**Proof.**
Let $f \in A[T]$ be a monic polynomial and let $f \bmod I = g_0 h_0$ be a factorization over $A/I$ with $g_0$, $h_0$ monic such that $g_0$ and $h_0$ generate the unit ideal of $A/I[T]$. This means that

\[ A/I[T]/(f) = A/I[T]/(g_0) \times A/I[T]/(h_0) \]

Denote $e \in A/I[T]/(f)$ the element correspoing to the idempotent $(1, 0)$ in the ring on the right. Write $f = (T - a_1) \ldots (T - a_ d)$ with $a_ i \in A$. For each $i \in \{ 1, \ldots , d\} $ we obtain an $A$-algebra map $\varphi _ i : A[T]/(f) \to A$, $T \mapsto a_ i$ which induces a similar $A/I$-algebra map $\overline{\varphi }_ i : A/I[T]/(f) \to A/I$. Denote $e_ i = \overline{\varphi }_ i(e) \in A/I$. These are idempotents. By our assumption (2) we can lift $e_ i$ to an idempotent in $A$. This means we can write $A = \prod A_ j$ as a finite product of rings such that in $A_ j/IA_ j$ each $e_ i$ is either $0$ or $1$. Some details omitted. Observe that $A_ j$ is absolutely integrally closed as a factor ring of $A$. It suffices to lift the factorization of $f$ over $A_ j/IA_ j$ to $A_ j$. This reduces us to the situation discussed in the next paragraph.

Assume $e_ i = 1$ for $i = 1, \ldots , r$ and $e_ i = 0$ for $i = r + 1, \ldots , d$. From $(g_0, h_0) = A/I[T]$ we have that there are $k_0, l_0 \in A/I[T]$ such that $g_0 k_0 + h_0 l_0 = 1$. We see that $e = h_0 l_0$ and $e_ i = h_0(a_ i) l_0(a_ i)$. We conclude that $h_0(a_ i)$ is a unit for $i = 1, \dots ,r$. Since $f(a_ i) = 0$ we find $0 = h_0(a_ i)g_0(a_ i)$ and we conclude that $g_0(a_ i) = 0$ for $i = 1, \ldots , r$. Thus $(T - a_1)$ divides $g_0$ in $A/I[T]$, say $g_0 = (T - a_1) g_0'$. Set $f' = (T - a_2) \ldots (T - a_ d)$ and $h'_0 = h_0$. By induction on $d$ we can lift the factorization $f' \bmod I = g'_0 h'_0$ to a factorization of $f' = g' h'$ over over $A$ which gives the factorization $f = (T - a_1) g' h'$ lifting the factorization $f \bmod I = g_0 h_0$ as desired.
$\square$

## Comments (0)