8
$\begingroup$

I apologize if this question is elementary: Let $A$, $B$, and $A^{\prime}$ be groups such that $A^{\prime}$ is an elementary extension of $A$. Is it true that $A^{\prime}\times B$ is an elementary extension of $A\times B$? Clearly this is true for ultrapowers.

$\endgroup$
4
  • $\begingroup$ It's better to ask in math.stackexchange.com $\endgroup$
    – Marryamry
    Feb 11, 2017 at 9:17
  • 4
    $\begingroup$ I think the question is fine here. It is on-topic for MO. $\endgroup$ Feb 11, 2017 at 14:11
  • 3
    $\begingroup$ The result also holds for reduced products and more general constructions by the Feferman-Vaught theorem (the Feferman-Vaught theorem is proven in the model theory book by Chang and Kiesler). $\endgroup$ Feb 11, 2017 at 22:30
  • 1
    $\begingroup$ @JosephVanName, perhaps it is also worth to mention Mostowski's general result On direct products of theories (JSL 1952). $\endgroup$ Feb 12, 2017 at 0:06

1 Answer 1

8
$\begingroup$

Yes. Let $T$ be the theory of two disjoint groups, in the language $(\cdot_1, \cdot_2, U_1, U_2)$. Note that if $(G_1, G_2) \models T$ then the group operation on $G_1 \times G_2$ is definable without parameters. Thus we can recover the theory of $G_1 \times G_2$ from the theory of $(G_1, G_2)$, which is clearly determined by $(\mbox{Th}(G_1), \mbox{Th}(G_2))$. Thus if $G_1, G_2, G_1', G_2'$ are any groups with each $G_i \equiv G_i'$, then $G_1 \times G_2 \equiv G_1' \times G_2'$.

EDIT: this just shows elementary equivalence. In order to get elementary extensions: let $G_1, G_2$ be given groups, and consider the language $(\cdot_1, \cdot_2, U_1, U_2,c_g: g\in G_1, d_h: h \in G_2)$ where we add constant symbols for elements of $G_1$ and $G_2$. Let $T_*$ assert that $T$ holds, and the elementary diagram of $G_1$ holds in $U_1$, and the elementary diagram of $G_2$ holds in $U_2$. Note that $T_*$ is actually complete. Now let $(G_1', G_2', g, h)_{g \in G_1, h \in G_2} \models T_*$. Then the group operation on $G_1' \times G_2'$ is definable without parameters, and further every element of $G_1 \times G_2$ is definable without parameters. Thus $(G_1 \times G_2, (g, h))_{g \in G_1, h \in G_2} \equiv (G_1' \times G_2', (g, h))_{g \in G_1, h \in G_2}$, since each sentence must be decided by $T_*$; this is the same as saying $G_1 \times G_2 \preceq G_1' \times G_2'$.

$\endgroup$
5
  • $\begingroup$ Thank you very much Douglas Ulrich. I am not a Model Theorist, so I need a bit more explanation. What do you mean by $U_1$ and $U_2$? Actually, I can't understand the above language which has two binary functional and two (maybe) unary symbols. $\endgroup$
    – Sh.M1972
    Feb 12, 2017 at 7:34
  • 4
    $\begingroup$ You probably also want some constants in the language to add the complete diagrams of $G_1$ and $G_2$, since the OP asks for elementary embeddings not just elementary equivalence. $\endgroup$ Feb 12, 2017 at 11:33
  • 2
    $\begingroup$ @M. Shahryari $U_1$ and $U_2$ are unary symbols, and $T$ will say that $\cdot_1$ is a group operation on $U_1$ (and trivial on $U_2$) and $\cdot_2$ is a group operation on $U_1$. So a model of $T$ is the disjoint union of two groups. $\endgroup$ Feb 12, 2017 at 15:55
  • $\begingroup$ @Ramiro de la Vega fixed. $\endgroup$ Feb 12, 2017 at 16:37
  • 1
    $\begingroup$ Also, in the previous comment I meant to say $\cdot_2$ is a group operation on $U_2$. $\endgroup$ Feb 12, 2017 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.