On representing Turing Degrees relative to iterates of the jump Consider the following definition. We define the 'index' of a Turing degree $A$, (denoted $i(A)$) as the smallest ordinal $\alpha\in \omega_1$ such that $A \ngeq 0^{(\alpha)}$. [is this concept found anywhere in the literature, and if so, under what name?]
I have 2 questions relating to this concept:
1) Does the following property hold: $i(A\vee B) = \max\{i(A), i(B)\}$?
2) If $A$ is a Turing degree, and $i(A)=\alpha+1$ is a successor ordinal, then can we always write $A = X \vee 0^{(\alpha)}$ where $X$ is a Turing degree satisfying $i(X) = 1$? Is there a natural way to extend this idea to limit ordinals?
I'm suspecting that the first question is false, and the second is true, but I don't know where to start.
 A: First, note that "$0^{(\alpha)}$" doesn't make sense unless $\alpha$ is a computable ordinal. We can do a bit better, especially via mastercodes, but that still only reaches up to $\omega_1^L$ (which might be vastly smaller than $\omega_1$).
So let's restrict attention to those Turing degrees which are not above every set of the form $0^{(\alpha)}$ for $\alpha$ a computable ordinal (that is, those Turing degrees not bounding every hyperarithmetic degree). 
The answer to your first question is extremely "no:" for any computable $\alpha$, we can find sets $A, B$ neither of which computes $0'$ but such that $A\oplus B$ computes $0^{(\alpha)}$. 
Proof: let $\mathbb{P}$ be the set of pairs $(p, q)$ of finite partial maps $\omega\rightarrow 2$ with the same domain $D$, such that $\{a\in D: p(a)=q(a)\}\subseteq 0^{(\alpha)}$ and $\{a\in D: p(a)\not=q(a)\}\subseteq \overline{0^{(\alpha)}}$. Now taking a sufficiently generic filter through $\mathbb{P}$ yields a pair of sets which agree precisely on $0^{(\alpha)}$ (hence whose join computes $0^{(\alpha)}$) but neither of which computes $0'$ (since each of the two sets will be sufficiently Cohen-generic).
A: Even if you fix a meaning for $0^{(\alpha)}$ for every countable ordinal $\alpha$, by fixing a code for $\alpha$, say, then it still isn't necessarily the case in ZFC that every real is below one of them, and so you haven't really defined a degree notion on the degrees. The reason is that this would imply the continuum hypothesis, as there are only $\omega_1$ many countable ordinals $\alpha$, and each $0^{(\alpha)}$ would compute only countably many reals. 
In this sense, the existence of a backbone sequence of length $\omega_1$ cofinal in the Turing degrees is equivalent to CH. 
