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Be a bit more precise with ranks of modules

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aisejohan committed Sep 7, 2019
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  1. +108 −81 algebra.tex
  2. +1 −43 examples.tex
  3. +1 −1 tags/tags
@@ -2654,6 +2654,62 @@ \section{Miscellany}
in the statement of the lemma.
\end{proof}

\begin{lemma}
\label{lemma-map-cannot-be-injective}
\begin{slogan}
A map of finite free modules cannot be injective if the source has
rank bigger than the target.
\end{slogan}
Let $R$ be a nonzero ring. Let $n \geq 1$. Let $M$ be an $R$-module generated
by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a
nonzero kernel.
\end{lemma}

\begin{proof}
Choose a surjection $R^{\oplus n - 1} \to M$.
We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$
(Lemma \ref{lemma-lift-map}).
It suffices to prove $f'$ has a nonzero kernel.
The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a
matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say
$a = a_{ij}$ is not, then we can replace $R$ by the localization $R_a$
and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel
after localization then there was a nonzero kernel to start with as
localization is exact, see Proposition \ref{proposition-localization-exact}.
In this case we can do a base change on both $R^{\oplus n}$
and $R^{\oplus n - 1}$ and reduce to the case where
$$
A =
\left(
\begin{matrix}
1 & 0 & 0 & \ldots \\
0 & a_{22} & a_{23} & \ldots \\
0 & a_{32} & \ldots \\
\ldots & \ldots
\end{matrix}
\right)
$$
Hence in this case we win by induction on $n$. If not then each
$a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that
$I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer
such that $I^m \not = 0$. Then we see that $(I^m)^{\oplus n}$ is
contained in the kernel of the map and we win.
\end{proof}

\begin{lemma}
\label{lemma-rank}
\begin{slogan}
The rank of a finite free module is well defined.
\end{slogan}
Let $R$ be a nonzero ring. Let $n, m \geq 0$ be integers.
If $R^{\oplus n}$ is isomorphic to $R^{\oplus m}$ as
$R$-modules, then $n = m$.
\end{lemma}

\begin{proof}
Immediate from Lemma \ref{lemma-map-cannot-be-injective}.
\end{proof}




@@ -12501,32 +12557,40 @@ \section{K-groups}
\end{example}

\begin{example}
\label{example-K0-polynomial-ring}
Let $k$ be a field. Then $K_0(k[x]) = K'_0(k[x]) = \mathbf{Z}$.

\medskip\noindent
Since $R = k[x]$ is a principal ideal domain, any finite projective
$R$-module is free. In a short exact sequence of modules
\label{example-K0-PID}
Let $R$ be a PID. We claim $K_0(R) = K'_0(R) = \mathbf{Z}$.
Namely, any finite projective $R$-module is finite free.
A finite free module has a well defined rank by Lemma \ref{lemma-rank}.
Given a short exact sequence of finite free modules
$$
0 \to M' \to M \to M'' \to 0
$$
we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$,
which gives $K_0(k[x]) = \mathbf{Z}$.
we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$
because we have $M \cong M' \oplus M'$ in this case (for example
we have a splitting by Lemma \ref{lemma-lift-map}).
We conclude $K_0(R) = \mathbf{Z}$.

\medskip\noindent
As for $K'_0$, the structure theorem for modules of a PID says that
The structure theorem for modules of a PID says that
any finitely generated $R$-module is of the form
$M = R^r \times R/(d_1) \times \ldots \times R/(d_k)$.
$M = R^{\oplus r} \oplus R/(d_1) \oplus \ldots \oplus R/(d_k)$.
Consider the short exact sequence
$$
0 \to (d_i) \to R \to R/(d_i) \to 0
$$
Since the ideal $(d_i)$ is isomorphic to $R$ as a module
(it is free with generator $d_i$), in $K'_0(R)$ we have
$[(d_i)] = [R]$. Then $[R/(d_i)] = [(d_i)]-[R] = 0$. From this it
follows that any torsion part ``disappears'' in $K'_0$.
Again the rank of the free part determines that $K'_0(k[x]) = \mathbf{Z}$,
and the canonical homomorphism from $K_0$ to $K'_0$ is an isomorphism.
follows that a torsion module has zero class in $K'_0(R)$.
Using the rank of the free part gives an identification $K'_0(R) = \mathbf{Z}$
and the canonical homomorphism from $K_0(R) \to K'_0(R)$ is an isomorphism.
\end{example}

\begin{example}
\label{example-K0-polynomial-ring}
Let $k$ be a field. Then $K_0(k[x]) = K'_0(k[x]) = \mathbf{Z}$.
This follows from Example \ref{example-K0-PID} as
$R = k[x]$ is a PID.
\end{example}

\begin{example}
@@ -12560,56 +12624,30 @@ \section{K-groups}

\begin{lemma}
\label{lemma-K0-local}
Let $R$ be a local ring. Every finite projective $R$-module
Let $(R, \mathfrak m)$ be a local ring. Every finite projective $R$-module
is finite free. The map $\text{rank}_R : K_0(R) \to \mathbf{Z}$
defined by $[M] \to \text{rank}_R(M)$ is well defined
and an isomorphism.
\end{lemma}

\begin{proof}
Let $P$ be a finite projective $R$-module.
The $n$ generators of $P$ give a surjection
$R^n \to P$, and since $P$ is projective it
follows that $R^n \cong P \oplus Q$ for some
projective module $Q$.

