About: http://webisa.webdatacommons.org/82257315

Property	Value
http://www.w3.org/1999/02/22-rdf-syntax-ns#type	http://www.w3.org/ns/prov#Entity
http://www.w3.org/ns/prov#value	So my point is that $\left( x^{m}y^{n}=x^{p}y^{q} \rightarrow (m,n)=(p,q)\right)$ follows if we know the ring is a UFD, but it also follows trivially from linearly independence. share|improve this answer answered Mar 6 '11 at 4:42
http://www.w3.org/ns/prov#wasQuotedFrom	mathoverflow.net