TITLE: subalgebra of a simple forcing
QUESTION [7 upvotes]: Let $\alpha > 0$ be any ordinal.  Consider the forcing $Add(\omega,\alpha) * Add(\omega_1,1)$.  Let $B$ be its boolean completion.  Let $\dot{X}$ be the canonical $B$-name for the generic subset of $\omega_1$ added by the second iterand.  By forcing folklore, there is a complete subalgebra $A \subseteq B$ that adds $X$, completely generated by the boolean values $|| \check{\beta} \in \dot{X} ||$  (see Jech p. 247).  One can check that for each $\beta < \omega_1$, $|| \check{\beta} \in \dot{X} || = (1,\check{ \lbrace (\beta,1) \rbrace } ) $.
Questions:
(1) What is the nature of $A$ in relation to $B$?  Although $A$ adds all the reals of $B$, I think I can show that it is always a proper subset of $B$ (even for $\alpha = 1$).  This leads to the second question.
(2) If $G$ is generic for $A$, what is the nature of the quotient algebra $B/G$?  One can show that it is $(\omega,\infty)$-distributive.  In the case where $\alpha$ is countable, it is atomic.  Is it nonatomic for uncountable $\alpha$?  Is it strategically closed?
(3) If $G$ is generic for $A$, does $V[G]$ always contain some $Add(\omega,\alpha)$ generic, like in the case $\alpha$ is countable?
Edit: I thank J.D. Hamkins for his answer, but I already knew about the case $\alpha < \omega_1$, as stated in the original post.  I would really like some insight on the case $\alpha \geq \omega_1$.

NEW RELATED QUESTION
Consider instead $Col(\mu,<\kappa)*Add(\kappa,1)$, where $\mu< \kappa$ are regular.  Let $B$ be the completion, and let $A$ be the complete subalgebra generated by the canonical name for the Cohen subset of $\kappa$ added by the second part.  Let $H \subseteq A$ be generic.
Does $B/H$ have a $\mu$-closed dense subset in $V[H]$?  Any proof or refutation would be great, or a discussion of other related structural aspects of $A,B$, and $B/H$.  The properties of this forcing in relation to large cardinals are important to my research, so all nontrivial information is welcome.

REPLY [2 votes]: I have answers the original questions, by way of counterexample, which I will sketch.  The "new related question" remains unsolved.  I am grateful to Mohammad Golshani for his answer to this question, which I realized helps to answer the present question.
Assume CH holds in the ground model $V$.  Let $B_0$ be the completion of $Col(\omega,<\omega_2) * Add(\omega_2^V,1)$.  Let $\mathbb{P}$ be the countable support product of $Col(\omega,\alpha)$ for $\alpha < \omega_2$, and let $B_1$ be the completion of $\mathbb{P} * Add(\omega_2^V,1)$.
Let $A_0 \subseteq B_0$ be the complete subalgebra generated by the name for the Cohen subset of $\omega_2^V$ from the second iterand, and let $A_1 \subseteq B_1$ be the same with respect to $B_1$.  The following is a special case of a more general lemma that is part of my work in progress, to be submitted soon.  (CH is used)

Claim: $A_0 \cong A_1$, and $X \subseteq \omega_2^V$ is generic for $A_0$ iff it is generic for $A_1$.

Now, Mohammad Golshani gives examples of two $\Sigma_1$ properties, $\varphi_0, \varphi_1$, in parameters from $V$, such that whenever $G_0 \subseteq Col(\omega,<\omega_2)$ is generic, and $G_1 \subseteq \mathbb{P}$ is generic, $V[G_0] \models \varphi_0 \wedge \neg \varphi_1$, and $V[G_1] \models \varphi_1 \wedge \neg \varphi_0$.  By the Claim, if $X \subseteq \omega_2^V$ is generic for $A_0$ (and $A_1$), then we necessarily have $V[X] \models \neg \varphi_0 \wedge \neg \varphi_1$.
Now we note that $Col(\omega,<\omega_2)$ is equivalent to $Col(\omega,\omega_1) \times Add(\omega,\omega_2^V)$.  Let $g: \omega \to \omega_1^V$ be generic for the left side.  Then $B_0 / g$ is equivalent to $Add(\omega,\omega_1^{V[g]}) * Add(\omega_1^{V[g]},1)$, like in the original question.  For any $G_0 * X$ generic for $B_0$ extending $g$, we have $g \in V[X]$, and $V[X]$ contains no filter over $Add(\omega,\omega_2^V)$ which is generic over $V[g]$, using Mohammad's lemma and the above factoring.  This answers question (3).
It also answers question (2).  The quotient in this case is indeed atomless, and it is not $<\omega_1$-strategically closed.  If it were, then we could construct an $Add(\omega,\omega_1^{V[g]})$-generic (over $V[g]$) filter, by recursively deciding initial segments.  This also sheds some light on question (1), although that is a broad and vague question.