Elementary suborders of Cohen forcing My question is basically whether being a Cohen poset is a first-order statement within the order itself. 
More specifically, let $\mathbb{P}$ be an elementary suborder of $\mathrm{Add}(\omega,\lambda)$, the poset adding $\lambda$ many Cohen reals. Is $\mathbb{P}$ forcing equivalent (or maybe even isomorphic) to $\mathrm{Add}(\omega,|\mathbb{P}|)$? 
 A: Yes, we in fact get full isomorphism; here's a sketch of the proof.
Let $M$ be an elementary subposet of $Add(\omega, \lambda)$. We'll show $M$ is in fact isomorphic to $Add(\omega, \lambda)$. We begin by looking at the atomic conditions: conditions corresponding to finite maps with domain a singleton. These are characterized by the following property:

$p$ is atomic if there is some $\overline{p}$ such that $r\perp p$ iff $r\le \overline{p}$.

This is first-order, so the atomic conditions in $M$ are exactly those atomic conditions in $Add(\omega, \lambda)$ which are in $M$, and the "complement" operation $p\mapsto\overline{p}$ is again the restriction of that from $Add(\omega,\lambda)$.
Now $M$ also satisfies that every element is covered by finitely many atomic conditions; and so by counting, $M$ contains $\lambda$-many atomic conditions - in fact, $M$ consists precisely of a set of $\lambda$-many complementary pairs of atomic conditions, as well as the meets of all compatible finite sets of these. From here it's easy to show that $M$ is in fact isomorphic to all of $Add(\omega, \lambda)$.

This assumes that we're viewing $Add(\omega, \lambda)$ as the set of partial maps $\lambda\rightarrow 2$ with finite domain. If we're viewing this as the set of partial maps $\lambda\rightarrow \omega$ with finite domain, then obviously the characterization of atomic conditions doesn't work; instead, we use

$p$ is atomic if there is no $q$ with $\emptyset\ge q\ge p$, $q\not=\emptyset, p$. 

Then - again - $M$ is generated by a set $X$ of atomic conditions of size $\lambda$, and $X$ is carved into $\lambda$-many "blocks" of size $\omega$: two elements of $X$ being compatible iff they lie in different blocks. This, again, establishes that $M\cong Add(\omega,\lambda)$.
