## 109.25 Fitting ideals

Exercise 109.25.1. Let $R$ be a ring and let $M$ be a finite $R$-module. Choose a presentation

$\bigoplus \nolimits _{j \in J} R \longrightarrow R^{\oplus n} \longrightarrow M \longrightarrow 0.$

of $M$. Note that the map $R^{\oplus n} \to M$ is given by a sequence of elements $x_1, \ldots , x_ n$ of $M$. The elements $x_ i$ are generators of $M$. The map $\bigoplus _{j \in J} R \to R^{\oplus n}$ is given by a $n \times J$ matrix $A$ with coefficients in $R$. In other words, $A = (a_{ij})_{i = 1, \ldots , n, j \in J}$. The columns $(a_{1j}, \ldots , a_{nj})$, $j \in J$ of $A$ are said to be the relations. Any vector $(r_ i) \in R^{\oplus n}$ such that $\sum r_ i x_ i = 0$ is a linear combination of the columns of $A$. Of course any finite $R$-module has a lot of different presentations.

1. Show that the ideal generated by the $(n - k) \times (n - k)$ minors of $A$ is independent of the choice of the presentation. This ideal is the $k$th Fitting ideal of $M$. Notation $Fit_ k(M)$.

2. Show that $Fit_0(M) \subset Fit_1(M) \subset Fit_2(M) \subset \ldots$. (Hint: Use that a determinant can be computed by expanding along a column.)

3. Show that the following are equivalent:

1. $Fit_{r - 1}(M) = (0)$ and $Fit_ r(M) = R$, and

2. $M$ is locally free of rank $r$.

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