Re: Irreducible polynomial

It has been 5 days now so you probably have figured it out by now, but still, here is my take. First, we find the units $\displaystyle R^{\times}$. Since $\displaystyle R$ is a subring of $\displaystyle k[x,y]$ any unit of $\displaystyle R$ is also a unit of $\displaystyle k[x,y]$. But we know (from a theorem) that the units of $\displaystyle k[x,y]$ are the non-zero elements $\displaystyle k^*$. Thus we see that $\displaystyle R^{\times} = k^*$.

Thus, to show that $\displaystyle xy$ is irreducible in $\displaystyle R$, we need to show that if $\displaystyle xy = pq$ for $\displaystyle p,q\in R$, then $\displaystyle p$ or $\displaystyle q$ is in $\displaystyle k^*$. So, let $\displaystyle xy = pq$ for some $\displaystyle p,q \in R$. We can view any polynomial of $\displaystyle R$ as a polynomial in $\displaystyle k[x,y]$ and thus we'll work over $\displaystyle k[x,y]$ instead. There, we have that

$\displaystyle p(x,xy,\dots) = p_1(x) + xy p_2(x,y)$

and

$\displaystyle q(x,xy,\dots) = q_1(x) +xy q_2(x,y)$

for some $\displaystyle p_1,p_2,q_1,q_2$. Multiplying these out gives us

$\displaystyle xy = pq = (p1+xyp_2)(q_1+xyq_2) = p_1q_1 + xy(p_2q_1+p_1q_2+xyp_2q_2)$

Since $\displaystyle p_1q_1 \in k[x]$, we see that it must be zero for the LHS and RHS to agree. Since $\displaystyle k[x,y]$ is an integral domain, $\displaystyle p_1 = 0$ or $\displaystyle q_1 = 0$. We choose $\displaystyle p_1 = 0$, that is

$\displaystyle xy = xy p_2(x,y) (q_1(x) + xy q_2(x,y))$

Since $\displaystyle k[x,y]$ is integral, we can cancel out the $\displaystyle xy$. We are left with

$\displaystyle p_2(x,y)(q_1(x)+xy q_2(x,y)) = 1$

and thus $\displaystyle p_2(x,y),q_1(x)+xy q_2(x,y) \in (k[x,y])^{\times} = k^*$. Thus we see that $\displaystyle p = c xy$ and $\displaystyle q = d$ for some $\displaystyle c,d \in k^*$. Therefore, $\displaystyle q$ is a unit, and we are done.