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    The structure theorem over a PID

    Let \(R\) be a principal ideal domain and \(M\) a finitely generated \(R\)-module. Then \(M\cong R^r\oplus R/(d_1)\oplus\cdots\oplus R/(d_s)\) for nonzero nonunits \(d_1\mid d_2\mid\cdots\mid d_s\) (invariant factors), and \(M\cong R^r\oplus\bigoplus_iR/(p_i^{e_i})\) for primes \(p_i\) (elementary divisors); the rank \(r\), the ideals \((d_j)\), and the multiset of ideals \((p_i^{e_i})\) are determined by \(M\). The Smith normal form of a presentation matrix of \(M\) computes the invariant factors. For \(R=\ZZ\) the theorem classifies finitely generated abelian groups. For \(R=F[x]\) with \(F\) a field, a finite-dimensional \(F\)-vector space with an operator \(T\) is an \(F[x]\)-module on which \(x\) acts as \(T\), and the theorem gives the canonical forms of \(T\).

    Structure theorem for finitely generated modules over a PID

    Finitely generated PID-modules split uniquely into a free part and cyclic torsion factors.Suggest a slogan

    Let \(R\) be a principal ideal domain and \(M\) a finitely generated \(R\)-module. Then there are an integer \(r\geq0\), nonzero nonunits \(r_1 \divides r_2 \divides \cdots\divides r_m\) of \(R\), and primes \(p_1,\ldots,p_s\) of \(R\), not necessarily distinct, with exponents \(e_i\geq1\), such that \[ M \cong R^r\oplus \bigoplus_{i=1}^m R/(r_i) \qquad\text{and}\qquad M \cong R^r\oplus \bigoplus_{i=1}^s R/(p_i^{e_i}). \] The first is the invariant factor decomposition and the second the elementary divisor decomposition. The rank \(r\) is unique, the \(r_i\) are unique up to units, and the \(p_i^{e_i}\) are unique up to units and reordering.

    Invariant factor decomposition

    Let \(G\) be a finitely generated abelian group, and for \(n\geq 1\) let \(C_n\) denote the cyclic group of order \(n\). An invariant factor decomposition of \(G\) is an isomorphism \[ G \cong \ZZ^r \times \prod_{k=1}^m C_{n_k} \] with \(r,m\geq 0\) and integers \(n_1,\ldots,n_m\geq 2\) such that \(n_1 \divides n_2\divides \cdots \divides n_m\). The integers \(n_1,\ldots,n_m\) are the invariant factors of \(G\).

    Every finitely generated abelian group \(G\) has an invariant factor decomposition, the direct product of the free abelian group \(\ZZ^r\) and finitely many finite cyclic groups. The integers \(r\) and \(m\) and the invariant factors \(n_1,\ldots,n_m\) are uniquely determined by \(G\).

    Elementary divisor decomposition

    Let \(G\) be a finitely generated abelian group, and for \(n\geq 1\) let \(C_n\) denote the cyclic group of order \(n\). An elementary divisor decomposition of \(G\) is an isomorphism \[ G \cong \ZZ^r \times \prod_{k=1}^m C_{p_k^{e_k}} \] with \(r,m\geq 0\), primes \(p_1,\ldots,p_m\) that need not be distinct, and exponents \(e_1,\ldots,e_m\geq 1\). The prime powers \(p_1^{e_1},\ldots,p_m^{e_m}\) are the elementary divisors of \(G\).

    Every finitely generated abelian group \(G\) has an elementary divisor decomposition. The integer \(r\) and the list \(p_1^{e_1},\ldots,p_m^{e_m}\), up to reordering, are uniquely determined by \(G\).

    Invariant factors of a module over a PID

    Let \(R\) be a principal ideal domain and \(M\) a finitely generated \(R\)-module. By the structure theorem, there are an integer \(r\geq 0\) and nonzero nonunits \(r_1,\ldots,r_m\in R\) with \[ M\cong R^r\oplus\bigoplus_{i=1}^m R/(r_i) \qquad\text{and}\qquad r_1 \divides r_2 \divides \cdots \divides r_m, \] and \(r_1,\ldots,r_m\) are unique up to units. The elements \(r_1,\ldots,r_m\) are the invariant factors of \(M\).

    What divisibility condition do the invariant factors satisfy?

    Elementary divisors of a finite abelian group

    Let \(G\) be a finite abelian group. By the elementary divisor decomposition, there are primes \(p_1,\ldots,p_m\), not necessarily distinct, and integers \(\alpha_1,\ldots,\alpha_m\geq 1\) with \[ G \cong \bigoplus_{i=1}^m \ZZ/p_i^{\alpha_i} \ZZ, \] and the list \(p_1^{\alpha_1},\ldots,p_m^{\alpha_m}\) is unique up to order. These prime powers are the elementary divisors of \(G\).

    What are the elementary divisors of a finite abelian group?

    Converting between elementary divisors and invariant factors

    Elementary divisors split prime powers; invariant factors recombine matching prime-power columns.Suggest a slogan

    Let \(G\) be a finite abelian group with \(G\cong\bigoplus_{i=1}^s \ZZ/m_i\ZZ\) for integers \(m_i\geq 2\), not necessarily distinct.

