Let $R=\mathbb R[x, y]/(x^3-y^5)$. Why is $\bar x$ irreducible in $R$?
Here's my attempt. Suppose that $\bar x$ is reducible. Then $\bar x = \bar p \bar q$ for some nonunits $\bar p, \bar q$ in $R$. So in $\mathbb R[x, y]$, $pq = x - (x^3-y^5)r$ for some $r\in \mathbb R[x, y]$. I tried to find a contradiction here.
$R$ is isomorphic to the subring of the polynomial ring $\Bbb R[t]$ with $\overline x$ corresponding to $t^5$ and $\overline y$ to $t^3$. This contains polynomials of the form $$a_0+a_3t^3+a_5t^5+a_6t^6+a_8t^8+a_9t^9+\cdots+a_nt^n.$$ The only nontrivial factors of $t^5$ in $\Bbb R[t]$ which also lie in $R$ are the associates of $t^3$, but $t^3$ isn't a factor of $t^5$ in $R$, since $t^5/t^3\notin R$.