In Proposition 2.9. on page 304 of Silverman's textbook Advanced Topics in the Arithmetic of Elliptic Curves:
Let $R$ be a DVR with field of fractions $K$, residue field $k$, and maximal ideal $\mathfrak{p}$. Let $X$ be an integral (i.e., reduced and irreducible) $R$-scheme of finite type over $R$ whose generic fiber $X_{\eta}/K$ is non-empty. Then $X$ is a smooth $R$-scheme if and only if $X_{\eta}(\overline{K})$ and $X_{\mathfrak{p}}(\overline{k})$ contain no singular points.
In the begining of its proof, Silverman claims
The scheme $X$ is irreducible, so it has a unique generic point. Our assumption that the generic fiber $X_{\eta}$ is nonempty shows that the generic point of $X$ maps to the generic point of $\text{Spec}\; R$, so $X$ is flat over $\text{Spec}\; R$ from Hartshorne [1, III.9.7].
My question is that, why the generic fiber $X_{\eta}$ being nonempty shows that the generic point of $X$ maps to the generic point of $\text{Spec}\; R$? Moreover, how does the flatness criterion (III.9.7.) in Hartshorne implies that $X$ is flat over $\text{Spec}\; R$? Proposition III.9.7. in Hartshorne stats that
Given a morphism of schemes $f: X \rightarrow Y$ with $Y$ integral and regular of dimension $1$. Then $f$ is flat if and only if every associated point $x \in X$ maps to the generic point of $Y$.
In our case, we have $Y = \text{Spec}\; R$ which has two points $0$ and $\mathfrak{p}$. But then what even are the associated points of $X$? By reading Silverman's claim, it seems like the generic point of $X$ should be the only associated point of $X$?
Also in Remark 7.7 on page 345:
Let $R$ be a DVR and $C \subset \mathbb{A}_R^2$ an arithmetic surfaced defined by a single equation $$f(x,y) = 0\;\;\;\;\;\text{for some polynomial $f \in R[x,y].$}$$
i.e., $C = \text{Spec}\; R[x,y]/(f)$. In order to ensure that $C$ is a two dimensional scheme whose special fiber has dimension one, we will assume
that $f$ is not a constant polynomial and that at least one coefficient of $f$ is a unit in $R$. In fancier terminology, this means that $C$ is flat over $R$.
I'd like to ask how does the condition Silverman imposes imply that $C$ is flat over $R$?
Best Answer
Claim. If $f:X\to Y$ is a morphism of irreducible schemes with generic points $\eta_X$ and $\eta_Y$ respectively, then either $f(\eta_X)=\eta_Y$ or $f(X)$ is contained in a proper closed subset of $Y$.
Proof. If $f(\eta_X)=\eta_Y$, then $\overline{f(X)} \supset \overline{\{f(\eta_X)\}} \supset \overline{\{\eta_Y\}} = Y$. On the other hand, $f(X)=f(\overline{\{\eta_X\}})\subset \overline{f(\eta_X)}$, so if $f(\eta_X)\neq \eta_Y$ then $f(X)$ is contained in the closed subset $\overline{f(\eta_X)}$, which is proper because $\eta_Y$ is the only point with closure $Y$. $\blacksquare$
This deals with the first question: if the image of the generic point of $X$ is not the generic point of $\operatorname{Spec} R$, then $f(X)$ would be contained in a closed subset and the fiber over the generic point would be empty.
The flatness criterion has already been mentioned in the comment. The key idea is that the only associated point on a noetherian integral scheme is the generic point. Here's how to see this: for a noetherian scheme, the definition of an associated point is one where the maximal ideal of the local ring consists entirely of zero-divisors. For an integral scheme, this condition is only satisfied at the generic point, so the unique associated point of $X$ is $\eta_X$ which maps to the generic point of $\operatorname{Spec} R$ and therefore we may apply the criteria for flatness from Hartshorne.
The final claim follows from the above observation with a little bit more work. If all the coefficients of $f$ are in the maximal ideal, then there is an irreducible component of $V(f)\subset\Bbb A^2_R$ which maps only to the special point: if $r$ is a uniformizer for $R$, then the statement that all coefficients of $f$ are in the maximal ideal implies that $f=r^ng$ for $n>0$ and some $g\in R[x,y]$ which has a coefficient which is a unit. Then $V(f)=V(r^n)\cup V(g)$, and the generic point of $V(r^n)$, which is an associated point of $V(f)$, does not map to the generic point of $\operatorname{Spec} R$. Thus $V(f)\to\operatorname{Spec} R$ cannot be flat if $r\mid f$, and in fact one can show by running this argument backwards that this is an equivalence.