Minimal polynomials of integral elements over an integrally closed domain have coefficients in the domain

By [L3], the minimal polynomial $m_u(T)\in K[T]$ of $u$ is monic and irreducible. Since $u$ is integral over $A$, it satisfies some monic polynomial in $A[T]$, so [L3] implies that $m_u$ is nonzero. By [L4], choose a splitting field $E\mathrm{/}K$ for $m_u$ and write $m_u(T)={\prod }_{i=1}^n(T-{\alpha }_i)$ with ${\alpha }_1=u$.

Let $g(T)\in A[T]$ be a monic polynomial with $g(u)=0$. For each root ${\alpha }_i$ of $m_u$, the universal property [L5] gives a $K$-homomorphism $K[u]\to E$ sending $u$ to ${\alpha }_i$. Applying that homomorphism to the identity $g(u)=0$ shows $g({\alpha }_i)=0$. Thus every ${\alpha }_i$ is integral over $A$.

The coefficients of $m_u$ are, up to sign, the elementary symmetric polynomials in the integral elements ${\alpha }_1,\dots ,{\alpha }_n$. Since $A$ is a domain and therefore nonzero, [L2] shows that these symmetric polynomials are integral over $A$. But the coefficients also lie in $K$, so [L1] forces them to lie in $A$. This proves the first statement.

For the factor statement, choose a splitting field of $f$ over $K$ and write $f(T)=(T-a){\prod }_{i=2}^n(T-{\beta }_i)$. Each ${\beta }_i$ is integral over $A$ because it is a root of the monic polynomial $f\in A[T]$. Therefore the coefficients of $h(T)={\prod }_{i=2}^n(T-{\beta }_i)$ are integral over $A$ by the same symmetric-polynomial argument as in step 3.1, and they lie in $K$ because $h\in K[T]$. Hence [L1] forces all coefficients of $h$ to lie in $A$.