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I am trying to solve Dummit Foote, Abstract Algebra, 15.1 Exercise 40. It reads,

Suppose $R$ and $S$ are commutative rings, $\phi:R \to S$ is a ring homomorphism, $M$ is a finitely generated $R$-module, and $M'=S\otimes_R M$ is the $S$-module obtained by extending scalars from $R$ to $S$. Prove that the Fitting ideal $\mathcal{F}_S(M')$ for $M'$ over $S$ is the extension to $S$ of the Fitting ideal $\mathcal{F}_R(M)$ for $M$ over $R$.

Below is my attempt: The previous exercise implies that the Fitting ideal is actually the image of $\wedge^n\ker f$ in to $R$ where $f$ is a surjection $f:R^n\to M$. In other word, I have a surjection $\wedge^n\ker f\to \mathcal F_R(M)$. Tensoring $S$, since tensoring is right exact functor, we have the following commutative diagram: $\require{AMScd}$ $$ \begin{CD} \wedge^n \ker f @>>> \mathcal{F}_R(M) @>>> 0 \\ @VVV @VVV \\ S\otimes \wedge^n \ker f @>>> S\otimes \mathcal{F}_R(M) @>>> 0 \\ @VVV @VVV \\ \wedge^n(\ker f_S) @>>> \mathcal{F}_S(M') @>>> 0 \end{CD} $$ where $f_s$ is $f\otimes \textrm{Id}_S: S\otimes R^n=S^n \to S\otimes M = M'$.

Now I am stuck here. I don't see why $S\otimes \mathcal{F}_R(M) \to \mathcal{F}_S(M')$ should be an isomorphism.

Thanks in advance.

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I don't know this is right or wrong, but anyway I dealt with it myself. $S$ is an $R$-algebra in which the multiplication $r\cdot s$ for $r\in R$ and $s\in S$ is defined to be $$ r\cdot s := \phi(r)s. $$ Our goal is to prove that $\mathcal F_R(M) S = \mathcal F_S(M')$. First we prove that $\mathcal F_R(M)S \subseteq \mathcal F_S(M')$. By the definition, the generators of $\mathcal{F}_R(M)$ are of the form $$ \det\begin{pmatrix} r_{11} & r_{12} & \cdots & r_{1n} \\ r_{21} & r_{22} & \cdots & r_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ r_{n1} & r_{n2} & \cdots & r_{nn} \end{pmatrix} $$ where $r_{i1}m_1 + r_{i2} m_2 + \cdots + r_{in} m_n = 0$. Hence the extension of $\mathcal F_R(M)$ to $S$ is generated by the elements of the form $$ \phi \det\begin{pmatrix} r_{11} & r_{12} & \cdots & r_{1n} \\ r_{21} & r_{22} & \cdots & r_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ r_{n1} & r_{n2} & \cdots & r_{nn} \end{pmatrix}. $$ Since $\phi$ is the ring homomorphism, and the determinant is just a polynomial of $n\times n$ variables, $\phi$ and $\det$ commutes. Thus, $$ \phi \det\begin{pmatrix} r_{11} & r_{12} & \cdots & r_{1n} \\ r_{21} & r_{22} & \cdots & r_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ r_{n1} & r_{n2} & \cdots & r_{nn} \end{pmatrix}=\det\begin{pmatrix} \phi(r_{11}) & \phi(r_{12}) & \cdots & \phi(r_{1n}) \\ \phi(r_{21}) & \phi(r_{22}) & \cdots & \phi(r_{2n}) \\ \vdots & \vdots & \ddots & \vdots \\ \phi(r_{n1}) & \phi(r_{n2}) & \cdots & \phi(r_{nn}) \end{pmatrix} $$ However, the right hand side is an element of $\mathcal F_S(M')$ since $$ \phi(r_{i1})1\otimes m_1+\phi(r_{i2})1\otimes m_2+\cdots+\phi(r_{in})1\otimes m_n = 1\otimes (r_{i1}m_1+ \cdots r_{in}m_n) = 0 $$ for all $i=1,2,\cdots,n$. Remind that since $R^n \to M$ is a surjection, and tensoring is right exact, so $S^n \to M'$ is still surjection. Thus $1\otimes m_1, \cdots , 1\otimes m_n$ generate $M'$. Hence we have the natural injection $$ \mathcal{F}_R(M)S \to \mathcal{F}_S(M'). $$

Now our goal is to prove that this is actually a surjection. There are two preceding lemmas.

