What is the kernel of $\varphi : R[u,v]\to R[x,1/x]$ defined by $\varphi(p(u,v))=p(x,1/x)$? Let $R[x,\frac{1}{x}]=\{\sum_{k=-m}^m a_kx^k\mid a_k\in R\}$. Let $\varphi:R[u,v]\to R[x,1/x]$ defined by $$\varphi(p(u,v))=p(x,1/x).$$
What is $\ker \varphi$ ? To me, $\varphi(p(u,v))=0\implies p(x,1/x)=0\implies p\in (u-x,v-\frac{1}{x})$
since $u=x$ and $v=\frac{1}{x}$ are solution. but this ideal look strange... (in fact, it's $v-\frac{1}{x}$ that looks strange).
 A: Clearly $I:=(1-uv) \subset \ker(\phi)$.

Now, look at the quotient ring $R[u,v] / I$. We want to show that any polynomial $P \in \ker(\phi)$ is $0$ modulo $I$, that is : $P \equiv 0 \pmod I$.
The trick is to notice that any polynomial $P(u,v)$ can be written as $P_1(u) + P_2(v)$ when working modulo $I$. Here is an example, first, say with $P(u,v) = 2+u+uv^2+3v^3-4u^2v^2+7u^5v$. Recall that $uv \equiv 1$ in $R[u,v]/I$ ! So we get:
$$P(u,v) \equiv
2+u+v+3v^3-4+7u^4 \pmod I$$
Let $P_1(u) = 7u^4+u-2,P_2(v)=4v^3+v$, then we have $P(u,v) \equiv P_1(u) + P_2(v) \pmod I$.
I let you the funny task to write this down in the general case
$$P(u,v) = \sum_{0 \leq i \leq n\\ 0 \leq j \leq m} a_{i,j} u^i v^j$$

But well, it works : let us write $P \in \ker(\phi)$ as $P_1(u) + P_2(v)$, modulo $I$. If $P_2$ has degree $m$, then
$X^m(P_1(X)+P_2(1/X))$ is the zero polynomial in $R[X]$, so you easily get that the coefficients of $P_1$ and of $P_2$ are zeros. Therefore $P \equiv P_1+P_2 = 0 \pmod I$, which precisely means $P \in I$ as desired.

From this, it is easy to show the following ring isomorphism:
$$R[u,v]/(uv-1) \cong R\left[x,\dfrac{1}{x}\right]$$
See also here : if $S$ is a commutative ring (not necessarily a domain) and $a \in S$, then $S_a \cong S[X]/(aX-1)$.
An isomorphism is given by $g : [P]_{(aX-1)} \mapsto P(1/a) \in S_a$.
The kernel of $f : S[X] \to S[X]/(aX-1) \cong S_a$ is $(aX-1)$.
In your case, $S=R[u],a=u$. The map $f$ is $$f : R[u][v] \to R[u][v]/(uv-1) \cong S_u = R[u,u^{-1}]$$
given by $f(P(v)) = P(1/u)$, i.e.
$$f : P(u,v) \in R[u,v] \;\longmapsto\; P(u,1/u)$$

We prove that
$$g : [P]_{(aX-1)} \mapsto P(1/a) \in S_a$$ provides an isomorphism $S[X]/(aX-1) \to S_a$.


*

*well-defined : OK

*$S$-algebras homomorphism : OK

*surjective : take $\alpha X^n$...

*injective : let $P(X) = \sum_{i=0}^m a_iX^i \in S[X]$ such that $P(1/a)=0$. Then $$a^mP(1/a)=a_m+a_{m-1}a+\dots+a_0a^m=0,$$
and
$$0 = a^mP(1/a) \cdot X^m \equiv a_m+a_{m-1}aX^m + \dots + a_0a^mX^m = P(X) \pmod{(aX-1)},$$
which precisely means that $P \in (aX-1)$.
$
\tag*{$\blacksquare$}
$