\medskip\noindent
If $\mathfrak m \subset R$ is the maximal ideal,
then $P/\mathfrak m$ and $Q/\mathfrak m$ are $R/\mathfrak m$-vector spaces,
with $P/\mathfrak m \oplus Q/\mathfrak m \cong (R/\mathfrak m)^n$.
Say that $\dim P = p$, $\dim Q = q$, so $p + q = n$.

\medskip\noindent
Choose elements $a_1, \ldots, a_p$ in $P$ and $b_1, \ldots, b_q$ in $Q$
lying above bases for $P/\mathfrak m$ and $Q/\mathfrak m$.
The homomorphism $R^n \to P \oplus Q \cong R^n$ given by
$(r_1, \ldots, r_n) \mapsto
r_1a_1 + \ldots + r_pa_p + r_{p + 1} b_1 + \ldots + r_nb_q$
is a matrix $A$ which is invertible over $R/\mathfrak m$. Let $B$
be a matrix over $R$ lying over the inverse of $A$ in $R/\mathfrak m$.
$AB = I + M$, where $M$ is a matrix whose entries all lie in $\mathfrak m$.
Thus $\det AB = 1 + x$, for $x \in \mathfrak m$, so $AB$ is invertible,
so $A$ is invertible.

\medskip\noindent
The homomorphism $R^p \to P$ given by
$(r_1, \ldots, r_p) \mapsto r_1a_1 + \ldots + r_pa_p$ inherits injectivity and
surjectivity from A. Hence, $P \cong R^p$.

\medskip\noindent
Next we show that the rank of a finite projective module over $R$ is
well defined: if $P \cong R^\alpha \cong R^\beta$, then $\alpha = \beta$.
This is immediate in the vector space case, and so it is true in the
general module case as well, by dividing out the maximal ideal on both sides.
If $0 \to R^\alpha \to R^\beta \to R^\gamma \to 0$
is exact, the sequence splits, so $R^\beta \cong R^\alpha \oplus R^\gamma$,
so $\beta = \alpha + \gamma$.

\medskip\noindent
So far we have seen that the map $\text{rank}_R : K_0(R) \to \mathbf{Z}$
is a well-defined homomorphism. It is surjective because
$\text{rank}_R[R] = 1$. It is injective because the element
of $K_0(R)$ with rank $\pm\alpha$ is uniquely $\pm [R^\alpha]$.
Let $P$ be a finite projective $R$-module. Choose elements
$x_1, \ldots, x_n \in P$ which map to a basis of $P/\mathfrak m P$.
By Nakayama's Lemma \ref{lemma-NAK} these elements generate $P$.
The corresponding surjection $u : R^{\oplus n} \to P$ has a splitting
as $P$ is projective. Hence $R^{\oplus n} = P \oplus Q$ with $Q = \Ker(u)$.
It follows that $Q/\mathfrak m Q = 0$, hence $Q$ is zero by
Nakayama's lemma. In this way we see that every finite projective
$R$-module is finite free.
A finite free module has a well defined rank by Lemma \ref{lemma-rank}.
Given a short exact sequence of finite free $R$-modules
$$
0 \to M' \to M \to M'' \to 0
$$
we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$
because we have $M \cong M' \oplus M'$ in this case (for example
we have a splitting by Lemma \ref{lemma-lift-map}).
We conclude $K_0(R) = \mathbf{Z}$.
\end{proof}