(i) $f:A\to B$ and $g:C\to D$ are surjections, then so is $f\wedge g:A\wedge C \to B\wedge D$. This is obvious since for every $b\wedge d\in B\wedge D$, there are $a\in A$ and $c\in C$ such that $f(a)=b$, $g(c)=d$, so $f\wedge g (a\wedge c) = b\wedge d$. Hence by the corollary, when $f:M \to N$ is a surjection between two $R$-modules, then so is $\det f: \wedge^n M \to \wedge^n N$ for every $n=1,2,\cdots$.

(ii) Suppose $L$ is an $R$-module and $S$ is an $R$-algebra. Then there is a cannonical $S$-module isomorphism $$ \bigwedge_S^n ( S\otimes L) \cong S \otimes \left(\bigwedge^n_R L\right). $$ The isomorphism is defined as $$ (s_1\otimes_R a_1) \wedge_S \cdots \wedge_S (s_n \otimes a_n) \mapsto s_1 \cdots s_n \otimes_R a_1 \wedge_R \cdots \wedge_R a_n $$ $$ (s\otimes_R a_1) \wedge_S (1\otimes_R a_2) \wedge_S \cdots \wedge_S (1\otimes a_n) \leftarrow s\otimes a_1 \wedge_R \cdots \wedge_R a_n $$ It is obvious that those two maps are homormophisms, and they are inverse to each other. This is actually just the commutativity of the tensor product.

Now we are ready to prove the original problem. First, we let $f_S:S^n \to M'$ the surjection given by tensoring $S$ to the surjection $f:R^n \to M$. Then we have the following commutative diagram

enter image description here

The first horizontal line is given by tensoring $S$ to the exact sequence $$ 0 \to \textrm{Ker} f \to R^n \to M \to 0. $$ The second horizontal sequence is obviously exact. For every $x\in \textrm{Ker} f_S$, it is actually in $S^n$, and is mapped to $0$ by $S^n \to M'$. Since the first horizontal sequence is exact, there should be $y\in S\otimes \textrm{Ker} f$ such that $y$ is mapped to $x$ by $S\otimes \textrm{Ker}f \to S^n$. This means $S\otimes \textrm{Ker}f \to \textrm{Ker}f_S$ is the surjection. By the first lemma, the following homomorphism $$ \bigwedge^n S\otimes \textrm{Ker}f \to \bigwedge^n \textrm{Ker}f_S $$ is surjection. And by the second lemma, $\bigwedge^n S\otimes \textrm{Ker}f$ is isomorphic to $S\otimes \bigwedge^n(\textrm{Ker}f)$. By the result of the previous exercise, $$ \bigwedge^n(\textrm{Ker}f) \to \mathcal F_R(M) $$ is a surjection. By tensoring $S$, we have another homomorphism $$ S\otimes \bigwedge^n(\textrm{Ker}f) \to S\otimes \mathcal F_R(M). $$ Also, there is a canonical homomorphism $$ S\otimes \mathcal F_R(M) \to \mathcal F_R(M)S. $$ Now we have the following diagram

enter image description here

If the diagram commutes, then $\mathcal F_R(M)S \to \mathcal{F}_S (M')$ is the surjection, so the proof ends. It is enough to observe how the element is mapped in the diagram to show that the diagram actually commutes. For every $s_1,\cdots, s_n \in S$ and $r_1,\cdots, r_n \in \mathrm{Ker}f$, let $s=s_1\cdots s_n$. First, along $$ \bigwedge^n (S\otimes \mathrm{Ker} f) \to S\otimes \bigwedge^n \mathrm{Ker} f \to S\otimes \mathcal F_R(M) \to \mathcal F_R(M)S \to \mathcal{F}_S (M') $$ we have \begin{align*} (s_1\otimes r_1)\wedge \cdots \wedge (s_n\otimes r_n) &\mapsto s \otimes r_1 \wedge \cdots \wedge r_n \\ &\mapsto s\otimes \det(r_1 \wedge \cdots \wedge r_n) \\ &\mapsto \det(r_1 \wedge \cdots \wedge r_n)\cdot s = \phi(\det(r_1 \wedge \cdots \wedge r_n)) s \end{align*} Next, along $$ \bigwedge^n (S\otimes \mathrm{Ker} f) \to \bigwedge^n \mathrm{Ker} f_S \to \mathcal{F}_S (M') $$ we have \begin{align*} (s_1\otimes r_1)\wedge \cdots \wedge (s_n\otimes r_n) &\mapsto s\phi(r_1) \wedge \cdots \wedge \phi(r_n) \\ &\mapsto \det(s\phi(r_1) \wedge \cdots \wedge \phi(r_n)) = \det(\phi(r_1) \wedge \cdots \wedge \phi(r_n))s \end{align*} Hence the diagram commutes decause $\det$ and $\phi$ commutes.

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