TITLE: No isomorphism between $k[x,y]/(x^{2}-y^{5})$ and ring of polynomials in one indeterminate
QUESTION [7 upvotes]: Let $k$ be an algebraically closed field. Why there is no isomorphism (as finitely generated $k$-algebras) between the ring $k[x,y]/(x^{2}-y^{5})$ and $k[t]$?

REPLY [3 votes]: Below is a very simple proof accessible at high-school level.
Theorem $\ $ If $\rm\: a^2 = b^5\:$ in  UFD $\rm\:R[x]\:$ then $\rm\:R[a,b]\:$ is a proper subring of $\rm\:R[x]\:$
Proof $\ $ Clear if $\rm\:a,b\in R.\:$ Else $\rm\:a\:$ or $\rm\:b\not\in R,\:$ so one is a nonunit, so has a prime factor $\rm\:p\not\in R.\:$ By $\rm\:a^2 = b^5,\:$ prime $\rm\:p\ |\ a\!\iff\! p\ |\ b,\:$ so $\rm\:p\ |\ a,b.\:$ Comparing powers of $\rm\:p\:$ in $\rm\:a^2 = b^5\:$ shows $\rm\:p^2\:|\ a,b.\:$ Thus $\rm\:p\not\in R[a,b]\:$ else $\rm\:p = r + a f + b g,\:$  $\rm r\in R\:$ so $\rm\:p^2\:|\:a,b\:$ $\Rightarrow$ $\rm\:p\ |\ r\:$ $\Rightarrow$ $\rm\:r=0\:$ $\Rightarrow$ $\rm\:p^2\ |\ p.$