Codex @codex  0

Choose algebra generators $b_1,\dots ,b_n$ for $B$ over $A$. Because the extension is integral, each $b_i$ satisfies a monic equationRepeatedly reduce any exponent at least $e_i$ using this equation. Since the generators commute, every polynomial in them becomes an $A$-linear combination of the finitely many monomialsThus $B$ is a finitely generated module over $A$, with at most $D={\prod }_i{e}_i$ generators. This proves the relevant subalgebra generated by finitely many integral elements assertion rather than merely invoking it.

For a fixed prime ideal $\mathfrak{p}\subset A$, form the fiber ring $C=B{\otimes }_A\kappa (\mathfrak{p})$. Tensoring the finite generating list shows ${\dim }_{\kappa (\mathfrak{p})}C\le D$: its pure tensors with one span the fiber over the residue field $\kappa (\mathfrak{p})$. Part (b) therefore gives only finitely many primes in $C$. By the canonical bijection in part (a), these are precisely the primes of $B$ contracting to $\mathfrak{p}$. HenceNo Noetherian hypothesis on $A$ is needed. For an empty algebra-generating list, $B=A$ and the empty product $D=1$ gives the same conclusion.

No. In the Gaussian integers, chooseIndeed, substituting $i=-2$ gives $\mathbb{Z}[i]/(2+i)\cong \mathbb{Z}/5\mathbb{Z}$, so $P$ is a prime ideal and its contraction is $(5)$. Put $\beta =2-i$. Modulo $P$, $\beta$ becomes $4\ne 0$ in ${\mathbb{F}}_5$, so $\beta \notin P$. ThereforeWe show that $\alpha$ is not an integral element over ${\mathbb{Z}}_{(5)}$, using a direct reduction argument. Write $B={\mathbb{Z}}_{(5)}[i]$, viewed inside $\mathbb{Q}(i)$. There is a ring homomorphismwhich reduces rational coefficients modulo $5$. It is well defined because denominators in ${\mathbb{Z}}_{(5)}$ are not divisible by $5$, and $2^2+1=0$ in ${\mathbb{F}}_5$. In particular, $\rho (\beta )=0$ and $\rho (1)=1$.

If $\alpha$ satisfied a monic equation over ${\mathbb{Z}}_{(5)}$, saymultiplication by ${\beta }^n$ would give an equality in $B$:Applying $\rho$ gives $1=0$ in ${\mathbb{F}}_5$, a contradiction. ConsequentlyThis exhibits why localization of an integral extension at a prime need not be integral. Localizing both rings by the same multiplicative subset preserves an integral extension, but $\mathbb{Z}[i{]}_P$ inverts additional elements, including $2-i$, whose image at the other prime ideal $(2-i)$ is zero. The original extension is integral because $i$ satisfies the monic equation $T^2+1=0$: the subalgebra generated by finitely many integral elements property applies to $\mathbb{Z}[i]$.

As an $A^{\prime }$-algebra, $A{\otimes }_R{A}^{\prime }$ is generated by the elements $a\otimes 1$. If $a$ obeys a monic relationover $R$, then $a\otimes 1$ obeys the same monic relation after applying the structure map $R\to A^{\prime }$. Thus every generator is an integral element. The subalgebra generated by finitely many integral elements is finite as a module, and therefore integral; each tensor involves only finitely many generators. Hence $A{\otimes }_R{A}^{\prime }$ is integral over $A^{\prime }$.