When is scalar multiplication decidable?
A real number is called quadratic, if it is a root of a quadratic polynomial with rational coefficients. The scalar multiplication by \(a\in\mathbb{R}\) is the function \(\lambda_a\colon\mathbb{R}\rightarrow\mathbb{R}\) defined by \(\lambda_a(x)=a\cdot x\). The question of the decidability of the structure \(\mathcal{S}_a=\langle\mathbb{R};<,+,\mathbb{Z},\lambda_a\rangle\) was left open in an earlier paper of the author [J. Symb. Log. 81, No. 3, 1007--1027 (2016; Zbl 1436.03208)] which is a prerequisite for reading the present paper. The main result (Theorem D) is that for \(d\in\mathbb{Q}\), the function \(\lambda_{\sqrt{d}}\) is definable in \(\mathfrak{A}_{\sqrt{d}}=\langle\mathbb{R};<,+,\mathbb{Z},\mathbb{Z}\sqrt{d}\rangle\). Since it was proved in [loc. cit.] that \(\mathfrak{A}_{\sqrt{d}}\) is decidable for \(d\in\mathbb{Q}\), then it follows that \(\mathcal{S}_a\) is decidable for quadratic real \(a\). An earlier result of the author with \textit{M. Tychonievich} [Proc. Am. Math. Soc. 142, No. 9, 3259--3267 (2014; Zbl 1337.03054)] states that \(\mathcal{S}_a\) is undecidable, when \(a\) is not quadratic. This implies that the structure \(\mathcal{S}_a\) is decidable, if and only if \(a\) is quadratic (Theorem A). The following theorems are obtained as corollaries of Theorem A: Theorem B: For \(S\subseteq\mathbb{R}\), the structure \(\langle\mathbb{R};<,+,\mathbb{Z},\{\lambda_b\}_{b\in S}\rangle\) defines the same sets as exactly one of the following structures: (i) \(\langle\mathbb{R},<,+,\mathbb{Z}\rangle\), (ii) \(\mathcal{S}_a\), for some quadratic \(a\in\mathbb{R}\setminus\mathbb{Q}\), (iii) \(\langle\mathbb{R},<,+,\cdot,\mathbb{Z}\rangle\). The three cases are exclusive. Theorem C: Let \(K\) be a subfield of \(\mathbb{R}\). The theory of the ordered \(K\)-vector space \(\mathbb{R}\) expanded by a predicate for \(\mathbb{Z}\) is decidable, if and only if \(K\) is a quadratic field. The paper finishes with some open questions, one of which is that whether \(\lambda_a\) is definable in \(\mathfrak{A}_a\), when \(a\in\mathbb{R}\) is arbitrary (even quadratic). One other interesting question is whether \(\mathfrak{A}_a\) is decidable, when \(a\) is non-quadratic (its decidability for quadratic \(a\) is proved in [the author, J. Symb. Log. 81, No. 3, 1007--1027 (2016; Zbl 1436.03208)].
expansions of real ordered additive groups
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