Let $K$ be a field and $R = K[x,y]/(xy^2)$.
Let me denote by $S^{*}$ the set of units of a generic set $S$. Moreover let $\pi : K[x,y] \to R $ be the quotient map.
I'm looking for the units of $R$, i.e. $R^{*}$.
- Clearly, $K[x,y]^{*} = K^{*} = K \setminus 0$ and using $\pi$ I know that $K^{*} \subset R^{*} $ (with this I mean that if $a \in K[x,y]^{*}$ then $[a]_{R} \in R^{*}$)
- $[1 \pm x^my]_{R} \in R^{*}$. Indeed $(1+x^my)(1-x^my) = 1 - x^{2m}y^2= 1 + xy^2(-x^{2m-1} ) $
- What else? I mean, maybe there could be oyher units, how can I detect them?
Moreover I know that $K[x,y]$ is UFD and so I have the notion of $gcd$: clearly, if $f(x,y) \in K[x,y]$ and $gcd(f(x,y),xy^2)=1$ then there exists $p(x,y)$ and $q(x,y)$ in $K[x,y]$ such that $p(x,y)f(x,y)+q(x,y)xy^2=1 $. Hence $f(x,y)p(x,y) = 1 + xy^2(-q(x,y))$. Now, how can I find an explicit definition of such polynomial $f(x,y)$?
Furthermore, if I call $X=[x]_{R}$ and $Y=[x]_{R}$, I know that if $f$ is a unit in $R$ then it is also in $R/(X)$ and $R/(Y^2)$. Can this fact help me?
Notice that there is an injective ring homomorphism $$K[X,Y]/(XY^2)\to K[X,Y]/(X)\times K[X,Y]/(Y^2)$$ given by $\hat f\mapsto(\bar f,\bar{\bar f})$.
If $\hat f$ is invertible in $R$, then
1) $\bar f$ is invertible in $K[X,Y]/(X)\simeq K[Y]$, that is, there exists $a\in K$, $a\ne 0$ such that $\bar f=\bar a$. We get $f=a+Xg(X,Y)$.
2) $\bar{\bar f}$ is invertible in $K[X,Y]/(Y^2)$. Now let us figure out which are the units of this factor ring. The elements of $K[X,Y]/(Y^2)$ write as $u(X)+v(X)y$, with $y^2=0$. It is easily seen that the invertible elements are $u+v(X)y$, with $u\in K$, $u\ne0$. We get $f=u+Yv(X)+Y^2h(X,Y)$.
Sending $X$ to $0$ in $a+Xg(X,Y)=u+Yv(X)+Y^2h(X,Y)$ we get $u=a$, $v(0)=0$, and $h(0,Y)=0$. This shows that $f=a+XYw(X)+XY^2k(X,Y)$, and therefore $\hat f=a+xyw(x)$ in $R$. (Here $x,y$ denote the residue classes of $X,Y$ in $R$.)
Now check that all elements of $R$ of this form are invertible.