\begin{lemma}
@@ -12629,27 +12667,13 @@ \section{K-groups}
\end{lemma}

\begin{proof}
By induction on the rank of $M$.
Suppose $\left[M\right] \in K_0(R)$.
Then $M$ is a finite projective $R$-module
over a local ring, so M is free;
$M \cong R^n$ for some $n$.
The claim is that
$\text{rank} (M) \text{length}_R (R) = \text{length}_R(M)$,
or equivalently that $n\text{length}_R(R) = \text{length}_R (R^n)$
for all $n \geq 1$. When $n = 1$, this is clearly true.
Suppose that $(n-1) \text{length}_R(R) =\text{ length}_R(R^{n-1})$.
Then since there is a split short exact sequence
$$
0 \to R \to R^n \to R^{n-1} \to 0
$$
by Lemma \ref{lemma-length-additive} we have
\begin{eqnarray*}
\text{length}_R(R^n) & = & \text{length}_R(R) + \text{length}_R(R^{n-1}) \\
& = & \text{length}_R(R) + (n-1) \text{length}_R(R) \\
& = & n\text{length}_R(R)
\end{eqnarray*}
as desired.
Let $P$ be a finite projective $R$-module. We have to show that
$\text{length}_R(P) = \text{rank}_R(P) \text{length}_R(R)$.
By Lemma \ref{lemma-K0-local} the module $P$ is finite free. So
$P \cong R^{\oplus n}$ for some $n \geq 0$. Then $\text{rank}_R(P) = n$ and
$\text{length}_R(R^{\oplus n}) = n \text{length}_R(R)$
by additivity of lenghts (Lemma \ref{lemma-length-additive}).
Thus the result holds.
\end{proof}


\end{definition}

\noindent
Note that a finite locally free $R$-module is
automatically finitely presented by
Lemma \ref{lemma-cover}.
Note that a finite locally free $R$-module is automatically
finitely presented by Lemma \ref{lemma-cover}. Moreover, if $M$ is a
finite locally free module of rank $r$ over a ring $R$ and if
$R$ is nonzero, then $r$ is uniquely determined by
Lemma \ref{lemma-rank} (because at least one of the
localizations $ R_{f_i}$ is a nonzero ring).

\begin{lemma}
\label{lemma-finite-projective}
@@ -2050,55 +2050,13 @@ \section{A projective module which is not locally free}
$R$-module.
\end{proof}

\begin{lemma}
\label{lemma-map-cannot-be-injective}
\begin{slogan}
A map of finite free modules cannot be injective if the source has
rank bigger than the target.
\end{slogan}
Let $R$ be a nonzero ring. Let $n \geq 1$. Let $M$ be an $R$-module generated
by $< n$ elements. Then any $R$-module map $f : R^{\oplus n} \to M$ has a
nonzero kernel.
\end{lemma}

\begin{proof}
Choose a surjection $R^{\oplus n - 1} \to M$.
We may lift the map $f$ to a map $f' : R^{\oplus n} \to R^{\oplus n - 1}$.
It suffices to prove $f'$ has a nonzero kernel.
The map $f' : R^{\oplus n} \to R^{\oplus n - 1}$ is given by a
matrix $A = (a_{ij})$. If one of the $a_{ij}$ is not nilpotent, say
$a = a_{ij}$ is not, then we can replace $A$ by the localization $A_a$
and we may assume $a_{ij}$ is a unit. Since if we find a nonzero kernel
after localization then there was a nonzero kernel to start with as
localization is exact, see
Algebra, Proposition \ref{algebra-proposition-localization-exact}.
In this case we can do a base change on both $R^{\oplus n}$
and $R^{\oplus n - 1}$ and reduce to the case where
$$
A =
\left(
\begin{matrix}
1 & 0 & 0 & \ldots \\
0 & a_{22} & a_{23} & \ldots \\
0 & a_{32} & \ldots \\
\ldots & \ldots
\end{matrix}
\right)
$$
Hence in this case we win by induction on $n$. If not then each
$a_{ij}$ is nilpotent. Set $I = (a_{ij}) \subset R$. Note that
$I^{m + 1} = 0$ for some $m \geq 0$. Let $m$ be the largest integer
such that $I^m \not = 0$. Then we see that $(I^m)^{\oplus n}$ is
contained in the kernel of the map and we win.
\end{proof}

\noindent
Suppose that $P \subset Q$ is an inclusion of $R$-modules with $Q$ a
finite $R$-module and $P$ locally free, see
Algebra, Definition \ref{algebra-definition-locally-free}.
Suppose that $Q$ can be generated by $N$ elements as an $R$-module.
Then it follows from
Lemma \ref{lemma-map-cannot-be-injective}
Algebra, Lemma \ref{algebra-lemma-map-cannot-be-injective}
that $P$ is finite locally free (with the free parts having rank
at most $N$). And in this case $P$ is a finite $R$-module, see
Algebra, Lemma \ref{algebra-lemma-finite-projective}.
@@ -7583,7 +7583,7 @@
05WF,stacks-properties-lemma-zariski-open-cover-stack-is-scheme
05WG,examples-section-projective-not-locally-free
05WH,examples-lemma-ideal-generated-by-idempotents-projective
05WI,examples-lemma-map-cannot-be-injective
05WI,algebra-lemma-map-cannot-be-injective
05WJ,examples-lemma-ideal-projective-not-locally-free
05WK,examples-lemma-chow-group-product
05WL,examples-lemma-projective-not-locally-free

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