On Unique factorization of polynomials

Question

I'm studying Lang's Linear Algebra and stumbled upon a lemma prior to the unique factorization of polynomials that says the following "Let p be irreducible in K[t]. Let f, g be non-zero polynomials in K[t]. Assume p divides fg, then p divides f or p divides g". I've come up with a counter example, let p(x)=x^2+1 (which is an irreducible polynomial over the Reals), f(x)=(x^2+1)^2 and g(x)=(x^2+1)^3, thus f(x)g(x)=(x^2+1)^5 in which p divides both f and g. I've also tried (dis)proving it in the following way "Let h,a,b,c,d be polynomials in K[t] such that ph=fg, pa+fb=1 and pc+gd=1, if we multiply the second equation by g and the third one by f to get pga+fgb=g and pfc+gfd=f but fg=gf=ph thus pga+fgb=pga+phb=p(ga+hb)=g and pfc+gfd=pfc+phd=p(fc+hd)=f meaning both f and g can by divided by p". What am I doing wrong??? Thanks

Answer 1

I repeat the answer for the sake of completeness: it's not an exclusive OR. $p$ can divide $f$ and $g.$

The interesting observation, however, is a different one. The statement $$ p(t)\,|\,f(t)\cdot g(t) \Longrightarrow p(t)\,|\,f(t)\, \vee \,p(t)\,|\,g(t) $$ for polynomials $p(t),f(t),g(t)\in K[t],$ where $K$ is a field, is exactly the definition for $p(t)$ being a prime element in the ring $K[t].$ The theorem above therefore proves:

Every irreducible polynomial in $K[t]$ is prime.

The converse is also true: every prime polynomial is irreducible.