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 $\alpha \geq \omega_1$.

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 $\text{Add}(\omega,1)$.