Is $\mathbb Z$ first-order definable in (the ring) $\mathbb{Z\times Z}$? Is $\mathbb Z$ first-order definable in $\mathbb{Z\times Z}$ (using sum and product but obviously not the concept of "component")? I believe no but how may I prove it? Is this standard?
 A: You are asking if the set $\Delta = \{(a, a) \mathrel{|} a \in \Bbb{Z}\}$
is definable using the first-order language of rings in the product ring $\Bbb{Z}\times\Bbb{Z}$.
I have not come across this before, but I believe the answer is no (as you suspected). To see this, choose variables $e$ and $f$ and
think of them as parameters such that $\{e, f\} = \{(1, 0), (0, 1)\}$ (i.e., we know
that $e$ is one of $(1, 0)$ and $(0, 1)$ and $f$ is the other one, but we don't
know which is which). Consider the the annihilators of $f$ and $e$, i.e., the
sets defined by the formulas:
$$
\begin{array}{rcl}
E(x) & {:=} & xf = 0 \\
F(x) & {:=} &xe = 0
\end{array}
$$
I will refer to these sets as the $e$-axis and the $f$-axis. The multiplication and addition
of $\Bbb{Z}\times\Bbb{Z}$ make the $e$-axis into a ring with $e$ as the multiplicative identity
and likewise for the $f$-axis and $f$.
Now assume $\delta(x)$ is a formula defining $\Delta$ and consider the formula:
$$
\begin{array}{rcl}
\rho(x, y) & {:=} & E(x) \land F(y) \land \exists z(\delta(z) \land z = x + y)
\end{array}
$$
Let $\phi$ be a formula parametrised by $e$ and $f$ and with no other free
variables that asserts that the $e$-axis and the $f$-axis are rings with
multiplicative identities $e$ and $f$ respectively  and that $\rho(x, y)$
defines a ring isomorphism between these rings. By easy algebra we can show
that  $\phi$ holds in $\Bbb{Z}\times\Bbb{Z}$ if $e$ and $f$ are interpreted so
that  $\{e, f\} = \{(1, 0), (0, 1)\}$. Hence the following sentence holds in
$\Bbb{Z}\times\Bbb{Z}$:
$$
\begin{array}{rcl}
\Phi & {:=} & \exists e\exists f (ef = 0 \land e + f = 1 \land \phi)
\end{array}
$$
since in $\Bbb{Z}\times\Bbb{Z}$, $ef = 0$ and $e + f = 1$ iff
$\{e, f\} = \{(1, 0), (0, 1)\}$.
Now take a non-standard model $A$ of the theory of $\Bbb{Z}$ and consider the
ring $\Bbb{Z}\times A$. By the Feferman-Vaught theorem, the first-order theory
of $\Bbb{Z}\times A$ is the same as the first-order theory of
$\Bbb{Z}\times\Bbb{Z}$. Hence, assuming there is a formula $\delta(x)$ defining
$\Delta$ in $\Bbb{Z}\times\Bbb{Z}$, the sentence $\Phi$ described above in terms of $\delta(x)$ holds
in $\Bbb{Z}\times A$.  But just as in $\Bbb{Z}\times\Bbb{Z}$, in $\Bbb{Z}\times
A$, $ef = 0$ and $e + f = 1$ imply that $\{e, f\} = \{(1, 0), (0, 1)\}$ and
hence that $E(x)$ and $F(x)$ define the rings $\Bbb{Z}\times \{0\}$ and $\{0\}
\times A$ (or vice versa). But for such $e$ and $f$, $\phi$ asserts that these rings
are isomorphic, which is false. Hence $\Phi$ is false in $\Bbb{Z}\times A$ and
we have a contradiction. So $\Delta$
cannot be definable.
