What are the closure properties of spectra? If $X$ and $Y$ are spectra, I can see that $X\cap Y$ and $X\cup Y$ are also spectra.
Is it always true that
$$X+Y:=\{n+m:n\in X,m\in Y\},$$
and
$$X\cdot Y:=\{n\cdot m:n\in X,m\in Y\}$$
are also spectra?
See also this related question for the special case $X^{2}=\{n^2:n\in X\}$.

My initial approach, say for $+$ was to use the construction in my answer here. Thus, we have the $\mathcal{L}$-sentence $A'$. Then we add two predicates to $\mathcal{L}$, one $Fx:\,x$ is in $X$, and and analogous $G$ for $Y$. Then define $B$ to be the sentence
$$((\neg\exists w\,Rzw)\wedge\exists x\,\exists y\,(Fx\wedge Gy\wedge Pxyz)).$$
Then consider $A'\wedge B$. But...

Edit: I found out this is related, answering closure under addition.
 A: The answer is yes. First let me give a helpful construction:

Definition: Let $L$ be a first-order language and $P$ a fresh unary predicate. For an $L$-formula $\varphi(x_1,\dots,x_n)$, we are going to define a new sentence $\varphi^P(\bar{x})$, which I will call the relativization of $\varphi(\bar{x})$ to $P$, inductively as follows. If $\varphi(\bar{x})$ is atomic, we define $\varphi^P(\bar{x})$ as $\varphi(\bar{x})$. We then let relativization to $P$ distribute over conjunction and negation, so that $(\varphi(\bar{x})\wedge\psi(\bar{x}))^P=\varphi^P(\bar{x})\wedge\psi^P(\bar{x})$ and $(\neg\varphi)^P(\bar{x})=\neg(\varphi^P(\bar{x}))$. Finally, if $\varphi(\bar{x})$ is of form $\exists y\psi(\bar{x},y)$, then we define $\varphi^P(\bar{x})=\exists y(P(y)\wedge\varphi^P(\bar{x},y))$. $\blacksquare$
Okay, so what's the point of this construction? Well you can prove the following by induction on formulas, which I will leave as an exercise; recall that a language is "relational" if it has no function or constant symbols.
Fact: Let $M$ be an $L\cup\{P\}$-structure where $L$ is a relational first-order language and $P$ is a fresh unary predicate. Since $L$ is relational, any subset of $M$ can be naturally made into an $L$-structure; let $N$ denote the $L$-structure obtained by making $P^M\subseteq M$ into an $L$-structure in this way. Then for any $L$-sentence $\varphi$, we have $M\models\varphi^P$ if and only if $N\models\varphi$. $\blacksquare$
This is the sense in which $\varphi^P$ is a "relativization" of $\varphi$ to $P$, and it is a very useful construction throughout model theory!

With this construction done we can proceed. Let $\varphi$ and $\psi$ be sentences with spectrum $X$ and $Y$ respectively; without loss of generality we may assume that $\varphi$ and $\psi$ are sentences in disjoint languages, say $K$ and $L$. Without loss of generality we may also assume these language are relational.
To obtain $X+Y$, we're going to construct a sentence $\theta$ such that (i) any model of $\theta$ is of form $M\sqcup N$, where $M$ is a model of $\varphi$ and $N$ is a model of $\psi$, and (ii) conversely any such disjoint union can be made into a model of $\theta$. This will give the desired result. To construct this sentence $\theta$, we will work in a new language $I=K\cup L\cup\{P,Q\}$ where $P$ and $Q$ are fresh unary predicates. Intuitively, our sentence $\theta$ will say that $P$ and $Q$ are disjoint, and that the realizations of $P$ give a model of $\varphi$ and that the realizations of $Q$ give a model of $\psi$. By our Fact above, we can do this by taking $\theta:=\forall x(P(x)\leftrightarrow\neg Q(x))\wedge\varphi^P\wedge\psi^Q$, and we are done.
To obtain $X\cdot Y$ as a spectrum, we need something slightly more elaborate. We're going to construct a sentence $\xi$ such that (i) every model of $\xi$ is of form $M\times N$ where $M$ is a model of $\varphi$ and $N$ is a model of $\psi$, and (ii) conversely every such product can be made into a model of $\xi$; this will again give the desired result. To do this, we're going to work in a language $J=K\cup L\cup\{P,Q\}\cup\{\pi,\sigma\}$, where $P$ and $Q$ are fresh unary predicates and $\pi$ and $\sigma$ are fresh unary function symbols. Morally $\pi$ and $\sigma$ are going to be "projection maps", and $P$ and $Q$ will be their respective images. Our sentence $\xi$ will thus say the following:

*

*$\forall x[P(x)\leftrightarrow \exists y(x=\pi(y))]$ and $\forall x[Q(x)\leftrightarrow \exists y(x=\sigma(y))]$: this says that $P=\operatorname{im}(\pi)$ and $Q=\operatorname{im}(\sigma)$.

*$\forall y\forall z[P(y)\wedge Q(z)\to\exists x(\pi(x)=y\wedge\sigma(x)=z)]$ and $\forall x,x'\left[x\neq x'\to\left(\pi(x)\neq\pi(x')\vee\sigma(x)\neq\sigma(x')\right)\right]$: this says that every pair $(y,z)\in P\times Q$ represents an element of the structure, and moreover that every element of the structure can be represented uniquely in this way.

*$\varphi^P\wedge\psi^Q$: by the Fact, this says that $P$ is a model of $\varphi$, hence that $|P|\in X$, and that $Q$ is a model of $\psi$, hence that $|Q|\in Y$.

Now we are done! The three hypothesis guarantee that every model of $\xi$ has size $xy$ for some $x\in X$ and $y\in Y$. Conversely if $M\models\varphi$ and $N\models\psi$, then we can turn $M\times N$ into a $J$-structure as follows: pick any $a_0\in M$ and $b_0\in N$ and take $\pi^{M\times N}((a,b))=(a,b_0)$ and $\sigma^{M\times N}((a,b))=(a_0,b)$ for each $(a,b)\in M\times N$. Then, for a relation symbol $R(x_1,\dots,x_n)\in K$, define $R^{M\times N}$ as $\{((a_1,b_1),\dots,(a_n,b_n)):(a_1,\dots,a_n)\in R^M\}$ and similarly for every relation symbol $S(x_1,\dots,x_n)\in L$. In this way $M\times N$ becomes a model of $\xi$, as needed.
A: Following the lead (and notation) of Atticus Stonestrom's answer above, I wanted to look at a language having only predicates for the case of multiplication.
Consider the language $J=K\cup L\cup\{P,Q,H\}$, where the predicates $P$ and $Q$ are as before, but $H$ is a new ternary predicate. Then $\xi$ will be the conjunction of the following clauses:

*

*$(\forall x\,(Px\Rightarrow\neg Qx)\wedge(Qx\Rightarrow\neg Px))\wedge\varphi^{P}\wedge\psi^{Q}$

*$\forall x\,\forall y\,((Px\wedge Qy)\Rightarrow\exists z\,\forall t\,(Hxyt\Leftrightarrow (z=t)))$

*$\forall z\,((\exists x\,\exists y\,\forall t\,((Pt\Leftrightarrow (x=t))\wedge(Qt\Leftrightarrow (y=t))))\wedge Hxyz)$
$\mathrm{Spectrum}(\xi)=X\cdot Y$, as desired.
