At a proof that I am reading there is the following:

$R$ is an integral domain,
$S$ satisfes the following conditions:
-$S \subset R$
-$S$ contains $\mathbb{Z}$
-$S$ is diophantine over $R[T]$

If $R$ contains $\mathbb{Q}$, then we define $S$ by:
$x \in S \leftrightarrow x \in R[T] \land (x=0 \lor \exists y \in R[T]: xy=1)$


I am facing some difficulties understanding the last implication, the definition of $S$.
Since $S$ is a subset of $R$, we have that $x \in S \rightarrow R[T]$. Does the parenthesis $(x=0 \lor \exists y \in R[T]: xy=1)$ have a relation with the fact that $S$ contains $…

@DanielFischer Hi. I have an operator theory related question. Do you mind having a look at this:
Define for any $ f \in \mathcal{H}$ol(a), where $a$ is an element of a Banach algebra $\mathcal{A}$, the function $$f(a) = \frac{1}{2 \pi i} \int_{\Gamma}f(z)(z-a)^{-1}dz \in \mathcal{A}$$

given that $f: \Omega \to \mathcal{C}$ is analytic and $\Gamma$ is any finite collection of closed ctonours with $Ran(\Gamma) \subset \Omega$ and $\text{Ind}_{\Gamma}(z) = 1$ for all $z$ in $\alpha(a)$, where $\alpha(a)$ is the spectrum of $a$. If we define $$\Phi(f) := f(a) := \frac{1}{2 \pi i}\int_{\Gamma}…

@Semiclassical The reduce program says that hypersum(\{a,a,n+1\}, \{1,n+2\},1,n) Has two solutions Odd and Even n:$\left\{ \frac{(- 1)^{\frac{n}{2}} (n + 1) !n!}{4^n \left( \frac{n}{2}
\right) !^2 \mathrm{pochhammer} \left( \frac{- a + 3}{2}, \frac{n}{2}
\right)^2} \hspace{0.17em} \mathrm{, \hspace{0.17em}} \hspace{0.17em} $\frac{(-
1)^{\frac{n}{2}} \left( \frac{n - 1}{2} \right) ! \left( \frac{n + 1}{2}
\right) !i (- 2 a^2 - 3)}{3 \mathrm{pochhammer} \left( \frac{- a + 4}{2},
\frac{n - 1}{2} \right)^2} \right\}$

@DanielFischer In $C^*$ we define that element $a \in A$ is positive if $a=a^*$ and $\sigma(a) \subset [0,\infty)$. And we can prove that in case of $B(H)$ (special case of $C^*$ algebra) that definition for some operator $T$ is equivalent with $\langle Tx,x \rangle \geqslant 0$ for all $x \in H$.

Then, I say, it awkward to me that definition in $C^*$ for strict positivity which says $a=a^*$ and $\sigma(A) \subset (0,\infty)$ is not equivalent (for case $B(H)$) with $\langle Tx,x \rangle >0$ for all $x \neq 0$.

@DanielFischer I have one question related to question that Moses asked you. Namely, we define holomorphic functional calculus as $\Psi(f)=\frac{1}{2\pi} \int_{\Gamma} .......$ for $f \in H(\sigma(a))$, and then we say (first prove for entire functions) $\Psi(f)=f(a)$. I have two questions:

1. when we say $f \in H(\sigma(a))$ we assume that $f$ maps to $\mathbb{C}$?

2. when we say $\int_{\Gamma} f(\lambda) (\lambda - a)^{-1} d\lambda$, okay, $\Gamma$ is contour in $\mathbb{C}$, so this is just line integral (??), but we have (formal) some element of algebra inside that integral and I don'…