Let $k$ be a field, and consider the $k$-algebra $A:=k[X,Y]/(XY)$. Give an explicit solution of this algebra using Noether normalization.
So the task is to find a free $k$-algebra $k[t]$ such that $k[t]\subseteq A$ and that $A$ is a finite $k[t]$-module. My idea is to write $A$ as $k[x,y]$ (where $xy=0$, i.e. $x,y$ are the images of $X,Y$ under the projection $k[X, Y]\to A$), and set $t:=x+y$. Clearly $x+y$ cannot be the solution of any $f(Z):=a_rZ^r+\dots +a_1Z+a_0$, with $a_r,\dots,a_0\in k$, for we will always have a term $a_r(x^r+y^r)\neq 0$ and $x^r, y^r$ aren't anywhere else in $f(x+y)$. Now comes the part that confuses me more, I must prove that $A$ is a finite $k[x+y]$-module. Indeed, I'd say that it is generated by $1,x,y$. In fact any power $x^n$, with $n\ge 2$, can be obtained as $(x+y)^{n-1}x$, and similarly for the powers of $y$; also $x,y$ and the elements of $k$ are trivial to obtain. Is the exercise done (correctly)?
Plus, do you confirm that $A$ is not a free module over $k[x+y]$? I'd say so because, for example, $$(x+y)^2\cdot 1-(x+y)\cdot x-(x+y)\cdot y=0$$ but the coefficients are non-zero.