Let $F$ be a field and let $R= F\left[x, \frac{1}{x+y}, \frac{1}{y} \right]$. Find $u, v \in R$ such that $R$ is integral over $F[u,v]$.
Let $y'=\frac{1}{y}$. I considered the polynomial $f$ $$f(t)=(t-x)(t-y')\left( t-\frac{x}{xy'+1}\right)\left( t-\frac{y'}{xy'+1}\right).$$ I couldn't think of $u, v$ such that the coefficients of $f$ (checked it using wolfram) are in $F[u,v]$. Can anyone help me on this?
Let $F$ be a field, and let $R=F[a,b,c]$ where $$ \begin{cases} a=x\\[4pt] b={\large{\frac{1}{x+y}}}\\[4pt] c={\large{\frac{1}{y}}}\\ \end{cases} \qquad\;\;\;\;\; $$ and let $u,v,w\in R$ be given by $$ \begin{cases} u=a+b+c\\[4pt] v=ab+bc+ca\\[4pt] w=abc\\ \end{cases} $$ Then $a,b,c$ satisfy the equation $p(t)=0$, where $$p(t)=t^3-ut^2+vt-w$$ hence $a,b,c$ are ntegral over $F[u,v,w]$.
But identically, for $a,b,c$ defined as above, we have $$ w^6 +(6v-2u^2)w^4 +(u^4+9v^2-6u^2v+27)w^2 +(4u^3-18uv)w +(4v^3-u^2v^2) = 0 $$ hence $w$ is integral over $F[u,v]$.
It follows that $R$ is integral over $F[u,v]$.