subalgebra of a simple forcing

Question

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 $\alpha \geq \omega_1$.

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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.

Answer 1

Update. If $\alpha$ is countable, then I claim that $A$ and $B$ are forcing equivalent, the quotient forcing is atomic, isomorphic to $P(\omega_1)$, and every extension by $A$ adds a generic for $\text{Add}(\omega,\alpha)$.

Suppose that $\alpha$ is countable and we have $V[g][X]$, the extension by $B=\text{Add}(\omega,\alpha)\ast\text{Add}(\omega_1,1)$. Thus, $g$ is an $\alpha$ sequence of Cohen reals, and we may string them together end-to-end and make a binary sequence of length $\omega\cdot\alpha$. In the model $V[g]$, it is dense that this sequence appears explicitly as a block in the generic set $X$, since any condition in $V[g]$ can be extended to include it. Thus, it follows that $g\in V[X]$ and at least in this situation, the set $X$ explicitly gives us a $V$-generic $\alpha$-sequence of Cohen reals, namely $g$.

But more to the point, it follows that $V[g][X]=V[X]$. Thus, forcing with $A$ or $B$ gives rise to the same extensions, and so $A$ and $B$ are forcing equivalent. As Andreas pointed out in a comment to my earlier answer (and as the OP seems to be aware), this does not necessarily mean that $A=B$, although it does mean that the quotient forcing is atomic.

Note that every ordinal $\beta\lt\omega_1$ has a nonzero Boolean possibility in $B$ of being the first ordinal where $g$ appears as a block in $X$. Thus, we have the Boolean value $b_\beta$ which is the Boolean value in $B$ that this is the case. It follows that $\{b_\beta\mid\beta\lt\omega_1\}$ forms a maximal antichain in $B$. Furthermore, I claim that $A$ together with the $b_\beta$'s generate all of $B$, because if we know the values in $A$ and we also know which $b_\beta$ holds then we can compute any Boolean value in $B$, which is determined by the information about $g$ and the information about $X$. Finally, notice that the OP's automorphism argument in the comments shows that no $b_\beta$ is in $A$, since we can know $X$ fully and still not know $g$ exactly, since we might have had $\pi(g)$ instead. So no information about $X$ (which is all $A$ knows about) can tell us anything definite about where the block of $g$ starts.

Thus, the quotient forcing is the atomic forcing using the atoms $b_\alpha$, which is isomorphic to the power set $P(\omega_1)$.

Lastly, let me just mention---although you probably know this already---that when $\alpha$ is countable, then $\text{Add}(\omega,\alpha)$ is isomorphic to $\text{Add}(\omega,1)$.

Answer 2

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.