Fitting ideals do not depend on a presentation

Put $F=A^n$ and $K=\ker (F\stackrel{\pi }{\to }M)=\mathrm{im}\phi$, and for a submodule $N\subseteq F$ and $r\ge 0$ let $J_r(N)$ be the ideal generated by the determinants of all $r\times r$ matrices whose columns are $r$ elements of $N$, with $J_r(N)=A$ for $r\le 0$ and $J_r(N)=0$ for $r>n$; multilinearity of the determinant in the columns shows that the ideal generated by the $r\times r$ minors of a matrix whose columns generate $N$ equals $J_r(N)$, so $I_r(\phi )=J_r(K)$ for every $r$ and the stated conventions match, and it suffices to prove $J_{n-k}(K)=J_{n^{\mathrm{\prime }}-k}(K^{\mathrm{\prime }})$ for the kernels of two surjections $F=A^n\twoheadrightarrow M$ and $F^{\mathrm{\prime }}=A^{n^{\mathrm{\prime }}}\twoheadrightarrow M$ from finite free modules.

Determinant calculus, three elementary facts. (a) Nestedness: $J_r(N)\subseteq J_s(N)$ for $s\le r$, by Laplace expansion of an $r\times r$ determinant along a row. (b) Cauchy--Binet: every $r\times r$ minor of a product of matrices $CA$ is an $A$-linear combination of $r\times r$ minors of $A$, obtained by expanding the determinant of the product and regrouping terms according to the $r$ rows of $A$ used; consequently $J_r(CN)=J_r(N)$ for every automorphism $C$ of $F$, by applying this inclusion to $C$ and to $C^{-1}$. (c) Block Laplace expansion: for a square submatrix of $\left(\begin{array}{cc}{M}_1& 0\\ M_2& M_3\end{array}\right)$ with $h$ selected lower rows, expand its determinant along those $h$ rows. Every term is a product of an $h\times h$ minor from the selected lower rows and a complementary minor from $M_1$; expanding the former further in columns separates contributions from $M_2$ and $M_3$. In particular, whenever a term uses $j$ columns from the $M_3$ block, its other columns contribute a minor of $M_1$ of size equal to the number of selected upper rows.

Augmentation identity: for $N\subseteq A^n$ and $m\ge 0$, with $N\oplus A^m\subseteq A^n\oplus A^m$, one has $J_r(N\oplus A^m)=J_{r-m}(N)$, and the same identity holds with the two blocks interchanged, $J_r(A^m\oplus N)=J_{r-m}(N)$. If $r\le m$, the right side is $A$ by convention and an $r\times r$ identity minor in the free block shows that the left side is $A$. If $r>m$, adjoining the $m$ standard free-block vectors to any $r-m$ vectors of $N$ gives a block identity minor, so $J_{r-m}(N)\subseteq J_r(N\oplus A^m)$. Conversely, expand any $r\times r$ minor of vectors $(v_i,c_i)\in N\oplus A^m$ along its selected free-block rows. Every resulting term is a free-block minor times a minor of the $v_i$ of size at least $r-m$, hence lies in $J_{r-m}(N)$ by nestedness in step 1.2(a); thus $J_r(N\oplus A^m)\subseteq J_{r-m}(N)$. Swapping the two blocks proves the second identity.

Now let $p:F\twoheadrightarrow M$ and $p^{\mathrm{\prime }}:F^{\mathrm{\prime }}\twoheadrightarrow M$ be two surjections from finite free modules, with kernels $K$ and $K^{\mathrm{\prime }}$ as in step 1.1; choose an $A$-linear lift $w:F^{\mathrm{\prime }}\to F$ with $p\circ w=p^{\mathrm{\prime }}$ by choosing values on the $n^{\mathrm{\prime }}$ standard basis elements of $F^{\mathrm{\prime }}$ one at a time, a finite and choice-free construction, and define $L=\{(x,y)\in F\oplus F^{\mathrm{\prime }}:p(x)=p^{\mathrm{\prime }}(y)\}\subseteq F\oplus F^{\mathrm{\prime }}$, which is the kernel of the $A$-linear map $p-p^{\mathrm{\prime }}\circ {\mathrm{p}\mathrm{r}}_2$ on $F\oplus F^{\mathrm{\prime }}$.

