I am studying for my qualifying exams and I just want to verify that I have done this problem correctly.
First, let $\mathbb{F}$ be a field with the property, $$ a^2 + b^2 = 0 \implies a = 0 \textrm{ and } b = 0.$$
Show that $x^2+1$ is irreducible in $\mathbb{F}\lbrack x \rbrack.$
So first, I basically justified that $\mathbb{F}\lbrack x \rbrack$ was a unique factorization domain because every field is a principal ideal domain, thus a unique factorization domain, which implies $\mathbb{F}\lbrack x \rbrack$ is as well. Now, assume to get a contradiction that $x^2 + 1$ is reducible. This implies there exists $a,b \in \mathbb{F}$ such that $$x^2 + 1 = (x+a)(x+b) = x^2+(a+b)x+ab.$$ The linear part of this equation implies $a = -b$, so the equation becomes, $$ x^2+(a+b)x + ab = x^2 -a^2.$$ This further implies that $a^2 = -1$. Now if we let $b = 1$ we see $a^2 + b^2 = -1+1 = 0$ which violates our field axiom. Thus no such $a$ exists and $x^2 + 1$ is irreducible.
I'm not quite sure that this is correct, and I would appreciate any feedback!