# $B$ is finitely generated as an $A$-module iff $B$ is finitely generated as an $A$-algebra and integral.

I am trying to prove that if $B/A$ is a commutative ring extension then $B$ is finitely generated as an $A$-module iff $B$ is finitely generated as an $A$-algebra and $B$ is integral over $A.$ The converse is I think easy.

Suppose $B$ is finitely generated as an $A$-algebra and is integral over $A.$ Then, $B = A[f_1, ..., f_n] = A[f_1, ..., f_{n - 1}][f_n]$ is finitely generated as an $A[f_1, ..., f_{n - 1}]$-module. Repeat for $f_{n - 1}, ..., f_1.$ We can see that $B$ is then finitely generated as an $A$-module. However, I am having trouble proving the other way. Can I have a hint please?

Edit for a question to Bernard:

Are you saying that because $$\begin{vmatrix} x - a_{11} & -a_{12} & \dots & -a_{1n} \\ -a_{21} & x - a_{22} & \dots & -a_{2n} \\ \vdots & & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \dots & x - a_{nn} \end{vmatrix} \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} = 0$$ and since the column matrix is not $0$ that $$\begin{vmatrix} x - a_{11} & -a_{12} & \dots & -a_{1n} \\ -a_{21} & x - a_{22} & \dots & -a_{2n} \\ \vdots & & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \dots & x - a_{nn} \end{vmatrix} = 0?$$

• More precisely, since $\det A \,b_i=0$ for all $i$ and the $b_i$s are a set of generators of $B$, there results that $\det A\,x=0$ for all $x\in B$, in particular for $x=1$. Commented Oct 29, 2017 at 11:30
• Ok so are you specifically taking $x = 1$ to rid the case of zero divisors? Commented Oct 29, 2017 at 20:33
• Yes. Actually, the argument is more general: it works with any faithful $A$-module, and a ring extension of $A$ is such a faithful $A$-module. That's the way the three characterisations of integral elements are shown to be equivalent in Bourbaki, Commutative Algebra, ch. 5 Integral elements. Commented Oct 29, 2017 at 21:21

Here it is: for any $x \in B$, consider the $A$-endomorphism of $B$ $\;m_x\colon b\longmapsto xb$, and let $b_1, \dots, b_n$ a system of generators of $B$ as an $A$-module. Each $xb_i$ is a linear combination of the $b_i$s ($1\le i\le n$). So we have a linear system: \begin{cases} xb_1=a_{11}b_1+a_{12}b_2+\dots +a_{1n}b_n\\ xb_2=a_{21}b_1+a_{22}b_2+\dots +a_{21}b_n\\ \qquad\vdots\\ xb_n=a_{n1}b_1+a_{n2}b_2+\dots +a_{nn}b_n \end{cases} where the $a_{i,j} \in A$ for all $i , j$, which we can rewrite matricially as $$\begin{bmatrix} x-a_{11}&-a_{12}&\dots&-a_{1n}\\ -a_{21}&x-a_{22}&\dots&-a_{2n}\\ \vdots&&\ddots&\vdots\\ -a_{n1}&-a_{n2}&\dots& x-a_{nn} \end{bmatrix} \begin{bmatrix} b_1\\b_2\\\vdots\\b_n \end{bmatrix}=0$$ Multiplying on the left by the adjugate of this matrix we deduce that its determinant kills all the generators of $b$, hence it kills $1$, so that $$\begin{vmatrix} x-a_{11}&-a_{12}&\dots&-a_{1n}\\ -a_{21}&x-a_{22}&\dots&-a_{2n}\\ \vdots&&\ddots&\vdots\\ -a_{n1}&-a_{n2}&\dots& x-a_{nn} \end{vmatrix}=0$$ and this determinant is a monic polynomial in $x$.