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Let $K$ be a commutative ring and $u, t$ be infinite sets of formal variables such that $u \subset t$. Prove that $K[u]$ is an elementary substructure of $K[t]$ with signature $\sigma = \{+, \cdot ,1, 0, =\}$.

I need an idea to start with. Tarski-Vaught test seems impractical here.

Thanks!

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  • $\begingroup$ What you want to use is the Tarski-Vaught test, not the Los-Vaught test. And the former works just fine in this case. $\endgroup$ – Stefan Mesken Nov 14 '18 at 19:22
  • $\begingroup$ @StefanMesken I meant Tarski-Vaught test of course. I still can't figure out how we can show that necessary condition of criterion holds for this two rings. $\endgroup$ – Gleb Chili Nov 14 '18 at 21:51
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Hint: We verify the Tarsk-Vaught criterion. Let $\vec{p} \in K[u]^n$ and let $\phi$ be a formula such that $$ K[t] \models \exists x \phi[x, \vec{p}]. $$ Fix some $x \in K[t]$ such that $$ K[t] \models \phi[x,\vec{p}]. $$ Consider the set of variables $\{v_0, \ldots, v_k\}$ that appear in $x$ and that are not in $u$. Let $\{u_0, \ldots,u_k\}$ be variables in $u$ that don't appear in $\phi[x, \vec{p}]$. Let $x^* \in K[u]$ be the result of replacing each occurance of $v_j$ in $x$ with $u_j$ for all $j \le k$. Show that $$ K[t] \models \phi[x^*, \vec{p}]. $$ One (and probably the easiest) way to see this is to verify that there is a unique automorphism $$ \pi \colon K[t] \to K[t] $$ such that $\pi(v_j) = u_j$, $\pi(u_j) = v_j$ for all $j \le k$ and $$\pi \restriction K \cup (t \setminus \{v_0, \ldots, v_k, u_0, \ldots, u_k \}) = \mathrm{id}.$$ Finally note that $\pi(\vec{p}) = \vec{p}$ and $\pi(x) = x^*$, so that $$ K[t] \models \phi[x, \vec{p}] \iff K[t] \models \phi[\underbrace{\pi(x)}_{= x^*}, \underbrace{\pi(\vec{p})}_{= \vec{p}}]. $$

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