Exact gcd squarefree decomposition preserves roots and multiplicities

By [F8], the monic gcd is constructed by Euclidean division over $\mathbb{Q}$, and the divisions terminate: each nonzero remainder has smaller nonnegative integer degree than the divisor by [F2]. The same argument defines $p_k$; if $\mathrm{deg}{p}_{k-1}>0$, then $p_{k-1}^{\mathrm{\prime }}\mathrm{\ne }0$ and has smaller degree by [F3] and characteristic zero, so $\mathrm{deg}{p}_k<\mathrm{deg}{p}_{k-1}$. Thus $p_d=1$, and each $s_k=p_{k-1}\mathrm{/}{p}_k$ is a polynomial.

By [F4], repeated application of [F5] factors $f$ over $\mathbb{C}$: $f(x)={\prod }_{\alpha }(x-\alpha {)}^{m_{\alpha }}$, where the finite list contains each distinct root once and $m_{\alpha }\ge 1$. The induction is finite because each division by $x-\alpha$ lowers degree by one by [F6].

Fix a current polynomial $p$ in the gcd chain and a root $\alpha$ of multiplicity $m>0$, so $p=(x-\alpha {)}^{m}u$ with $u(\alpha )\mathrm{\ne }0$. The product rule [F3] gives $p^{\mathrm{\prime }}=(x-\alpha {)}^{m-1}(mu+(x-\alpha )u^{\mathrm{\prime }})$, whose parenthesized factor is nonzero at $\alpha$ because the field has characteristic zero. Hence the gcd of $p$ and $p^{\mathrm{\prime }}$ has multiplicity exactly $m-1$ at $\alpha$; a root absent from $p$ is absent from their gcd. By [F9] for the field extension $\mathbb{Q}\subseteq \mathbb{C}$, the monic gcd computed over $\mathbb{Q}$ is exactly the monic gcd over $\mathbb{C}$. Thus this root-multiplicity computation applies to each $p_k$.

Induction on $k$ now shows that the multiplicity of $\alpha$ in $p_k$ is $\max (m_{\alpha }-k,0)$. Therefore $\alpha$ occurs once in $s_k=p_{k-1}\mathrm{/}{p}_k$ exactly when $k\le m_{\alpha }$, and it occurs in $s_k\mathrm{/}{s}_{k+1}$ exactly when $k=m_{\alpha }$. This root factorization proves $s_{k+1}$ divides $s_k$ over $\mathbb{C}[x]$; division over $\mathbb{Q}[x]$ has the same zero remainder, so $q_k=s_k\mathrm{/}{s}_{k+1}$ lies in $\mathbb{Q}[x]$.

Each complex root of $f$ occurs in exactly one $q_k$, with exponent one there and exponent $k$ in $q_k^k$. The factorization in step 1.2 therefore gives $f={\prod }_{k=1}^d{q}_k^k$; it also proves that each $q_k$ is squarefree and that distinct $q_k$ have no common root, hence are pairwise coprime. The degree-zero case has $f=1$ and an empty root list.