Let’s see how to classify modules over PIDs using the Smith Normal Form for their presentation matrix. First, we need a lemma on how to interpret the module presented by a matrix in Smith Normal Form.
Lemma14.26.Module Presented in Smith Normal Form.
Let \(R\) be a commutative ring with \(1 \neq 0\text{,}\) let \(m \geqslant n\text{,}\) let \(A=\left[a_{i j}\right] \in \mathrm{M}_{m, n}(R)\) be a matrix such that all nondiagonal entries of \(A\) are 0, and let \(M\) be the \(R\)-module presented by \(A\text{.}\) Then \(M \cong R^{m-n} \oplus R /\left(a_{11}\right) \oplus \cdots \oplus R /\left(a_{n n}\right)\text{.}\)
Theorem14.27.Classification of Finitely Generated Modules over a PID using Invariant Factors.
Let \(R\) be a PID and let \(M\) be a finitely generated module. Then there exist \(r \geqslant 0\text{,}\)\(k \geqslant 0\text{,}\) and nonzero nonunit elements \(d_{1}, \ldots, d_{k}\) of \(R\) satisfying \(d_{1}\left|d_{2}\right| \cdots \mid d_{k}\) such that
\begin{equation*}
M \cong R^{r} \oplus R /\left(d_{1}\right) \oplus \cdots \oplus R /\left(d_{k}\right).
\end{equation*}
Moreover \(r\) and \(k\) are uniquely determined by \(M\text{,}\) and the \(d_{i}\) are unique up to associates.
Proof.
By Corollary 14.13, \(M\) has a presentation matrix \(A\text{.}\) By Theorem 14.14, \(A\) can be put into Smith Normal Form \(B\text{,}\) where the diagonal entries of \(B\) are \(b_{1}, \ldots, b_{\ell}\) and satisfy \(b_{1}\left|b_{2}\right| \cdots \mid b_{k}\text{.}\) Moreover, \(k\) is unique and the \(d_{i}\) are uniquely determined up to associates (ie, up to multiplication by units) by \(A\text{,}\) hence by \(B\text{.}\) By Corollary 14.13, \(M\) is isomorphic to the module presented by \(B\text{.}\) By Lemma 14.26, this is isomorphic to
\begin{equation*}
M \cong R^{r} \oplus R /\left(b_{1}\right) \oplus \cdots \oplus R /\left(b_{\ell}\right)
\end{equation*}
Finally, some of these \(b_{i}\) might be units; let \(d_{1}|\cdots| d_{k}\) be the nonunits among the \(b_{i}\text{,}\) and note that if \(u\) is a unit, then \(R /(u) \cong(0)\text{.}\) We conclude that
\begin{equation*}
M \cong R^{r} \oplus R /\left(d_{1}\right) \oplus \cdots \oplus R /\left(d_{k}\right),
\end{equation*}
as desired
Definition14.28.Invariant Factors.
Let \(R\) be a PID, let \(r \geqslant 0, k \geqslant 0\text{,}\) and let \(d_{1}, \ldots, d_{k}\) be nonzero nonunit elements of \(R\) satisfying \(d_{1}\left|d_{2}\right| \cdots \mid d_{k}\text{.}\) Let \(M\) be any \(R\)-module such that
\begin{equation*}
M \cong R^{r} \oplus R /\left(d_{1}\right) \oplus \cdots \oplus R /\left(d_{k}\right) .
\end{equation*}
We say \(M\) has free rank\(r\) and invariant factors\(d_{1}, \ldots, d_{k}\text{.}\)
Remark14.29.
The classification theorem can be interpreted as saying that \(M\) decomposes into a free submodule \(R^{r}\) and a torsion submodule \(\operatorname{Tor}(M)=R /\left(d_{1}\right) \oplus \cdots \oplus R /\left(d_{k}\right)\text{.}\)
Example14.30.\(k[x]\)-Modules of Dimension \(4\).
Let \(k\) be a field and \(M\) a \(k[x]\)-module such that the dimension of \(M\) as a \(k\)-vector space is \(4.\) What are the possibilities for \(M\) up to isomorphism?
with \(r, s, f_1, \dots, f_s\) as in the Corollary. But \(r\) must be \(0\) since \(k[x]\) is infinite dimensional as a \(k\)-vector space. Moreover, \(\dim_F k[x]/g = \deg(g)\) for any non-zero polynomial \(g\text{.}\) So we must have \(\sum_i \deg(f_i) = 4\text{.}\) There are five possibilities:
\(s = 1\) and \(f_1\) is monic of degree \(4\text{.}\)
\(s = 2\text{,}\)\(f_1\) is linear, \(f_2\) is cubic and \(f_1 \mid f_2\text{.}\) (So if \(f_1 = x-a\text{,}\) then \(a\) must be a root of \(f_2\)).
\(s = 2\text{,}\)\(f_1 = f_2\) is quadratic.
