Computing Picard groups by showing invertible modules are uniquely determined I am going to have another go at showing that  

the Picard group of $k[x,y]/(xy)$ is trivial.

(see my previous stackexchange posts)  
Here I define the Picard group of a ring $R$ as the isomorphism classes of finite locally free modules of rank 1 (that is, invertible) over $R$.
Here are some useful things I showed.
Let $M$ with my choice of ring $k[x,y]/(xy)$ be an invertible module over $k[x,y]/(xy)$. I know that $M/xM$ is on the nose free of rank one over $k[x,y]/(x,xy)$ so $M/xM \cong k[y]$ as $k[x,y]$ modules.  Similarly $M/yM \cong k[x]$ as $k[x,y]$ modules.

Question:  Is there a way I can use Nakayama or something to show that
  I have a surjection  $k[x,y]/(xy) \to M$ from the fact that I have a
  surjection $(k[x,y]/(xy))/(x) \to M/(x)$ and another surjection
  $(k[x,y]/(xy))/(y) \to M/(y)$?

If I could get this information, then since $M$ must be $k[x,y]/(xy)$ torsion free by hypothesis, and $M$ is a cyclic $k[x,y]/(xy)$ module, I would know that $M \cong k[x,y]/(x,y)$ and thus that the Picard group is trivial.

From what you said, there exists $u,v\in M$ such that $u$ generates $M$ modulo $x$, $v$ generates $M$ modulo $y$. Then, they both generate the one dimensional $k$ vector space $M/(x,y)M$ and thus $u=av$ for some $0\neq a\in k$ modulo $(x,y)$. Clearly, we may replace $v$ by $av$ without changing anything and thus assume $u=v$ modulo $(x,y)$. So, write $u-v=xm+yn$ for $m,n\in M$. Then, easy to check that $u-xm=v+yn$ generates $M$.
