to non-trivial zero-divisors in $W(A)$.
Fact 2: If $p\cdot 1_A$ is invertible in $A$, then $W(A) \simeq A^\mathbb N$ (countable direct product) as rings. In particular, $W(A)$ has non-trivial zero-divisors.
Proof: (This is actually used to construct all the summation and multiplication polynomials.) See [Bou] no.2 Prop. 2 and no. 4 Theoreme 1, and [Haz] Theorem 5.14.
To construct "explicit" zero-divisors e.g. for $A=\mathbb Q$, take e.g. the obvious pair of zero-divisors $g_1:=(1,0,0,...), g_2=(0,1,0,0,...)$ in $\mathbb Q^\mathbb N$ and "pull them back" i.e. find actual Witt vectors $(x_0,x_1,...), (y_0,y_1,...) \in W(\mathbb Q)$ which have $g_1,g_2$ as ghost components, by setting $x_0=1$ and then recursively solving $$\Phi_{n-1}(x_0^p, ..., x_{n-1}^p) + p^n x_n =0 \qquad \text{for } x_n$$ as well as $y_0=0$,
$$\Phi_{0}(y_0^p) + p^1 y_1 = \color{red}{1} \qquad \text{i.e. } y_1=\frac{1}{p}$$
and then recursively
$$\Phi_{n-1}(y_0^p, ..., y_{n-1}^p) + p^n y_n =0 \qquad \text{for } y_{n \ge 2}.$$
I get $x=(1, -\frac{1}{p}, -p^2+p^{p-1}, ...), y= (0, \frac{1}{p}, -p^{p-1}, ...)$. See how all you need is to be able to invert $p$ in $A$, i.e. this works in any $\mathbb Z[\frac{1}{p}]$-algebra.
Fact 3: If $p\cdot 1_A =0_A$ and $A$ is an integral domain, then $W(A)$ is an integral domain.
Proof: [Haz] Cor. 6.4, or look into the proof of [Bou] no. 8 Prop. 8 (which is stated for fields, but this part does not need the full assumption). The idea is: Let $x',y' \in W(A)$ be non-zero; then they can be written as $V^m((x_0, x_1, ...))$ and $V^n((y_0,y_1,...))$ for certain $m,n \in \mathbb N_0$ and $x_0 \neq 0 \neq y_0$. Now because of the condition on $A$, one has $$V^m(w_1) \times V^n(w_2) = V^{m+n}(F^n(w_1) \times F^m(w_2))$$ ([Bou], no. 8 Prop. 5; [Haz], last of formulae 6.2) for all $w_1, w_2 \in W(A)$; in particular, the $(m+n)$-th component of $x' \times y'$ is $x_0^{p^n} \cdot y_0^{p^m} \neq 0$.