Lemma 42.68.41. Let $R$ be a Noetherian local ring. Let $\mathfrak q \subset R$ be a prime with $\dim (R/\mathfrak q) = 1$. Let $\varphi : M \to N$ be a homomorphism of finite $R$-modules. Assume there exist $x_1, \ldots , x_ l \in M$ and $y_1, \ldots , y_ l \in M$ with the following properties

1. $M = \langle x_1, \ldots , x_ l\rangle$,

2. $\langle x_1, \ldots , x_ i\rangle / \langle x_1, \ldots , x_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots , l$,

3. $N = \langle y_1, \ldots , y_ l\rangle$, and

4. $\langle y_1, \ldots , y_ i\rangle / \langle y_1, \ldots , y_{i - 1}\rangle \cong R/\mathfrak q$ for $i = 1, \ldots , l$.

Then $\varphi$ is injective if and only if $\varphi _{\mathfrak q}$ is an isomorphism, and in this case we have

$\text{length}_ R(\mathop{\mathrm{Coker}}(\varphi )) = \text{ord}_{R/\mathfrak q}(f)$

where $f \in \kappa (\mathfrak q)$ is the element such that

$[\varphi (x_1), \ldots , \varphi (x_ l)] = f [y_1, \ldots , y_ l]$

in $\det _{\kappa (\mathfrak q)}(N_{\mathfrak q})$.

Proof. First, note that the lemma holds in case $l = 1$. Namely, in this case $x_1$ is a basis of $M$ over $R/\mathfrak q$ and $y_1$ is a basis of $N$ over $R/\mathfrak q$ and we have $\varphi (x_1) = fy_1$ for some $f \in R$. Thus $\varphi$ is injective if and only if $f \not\in \mathfrak q$. Moreover, $\mathop{\mathrm{Coker}}(\varphi ) = R/(f, \mathfrak q)$ and hence the lemma holds by definition of $\text{ord}_{R/q}(f)$ (see Algebra, Definition 10.121.2).

In fact, suppose more generally that $\varphi (x_ i) = f_ iy_ i$ for some $f_ i \in R$, $f_ i \not\in \mathfrak q$. Then the induced maps

$\langle x_1, \ldots , x_ i\rangle / \langle x_1, \ldots , x_{i - 1}\rangle \longrightarrow \langle y_1, \ldots , y_ i\rangle / \langle y_1, \ldots , y_{i - 1}\rangle$

are all injective and have cokernels isomorphic to $R/(f_ i, \mathfrak q)$. Hence we see that

$\text{length}_ R(\mathop{\mathrm{Coker}}(\varphi )) = \sum \text{ord}_{R/\mathfrak q}(f_ i).$

On the other hand it is clear that

$[\varphi (x_1), \ldots , \varphi (x_ l)] = f_1 \ldots f_ l [y_1, \ldots , y_ l]$

in this case from the admissible relation (b) for symbols. Hence we see the result holds in this case also.

We prove the general case by induction on $l$. Assume $l > 1$. Let $i \in \{ 1, \ldots , l\}$ be minimal such that $\varphi (x_1) \in \langle y_1, \ldots , y_ i\rangle$. We will argue by induction on $i$. If $i = 1$, then we get a commutative diagram

$\xymatrix{ 0 \ar[r] & \langle x_1 \rangle \ar[r] \ar[d] & \langle x_1, \ldots , x_ l \rangle \ar[r] \ar[d] & \langle x_1, \ldots , x_ l \rangle / \langle x_1 \rangle \ar[r] \ar[d] & 0 \\ 0 \ar[r] & \langle y_1 \rangle \ar[r] & \langle y_1, \ldots , y_ l \rangle \ar[r] & \langle y_1, \ldots , y_ l \rangle / \langle y_1 \rangle \ar[r] & 0 }$

and the lemma follows from the snake lemma and induction on $l$. Assume now that $i > 1$. Write $\varphi (x_1) = a_1 y_1 + \ldots + a_{i - 1} y_{i - 1} + a y_ i$ with $a_ j, a \in R$ and $a \not\in \mathfrak q$ (since otherwise $i$ was not minimal). Set

$x'_ j = \left\{ \begin{matrix} x_ j & \text{if} & j = 1 \\ ax_ j & \text{if} & j \geq 2 \end{matrix} \right. \quad \text{and}\quad y'_ j = \left\{ \begin{matrix} y_ j & \text{if} & j < i \\ ay_ j & \text{if} & j \geq i \end{matrix} \right.$

Let $M' = \langle x'_1, \ldots , x'_ l \rangle$ and $N' = \langle y'_1, \ldots , y'_ l \rangle$. Since $\varphi (x'_1) = a_1 y'_1 + \ldots + a_{i - 1} y'_{i - 1} + y'_ i$ by construction and since for $j > 1$ we have $\varphi (x'_ j) = a\varphi (x_ i) \in \langle y'_1, \ldots , y'_ l\rangle$ we get a commutative diagram of $R$-modules and maps

$\xymatrix{ M' \ar[d] \ar[r]_{\varphi '} & N' \ar[d] \\ M \ar[r]^\varphi & N }$

By the result of the second paragraph of the proof we know that $\text{length}_ R(M/M') = (l - 1)\text{ord}_{R/\mathfrak q}(a)$ and similarly $\text{length}_ R(M/M') = (l - i + 1)\text{ord}_{R/\mathfrak q}(a)$. By a diagram chase this implies that

$\text{length}_ R(\mathop{\mathrm{Coker}}(\varphi ')) = \text{length}_ R(\mathop{\mathrm{Coker}}(\varphi )) + i\ \text{ord}_{R/\mathfrak q}(a).$

On the other hand, it is clear that writing

$[\varphi (x_1), \ldots , \varphi (x_ l)] = f [y_1, \ldots , y_ l], \quad [\varphi '(x'_1), \ldots , \varphi (x'_ l)] = f' [y'_1, \ldots , y'_ l]$

we have $f' = a^ if$. Hence it suffices to prove the lemma for the case that $\varphi (x_1) = a_1y_1 + \ldots a_{i - 1}y_{i - 1} + y_ i$, i.e., in the case that $a = 1$. Next, recall that

$[y_1, \ldots , y_ l] = [y_1, \ldots , y_{i - 1}, a_1y_1 + \ldots a_{i - 1}y_{i - 1} + y_ i, y_{i + 1}, \ldots , y_ l]$

by the admissible relations for symbols. The sequence $y_1, \ldots , y_{i - 1}, a_1y_1 + \ldots + a_{i - 1}y_{i - 1} + y_ i, y_{i + 1}, \ldots , y_ l$ satisfies the conditions (3), (4) of the lemma also. Hence, we may actually assume that $\varphi (x_1) = y_ i$. In this case, note that we have $\mathfrak q x_1 = 0$ which implies also $\mathfrak q y_ i = 0$. We have

$[y_1, \ldots , y_ l] = - [y_1, \ldots , y_{i - 2}, y_ i, y_{i - 1}, y_{i + 1}, \ldots , y_ l]$

by the third of the admissible relations defining $\det _{\kappa (\mathfrak q)}(N_{\mathfrak q})$. Hence we may replace $y_1, \ldots , y_ l$ by the sequence $y'_1, \ldots , y'_ l = y_1, \ldots , y_{i - 2}, y_ i, y_{i - 1}, y_{i + 1}, \ldots , y_ l$ (which also satisfies conditions (3) and (4) of the lemma). Clearly this decreases the invariant $i$ by $1$ and we win by induction on $i$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).