So I was studying some stuff a bout ring theory and ended up with the question in the title. This is what I have so far:

Suppose $\bar{x}$ is reducible, i.e. $\bar{x}=\bar{p}\bar{q}$, for some non trivial $\bar{p},\bar{q}\in A$, now \begin{eqnarray} \bar{x}=\bar{p}\bar{q} &\Longrightarrow& x=pq+t(x^2-y^3)\quad\text{for some }p,q,t\in B\\ &\Longrightarrow&1=deg(x)=deg(pq+t(x^2-y^3))\\ &\Longrightarrow&deg(pq)=deg(t(x^2-y^3))\\ &\Longrightarrow&deg(p)+deg(q)=3+deg(t) \end{eqnarray} but doesn't seem to get any better. I have also tried to apply this property: If $A$ is a domain and $(a)\triangleleft A$ is prime, then $a$ is irreducible. But the ideal $(\bar{x})$ doesn't seem to be prime so $A/(\bar{x})$ is not domain (unless there is something I'm missing). That's what I have, any hints or approaches will be very well received. Thanks a lot.

  • $\begingroup$ You are right that $(\bar x)$ isn't prime, since $\bar y\notin(\bar x)$, but $\bar y^3=\bar x^2\in(\bar x)$. Alternatively, $$A/(\bar x)\cong B/(I,x)\\\cong (B/(x))/(I/(x))\cong \Bbb R[y]/(y^3)$$ which is not a domain. $\endgroup$
    – Arthur
    May 24 '18 at 3:57
  • $\begingroup$ As a related exercise, you might want to try proving that $(X^2-Y^3)$ is prime in $\Bbb{R}[X,Y]$ so that you know $B$ is a integral domain. $\endgroup$
    – C Monsour
    May 25 '18 at 10:50

First define a special degree $d(f)=3deg_x(f)+2deg_y(f)$. Using this you can see that the only monomial in $pq$ with $d<6$ is $x$.

Consider that in some ring (like $\Bbb{R}[x^{1/6},y^{1/6}]$ you can specialize $y$ to be $x^{2/3}$. In that specialization you have $x=p(x,x^{2/3})q(x,x^{2/3})$. Thus, $deg_x(p(x,x^{2/3}))+deg_x(q(x,x^{2/3}))=1$. Using the previous paragraph, you can show that in this case the degrees must be integers, so, after moving a constant multiple from $p$ to $q$ we have $p(x,x^{2/3})=1$ and $q(x,x^{2/3})=x$ (or conversely). In order for this to be the case, you can show (by induction, that this is necessary to make higher degree terms cancel) we must have $p=1+r(x^2-y^3)$ and $q=x+s(x^2-y^3)$. But then $\bar{q}=\bar{x}$, so $x$ is irreducible.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.