Why does Dummit and Foote seem to imply that $x^2+1$ and $2x^2+2$ do not have common zeros?

Question

I am reading Dummit and Foote's Abstract Algebra, and in Chapter 13, Prop 9, the book says distinct irreducibles never have zeros in common. I think one irreducible can be got from another irreducible multiplied by a unit, such as in $\mathbb R[x]$, $x^2+1$ and $2x^2+2$ are both irreducibles but have common zeros. Where did I go wrong?

Answer 1

When talking divisibility, irreducibility, and similar, associate elements (i.e. one is a unit times the other) are usually considered non-distinct by convention.

Answer 2

By distinct irreducibles, the author means upto a unit. Consider an equivalence class where $f \sim g$ if and only if $f=ug$ for some unit $u$. Under this equivalence class $f$ and $g$ are equal.

Answer 3

Two irreducible with a common root would both be the minimal polynomial of that shared root and the minimal polynomial is unique up to multiplication by a constant factor.

Let $f,g\in F[x]$ be your two irreducible polynomials and suppose that in a field extension $F\hookrightarrow E$ they have a common root $\theta$, then $$I=\{p\in F[x]:p(\theta)=0\}\subseteq F[x]$$ is an ideal and is principal (since $F[x]$, for $F$ a field, is always a PID), that is $$I=(h)$$ for some polynomial $h$. Since $f(\theta)=g(\theta)=0$, there are $\alpha,\beta\in F[x]$ such that $$f=\alpha h,\qquad g=\beta h$$using that $f,g$ are irreducible, we conclude that $\alpha,\beta$ are both constants ($h$ cannot be a constant, as it would imply $(h)=I=F[x]$) and $$g=\frac{\beta}{\alpha}f$$