The endomorphisms $C(x,y)=(x-w(y),y)$ and $D(x,y)=(x,y-u(x))$ of $F\oplus F^{\mathrm{\prime }}$, where $u:F\to F^{\mathrm{\prime }}$ is an $A$-linear lift with $p^{\mathrm{\prime }}\circ u=p$ (which exists by choosing values on the finite standard basis of $F$), are automorphisms with inverses $(x,y)\mapsto (x+w(y),y)$ and $(x,y)\mapsto (x,y+u(x))$; for $(x,y)\in L$ one has $p(x-w(y))=p(x)-p^{\mathrm{\prime }}(y)=0$, so $C(L)\subseteq K\oplus F^{\mathrm{\prime }}$, while $p(k+w(y))=p^{\mathrm{\prime }}(y)$ for $k\in K$ shows $C(L)=K\oplus F^{\mathrm{\prime }}$, and likewise $p^{\mathrm{\prime }}(y-u(x))=p^{\mathrm{\prime }}(y)-p(x)=0$ on $L$ gives $D(L)=F\oplus K^{\mathrm{\prime }}$.

By step 1.2(b) applied to the automorphisms $C$ and $D$ of step 3.1, $J_{n+n^{\mathrm{\prime }}-k}(L)=J_{n+n^{\mathrm{\prime }}-k}(C(L))=J_{n+n^{\mathrm{\prime }}-k}(K\oplus F^{\mathrm{\prime }})$ and $J_{n+n^{\mathrm{\prime }}-k}(L)=J_{n+n^{\mathrm{\prime }}-k}(D(L))=J_{n+n^{\mathrm{\prime }}-k}(F\oplus K^{\mathrm{\prime }})$; the augmentation identity of step 2.1, with $m=n^{\mathrm{\prime }}$ and $m=n$ respectively, gives $J_{n+n^{\mathrm{\prime }}-k}(K\oplus F^{\mathrm{\prime }})=J_{n-k}(K)$ and $J_{n+n^{\mathrm{\prime }}-k}(F\oplus K^{\mathrm{\prime }})=J_{n^{\mathrm{\prime }}-k}(K^{\mathrm{\prime }})$, whence $J_{n-k}(K)=J_{n^{\mathrm{\prime }}-k}(K^{\mathrm{\prime }})$; combined with step 1.1, this proves $I_{n-k}(\phi )=I_{n^{\mathrm{\prime }}-k}(\psi )$ for any two presentations.

Base change: let $A\to B$ be a ring homomorphism and apply $-{\otimes }_{A}B$ to the presentation, so that by [F3] the sequence $B^{(J)}\to B^n\to M{\otimes }_{A}B\to 0$ is exact, hence is a presentation of $M{\otimes }_{A}B$ whose matrix has the images of the entries of $\phi$; its $(n-k)\times (n-k)$ minors are the images of those of $\phi$, and the two ideals are generated by these images, giving ${\mathrm{Fitt}}_k(M{\otimes }_{A}B)={\mathrm{Fitt}}_k(M)\cdot B$; for the localisation statement apply [F2] to the presentation, so that the localised sequence $A_f^{(J)}\to A_f^n\to M_f\to 0$ is exact, its matrix is the localisation of the matrix of $\phi$, and minors commute with the coefficient map, giving ${\mathrm{Fitt}}_k(M_f)={\mathrm{Fitt}}_k(M)\cdot A_f$; all steps used finitely many choices on standard bases and universal constructions only, so no choice principle is invoked.