Is $k[x,y]/(xy-1)$ an integral extension of $k[x]$? Consider the following result:

Let $A$ and $B$ be rings. Suppose that $B$ is an integral extension of $A$ and $J$ is an ideal of $B$. Then $B/J$ is an integral extension of $A/(J\cap A)$.

I wanted to express $k[x,y]/(xy-1)$ (where $k$ is a field) as an integral extension of some polynomial ring over $k$. The above result was the first thing that came to my mind.
We let $B=k[x,y]$, $J=(xy-1)$ and $A=k[x]$. Then $B/J$ is integral over $A/(J\cap A)$. Since $J\cap A$ is $(0)$ and $A/(0) \cong A$ we are done.
But is $k[x,y]$ an integral extension of $k[x]$? For every $f(x,y) \in k[x,y]$, we can write $f(x,y)=f_n(x)y^n+f_{n-1}(x)y^{n-1}+\cdots+f_1(x)y+f_0(x)$ where $f_i(x) \in k[x].$ We can assume $f_n(x) \neq 0$. Is $\frac{1}{f_{n}(x)} $ and inverse of $f_n(x)$?
If it is, then we can take the monic polynomial $p(y)=\frac{1}{f_{n}(x)}y-\frac{1}{f_n(x)}f(x,y) \in k[x][y]$ which has $f(x,y)$ as a root. So $k[x,y]$ is integral over $k[x]$.
Is there something wrong?
 A: $k[x,y] $ is certainly not integral over $k[x]$. 
Now back to your question . Observe that $$k[x,y](xy-1) \cong k[t,\frac{1}{t}]$$ under the isomorphism 
$$ x\mapsto t$$
$$y \mapsto \frac{1}{t}$$
So u have the ring extension $k[t]\subset k[t,\frac{1}{t}]$ with both having the same field of fractions.
Now observe that $k[t]$ being a UFD is integrally closed and hence any element of $k(t)$ integral over $k[t]$ must land in $k[t]$. In particular  $k[t]\subset k[t,\frac{1}{t}]$  is not integral. 
Since you want $k[x,y]/(xy-1)$ integral over "some" polynomial ring consider the extension $$k[x+y] \subset  k[x,y]/(xy-1)$$
Now this is an integral extension. 
A: Further to Soumik's answer, I'll explain why $k[x,y]$ is not integral over $k[x]$.
To see this, observe that $y$ is not an integral element over $k[x]$. For if $y$ was an integral element over $k[x]$, then there would have to exist an $n \geq 1$ and $f_0, f_1, \dots, f_{n-1} \in k[x]$ such that
$$ y^n + f_{n-1}(x) y^{n-1} + \dots + f_1(x) y + f_0(x) = 0,$$
which is absurd, since the zero element in $k[x,y]$ cannot possibly have a non-zero $y^n$ term.
Intuitively, there's no way that $y$ can be integral over $k[x]$ because $y$ is an indeterminate variable. $y$ doesn't satisfy any sort of "algebraic relationship".
[Contrast this with the element $\sqrt 2 \in \mathbb Z[\sqrt 2]$, which is integral over $\mathbb Z$, because it satisfies the relationship $(\sqrt 2)^2 - 2 = 0$, i.e. it is a root of the monic polynomial $T^2 - 2$.]
As for your argument about $k[x,y]$ being integral over $k[x]$, a few comments:


*

*$1 / f_n(x)$ doesn't exist in $k[x]$ (unless $f_n(x)$ is constant). It exists in $k(x)$, but not in $k[x]$.

*If you want to show that $f(x,y)$ is integral over $k[x]$, then you need to find a monic polynomial $p(T) \in k[x][T]$ such that $p(f(x,y)) = 0$ in $k[x,y]$. Notice that we're introducing a new indeterminate variable here, namely $T$. We can't reuse $y$ as the variable for this polynomial.

*A monic polynomial should have a leading term with a coefficient of $1$. Your $p$ doesn't look like a monic polynomial.

