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