Justifying the representation of a monic polynomial over a UFD

Question

Let $D$ be a UFD and $f(x) \in D[x]$ be monic. The book I'm reading from claims that $$f(x) = p_1(x)^{e_1} \cdots p_n(x)^{e_n}$$

where $p_i(x)$ are distinct, irreducible, and monic, and $e_i >0$.

I don't see why we can always choose monic ones. Any clue on this?

Answer 1

For $f \in D[x]$ let $LC(f) \in D$ denote the leading coefficient of $f$.

Assume $f \in D[x]$ is monic, and can be written as a product of distinct irreducibles $$ f = p_1(x)^{e_1} \cdot \ldots \cdot p_n(x)^{e_n}. $$ Then $$ 1 = LC(f) = LC(p_1)^{e_1} \cdot \ldots \cdot LC(p_n)^{e_n}. $$ In particular $LC(p_k)$ is a unit for every $k$. Define $q_k = p_k \cdot LC(p_k)^{-1}$. Polynomials $q_k$ are clearly monic and irreducible, and it's easy to check that $$ f = q_1(x)^{e_1} \cdot \ldots \cdot q_n(x)^{e_n}. $$

Answer 2

Hint $ $ If a product of polynomials $f_i$ is monic then each factor $f_i$ has unit leading coefficient $u_i\,$ hence scaling each $f_i$ by $u_i^{-1}$ makes each factor monic, and doesn't change their product since $\, \prod u_i^{-1} = (\prod u_i)^{-1} = 1^{-1} = 1$.