Elementary substructure in the ring of polynomials

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!

• What you want to use is the Tarski-Vaught test, not the Los-Vaught test. And the former works just fine in this case. – Stefan Mesken Nov 14 '18 at 19:22
• @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. – Gleb Chili Nov 14 '18 at 21:51

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 $$K[t] \models \phi[x, \vec{p}] \iff K[t] \models \phi[\underbrace{\pi(x)}_{= x^*}, \underbrace{\pi(\vec{p})}_{= \vec{p}}].$$