Proposition

Let $f: B \rightarrow B’$ be a homomorphism of $A$-algebras, and let $C$ be an $A$-algebra. If $f$ is integral, prove that $f \otimes 1: B \otimes_A C \rightarrow B’ \otimes_A C$ is integral.

Solution

Each element in $B’ \otimes_A C$ is a finite sum $\sum_{i=1}^{n} b_i’ \otimes c_i$ where each $b_i’ \in B’$ and $c_i \in C$. By Corollary 5.3[Atiyah], it suffices to show that each $b’ \otimes c$ is integral over $\Im(f \otimes 1)$.

Let $b’ \otimes c$ be given where $b’ \in B’$ and $c \in C$. Since $b’ \in B’$, $b’$ is integral over $f(B)$. Then there must exist $b_0, \cdots, b_n \in B$ such that $\sum_{i=0}^{n} f(b_i)(b’)^i = 0$ and $b_n = 1$.

\[\begin{align*} \sum_{i=0}^{n} f(b_i)(b')^i = 0 &\implies (\sum_{i=0}^{n} f(b_i)(b')^i) \otimes c^n = 0 \\ &\implies \sum_{i=0}^{n} [f(b_i)(b')^i \otimes c^n] = 0 \\ &\implies \sum_{i=0}^{n} [((b')^i \otimes c^i)(f(b_i) \otimes c^{n - i})] = 0 \\ &\implies \sum_{i=0}^{n} [f(b_i)(b' \otimes c)^i(f(b_i) \otimes c^{n - i})] = 0 \\ &\implies \sum_{i=0}^{n} [(f(b_i) \otimes c^{n - i})(b' \otimes c)^i] = 0. \end{align*}\]

Each $f(b_i) \otimes c^{n - i} \in \Im(f \otimes 1)$ and $f(b_n) = 1$ and $c^{n - n} = 1$, so $f(b_n) \otimes c^{n - n} = 1 \otimes 1$. Therefore, $b’ \otimes c$ is integral over $\Im(f \otimes 1)$.