    1. The elementary divisors of \(G\) are the prime powers \(p^e\), \(e\geq1\), occurring in the factorizations \(m_i=\prod_p p^{e_{p,i}}\), listed with multiplicity over all \(i\); that is, \(G\cong\bigoplus_{i}\bigoplus_{p} \ZZ/p^{e_{p,i}}\ZZ\).

    2. For each prime \(p\), list the exponents of the \(p\)-power elementary divisors in decreasing order, \(e_{p,1}\geq e_{p,2}\geq\cdots\), padding with zeros. Put \(d_j\coloneqq\prod_p p^{e_{p,j}}\) and let \(t\) be the largest \(j\) with \(d_j\neq 1\). Then \(d_t\divides d_{t-1}\divides\cdots\divides d_1\) and \[ G\cong \ZZ/d_t\ZZ\oplus\cdots\oplus\ZZ/d_1\ZZ \] is the invariant factor decomposition of \(G\).

    By the Chinese remainder theorem, \(\ZZ/m\ZZ\cong\bigoplus_p\ZZ/p^{e_p}\ZZ\) for \(m=\prod_p p^{e_p}\), which gives (a). For (b), the same theorem gives \(\ZZ/d_j\ZZ\cong\bigoplus_p\ZZ/p^{e_{p,j}}\ZZ\), so \(\bigoplus_j\ZZ/d_j\ZZ\) has the same elementary divisors as \(G\) and is isomorphic to \(G\). Since \(e_{p,j+1}\leq e_{p,j}\) for every \(p\), \(d_{j+1}\divides d_j\).

    Let \(G=\ZZ/12\ZZ\oplus\ZZ/18\ZZ\oplus\ZZ/5\ZZ\). Since \(12=2^2\cdot3\) and \(18=2\cdot3^2\), the elementary divisors are \(2^2, 2, 3^2, 3, 5\). The largest power of each prime gives \(d_1=2^2\cdot3^2\cdot5=180\), and the remaining powers give \(d_2=2\cdot3=6\). Hence \(G\cong\ZZ/6\ZZ\oplus\ZZ/180\ZZ\), with \(6\divides180\).

    The characteristic polynomial is the product of the invariant factors

    Multiply the invariant factors to recover the characteristic polynomial.Suggest a slogan

    Let \(k\) be a field, let \(A\in\Mat_n(k)\), and let \(f_1\divides f_2\divides\cdots\divides f_n\) be the invariant factors of \(A\): the monic diagonal entries of the Smith normal form of \(xI-A\) over \(k[x]\), some of which may equal \(1\). Then \[ \det(xI-A)=\prod_{j=1}^nf_j(x). \]

    The Smith normal form is \(P(xI-A)Q\) with \(P,Q\) invertible over \(k[x]\), so \(\det P\) and \(\det Q\) are nonzero constants and \(\det(xI-A)\) is a nonzero constant multiple of \(\prod_jf_j\). Both polynomials are monic, so they are equal.

    The minimal polynomial is the invariant factor of highest degree

    The largest invariant factor is the minimal polynomial.Suggest a slogan

    Let \(k\) be a field, let \(A\in\Mat_n(k)\), and let \(f_1\divides f_2\divides\cdots\divides f_n\) be the invariant factors of \(A\): the monic diagonal entries of the Smith normal form of \(xI-A\) over \(k[x]\). Then the minimal polynomial of \(A\) is \[ \min_A(x)=f_n(x). \]

    With \(x\) acting as \(A\), the \(k[x]\)-module \(k^n\) is isomorphic to \(\bigoplus_{j=1}^nk[x]/(f_j)\). A polynomial \(g\) satisfies \(g(A)=0\) if and only if \(g\) annihilates this module, that is, \(f_j\divides g\) for every \(j\); since \(f_j\divides f_n\) for all \(j\), this holds if and only if \(f_n\divides g\). So \(f_n\) is the monic generator of the ideal of polynomials vanishing at \(A\).

    Elementary divisors are the minimal polynomials of the Jordan blocks

    Elementary divisors record the sizes and eigenvalues of the Jordan blocks.Suggest a slogan

    Let \(k\) be a field and let \(A\in\Mat_n(k)\) have characteristic polynomial that splits over \(k\), so that \(A\) is similar to its Jordan canonical form \(\JCF(A)\). Then the elementary divisors of \(A\), the prime-power factors in the decomposition of the \(k[x]\)-module \(k^n\) with \(x\) acting as \(A\), are the minimal polynomials \((x-\lambda)^s\) of the Jordan blocks \(J_s(\lambda)\) of \(\JCF(A)\), listed with multiplicity.

    A Jordan block \(J_s(\lambda)\) is the matrix of multiplication by \(x\) on \(k[x]/((x-\lambda)^s)\) in the basis \((x-\lambda)^{s-1},\ldots,(x-\lambda),1\), and its minimal polynomial is \((x-\lambda)^s\). So the block decomposition of \(\JCF(A)\) is a decomposition \(k^n\cong\bigoplus k[x]/((x-\lambda)^s)\), one summand for each block, and the elementary divisors are unique up to order.