It is wellknown that the set $$ \mathcal{A} = \bigg\{ x\in \Bbb C: \sum_{i=1}^n a_i x^i = 0, a_i \in \mathbb{Z}\text{ and } n \text{ is a positive integer}\bigg\}, $$ is the set of all algebraic numbers, so by a heuristic argument, now let's consider the set $$ \mathcal{P} = \bigg\{ x\in \Bbb C : \sum_{i=1}^n a_i x^i = 0, a_i \in \pm \mathbb{P}\cup\{ 0\}\text{ and }n\text{ is a positive integer}\bigg \}, $$ where $\mathbb{P}$ is the set of all prime numbers. Now the question is: does the set $\mathcal{P}$ has a good algebraic structure?
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1$\begingroup$ By "algebraic structure" I guess you mean to say that the first set $\mathcal{A}$ is also a field. The operations on $\mathcal{A}$ transfer computatively to the $a_i$'s. Hence you can check whether $\mathcal{P}$ has an algebraic structure (group, ring, ...) by hand. Or, more likely, you can compute explicit counterexamples to any such structure induced by the usual operations. $\endgroup$– Loïc TeyssierDec 28 '20 at 10:36

3$\begingroup$ Out of curiosity, is there a particular motivation behind considering $\mathcal{P}$? $\endgroup$– Loïc TeyssierDec 28 '20 at 10:56

1$\begingroup$ Can you prove $\mathcal P$ is not the set of all algebraic numbers? For example is $\sqrt{6} \in \mathcal P$? $\endgroup$– Gerald EdgarDec 28 '20 at 12:59
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