Condition for the generic fiber to be dense This is an assertion in the article Groupes Reductifs sur un Corps Locale II by Bruhat and Tits.  Here $A$ is an integral domain with field of fractions $K$, and $\mathfrak X$ is an affine $A$-scheme with coordinate ring $B := A[\mathfrak X]$.

I don't think this assertion that density is equivalent to torsion free is true in the generality they state.  I can show $\Rightarrow$ if $B$ is reduced.  
Density means that for any $b\in B$ which is not in the nilradical of $B$, there exists a prime $\mathfrak p$ of $B$ such that $b \not\in \mathfrak p$ and $\phi^{-1}\mathfrak p = (0)$.  
Scratchwork: Assume $B$ is reduced.  For the implication $\Rightarrow$, assume $a \in A$ and $b \in B$ are both not zero.  We need to show that $\phi(a) b \neq 0$.   Then $b$ is not nilpotent, which means the open set $D(b)$ is nonempty.  By density, there exists a prime $\mathfrak p$ of $B$ such that $b \not\in \mathfrak p$ and $\phi^{-1}\mathfrak p = (0)$.  Since $\phi(a)$ and $b$ each do not lie in $\mathfrak p$, neither does $\phi(a)b$, so $\phi(a)b \neq 0$.
The implication $\Leftarrow$ is a standard result: an injective ring homomorphism makes for a dominant morphism of schemes.

You are correct: this result is not true without assuming $B$ is reduced.  For a simple example, take $A=\mathbb{Z}$ and $B=\mathbb{Z}[x]/(x^2,2x)$.  Then $A$ is $B$ mod its nilradical, so the map $\operatorname{Spec} B\to \operatorname{Spec} A$ is a homeomorphism and so the generic fiber is dense in $\operatorname{Spec} B$ since the generic point is dense in $\operatorname{Spec} A$.  But $B$ is not torsion-free: as a $\mathbb{Z}$-module, it is isomorphic to $\mathbb{Z}\oplus\mathbb{Z}/(2)$.
