# Irreducible polynomial

• Oct 9th 2012, 10:31 PM
jackie
Irreducible polynomial
Why is the polynomial $XY$ is irreducible in $R=k[X,XY, XY^2,XY^3,...]$? Assume $XY=pq$ for some $p,q \in R$. I want to show that either $p$ or $q$ are units. Would anyone help me on this?
• Oct 14th 2012, 02:21 AM
Vlasev
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 $R^{\times}$. Since $R$ is a subring of $k[x,y]$ any unit of $R$ is also a unit of $k[x,y]$. But we know (from a theorem) that the units of $k[x,y]$ are the non-zero elements $k^*$. Thus we see that $R^{\times} = k^*$.

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

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

and

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

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

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

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

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

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

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

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