\(s = 3\text{,}\)\(f_1 = f_2\) is linear and \(f_3\) is quadratic with \(f_1 \mid f_3\text{.}\)
Now suppose \(k = \Z/p\text{.}\) What is the total number of possibilities? For case \(1\text{,}\) there are \(p^4\) monic polynomial of degree \(4\text{.}\) For case \(2\text{,}\) there are \(p\) choices for \(f_1\) and \(p^2\) choices for \(f_2\) since \(f_2 = f_1 \cdot q\) for a unique quadratic \(q\text{,}\) for a total of \(p^3\) possibilities. For case \(3\) there are \(p^2\) choices. For case \(4\) there are \(p^2\) choices since there are \(p\) choices for \(f_1 = f_2\) and \(f_3 = f_1 \cdot l\) for a unique linear \(l\text{.}\) For case \(5\text{,}\) there are \(p\) choices for \(f_1\text{.}\) In total there are \(2p + 2p^2 + p^4\) such modules up to isomorphism.
SubsectionElementary Divisors
Here is a spinoff of the classification theorem.
Theorem14.31.Classification of Finitely Generated Modules over a PID using Elementary Divisors.
Let \(R\) be a PID and let \(M\) be a finitely generated module. Then there exist \(r \geqslant 0\text{,}\)\(s \geqslant 0\text{,}\) prime elements \(p_{1}, \ldots, p_{s}\) of \(R\) (not necessarily distinct), and \(e_{1}, \ldots, e_{s} \geqslant 1\) such that
\begin{equation*}
M \cong R^{r} \oplus R /\left(p_{1}^{e_{1}}\right) \oplus \cdots \oplus R /\left(p_{s}^{e_{s}}\right) .
\end{equation*}
Moreover, \(r\) and \(s\) are uniquely determined by \(M\text{,}\) and the list \(p_{1}^{e_{1}}, \ldots, p_{s}^{e_{s}}\) is unique up to associates and reordering.
Proof.
First, using Theorem 14.27 write \(M\) in invariant factor form \(M \cong R^{r} \oplus R /\left(d_{1}\right) \oplus \cdots \oplus R /\left(d_{k}\right)\text{.}\) Then write each invariant factor as a product of prime powers
\begin{equation*}
R /\left(d_{i}\right) \cong R /\left(p_{n_{i}}^{e_{n_{i}}}\right) \oplus \cdots \oplus R /\left(p_{n_{i+1}}^{e_{n_{i+1}}}\right) .
\end{equation*}
Substituting into the invariant factor form gives the desired result. Uniqueness follows from the uniqueness of the invariant factor form and of the prime factorizations of each \(d_{i}\text{.}\)
Theorem14.32.Sunzi’s Remainder Theorem (Rings).
Suppose \(I\) and \(J\) are ideals in a commutative ring \(R\) such that \(I + J = R\text{.}\) Then \(I \cap J = I \cdot J\) (where \(I \cdot J\) is defined as the set of all sums of products of the form \(ab\) with \(a \in I\) and \(b \in J\)) and there is an isomorphism of \(R\)-modules
Note that \(I \cdot J \subseteq I \cap J\) holds in general for any pair of ideals. If \(I + J = R\) then \(a + b = 1\) for some \(a \in I\) and \(b \in J\text{.}\) Given \(x \in I \cap J\) we have \(xa + xb = x\) with \(xa \in I \cdot J\) and \(xb \in I \cdot J\text{,}\) which proves \(x \in I \cdot J\text{.}\)
For the second assertion define \(g: R \to R/I \oplus R/J\) to be the \(R\)-module homomorphism \(g(r) = (r +I, r + J)\text{.}\) Note that the kernel of \(g\) is \(I \cap J\) which equals \(I\cdot J\) by the first assertion. I claim \(g\) is onto: Pick \((x + I, y +J) \in R/I \oplus R/J\text{.}\) With \(a,b\) chosen as above, we have
\begin{equation*}
g(xb + ya) = (xb + ya + I, xb + ya + J) = (xb + I, ya + J) = (x + I, y +J).
\end{equation*}
The last equation holds since \(xa + xb = x\) and thus \(xb + I = xb + xa + I = x + I\) and similarly \(ya + J = y + J\text{.}\)
For the final assertion, recall that when \(R\) is a UFD, two principle ideals \(Rx\) and \(Ry\) satisfy \(Rx + Ry = R\) if and only if \(x\) and \(y\) are relatively prime. Also, for \(I = Rx\) and \(J = Ry\text{,}\) we have \(I\cdot J = R(xy)\text{.}\)
Definition14.33.Elementary Divisor (Module).
With the notation in Theorem 14.31 above, the elements \(p_1^{e_1},\ldots, p_s^{e_s}\) of \(R\) are the elementary divisors of \(M\text{.}\)
Example14.34.\(\Z/90\).
Since \(\Z/90 \cong \Z/9 \oplus \Z/2 \oplus \Z/5\text{,}\) so the elementary divisors of \(\Z/90\) are \(9\text{,}\)\(2\) and \(3\text{.}\) The only invariant factor of \(\Z/90\) is \(90\text{.}\)
Example14.35.Direct Sums and \(\Z\).
Consider the group \(G = \Z/90 \oplus \Z/9 \oplus \Z/6 \oplus \Z/3 \oplus \Z/4\text{.}\) This is neither in IFF nor in EDF. Applying the Sunzi Remainder Theorem, we have
and this gives the EDF. The elementary divisors of \(G\) are \(9,2,5,9,2,3,3,4\) (ordering does not matter).
To find the IFF we start by finding the largest prime power order for each prime in the list of orders of the summands. These are \(9, 5, 4\text{.}\) The CRT gives
