Modules over a Principal Ideal Domain and the Canonical Forms — Examples

If the matrix is empty or zero, return the empty or zero Smith form. Otherwise move a nonzero entry of least Euclidean value among the current entries into the upper-left pivot position by row and column swaps.

Divide every entry in the pivot row and column by the pivot and use elementary operations to replace a nonzero remainder into the pivot position. If an entry outside that row and column is not divisible by the pivot, add its row to the pivot row and repeat. Each nonzero remainder has strictly smaller Euclidean value, so this descent terminates with a pivot dividing every entry.

Clear the rest of the pivot row and column using exact division, then apply the same procedure to the smaller remaining rectangle. The row or column count strictly decreases at each recursion, and zero or rank-deficient tails stop immediately.

Because each chosen pivot divides every entry in the remaining rectangle, successive nonzero pivots satisfy $d_1\mid d_2\mid \cdots$. Unit normalization gives Smith form; equivalently, adjacent diagonal factors may be replaced by their gcd and lcm through elementary operations without changing equivalence. The output agrees with the existence theorem [L1], but termination here uses the Euclidean function and is not a proof over an arbitrary PID.

Swap the first two columns, replace the second row by $R_2-4R_1$ and the third by $R_3-2R_1$, and replace the second column by $C_2-2C_1$. The remaining lower block is $\left(\begin{array}{cc}-4& 6\\ -2& 6\end{array}\right)$. Swap its rows, replace the last row by the last row minus twice the preceding row, and clear the off-diagonal $6$ by adding three times the pivot column. Unit row scalings give $\mathrm{diag}(1,2,6)$.

The gcd of all entries is $1$; the minor from rows $1,3$ and columns $1,2$ is $2$, and every two-by-two minor is even, so their gcd is $2$; the determinant is $12$. Thus the determinantal divisors are $1,2,12$, whose successive quotients are $1,2,6$, confirming step 1.1 by [L1].

Quotienting the diagonal presentation gives $\mathbb{Z}\mathrm{/}(1)\oplus \mathbb{Z}\mathrm{/}(2)\oplus \mathbb{Z}\mathrm{/}(6)$; the first summand is zero, yielding the displayed cokernel with no free part because the determinant is nonzero.

The exponent $3$ of $2$ has partitions $3$, $2+1$, and $1+1+1$; the exponent $2$ of $3$ has partitions $2$ and $1+1$; the exponent $1$ of $5$ has partition $1$. By [L1], independently combining these choices gives exactly the elementary-divisor rows displayed.

Align prime-power factors on the right and multiply columns. The resulting columns are respectively $(360)$, $(3,120)$, $(2,180)$, $(6,60)$, $(2,2,90)$, and $(2,6,30)$, giving the displayed invariant forms by [L1].

In every row the product of the invariant factors is $360$, and each factor divides the next. Factoring those factors back into prime powers reproduces its elementary-divisor row, so no regrouping is duplicated.

The relation matrix is $\mathrm{diag}(2,3)$. Since $2$ and $3$ are coprime, elementary integer row and column operations using $3-2=1$ transform it to $\mathrm{diag}(1,6)$, which is its Smith form.

Directly, the quotient is $\mathbb{Z}\mathrm{/}2\oplus \mathbb{Z}\mathrm{/}3$, and [L1] identifies this with $\mathbb{Z}\mathrm{/}6$; the unit Smith factor in step 1.1 contributes no summand.

The class of $(1,1)$ has order the least common multiple of $2$ and $3$, namely $6$, so it generates the order-six quotient. Its lower positive multiples are nonzero because neither coordinate has simultaneously reached its relation.

The lower three-by-three block is the companion matrix of $x^3-x^2+x-1=(x-1)(x^2+1)$, while the first block is $C(x-1)=(1)$. Since $x-1$ divides the cubic factor, [L1] gives the displayed polynomial-module summands and confirms that $A$ is in rational canonical form with the stated invariant factors.

The largest invariant factor in step 1.1 is $(x-1)(x^2+1)$, so the minimal-polynomial dictionary gives the displayed ${\mu }_A$.

The product of the two invariant factors is $(x-1{)}^2(x^2+1)$, so the characteristic-polynomial dictionary gives the displayed ${\chi }_A$; their degrees $1$ and $3$ sum to the matrix dimension.

Over $\mathbb{Q}$, [L1] gives the two rational companion blocks and the divisibility chain, so no splitting field is needed for rational canonical form.

After adjoining $i$, the second invariant factor splits as $(x-1)(x-i)(x+i)$ into pairwise coprime factors, while the first contributes another $x-1$. The elementary divisors are therefore $x-1,x-1,x-i,x+i$.

Each elementary divisor in step 2.1 is linear, so the module Jordan theorem gives four one-dimensional blocks with entries $1,1,i,-i$, and the agreement theorem confirms the same Jordan-string block data. Their product remains $(x-1{)}^2(x^2+1)$ and their least common multiple remains $(x-1)(x^2+1)$, preserving the characteristic and minimal polynomial identities.

With $e_1=(1,0)$ one has $A{e}_1=e_2$ and $A^2{e}_1=-e_1$, so $e_1,A{e}_1$ is a cyclic basis with annihilator $x^2+1$. The displayed matrix is exactly $C(x^2+1)$ and is its one-block rational canonical form.

The polynomial $x^2+1$ has no real root and does not split over $\mathbb{R}$. By [L1], the real operator has no real Jordan form.

After adjoining $i$, the polynomial splits as $(x-i)(x+i)$ with distinct roots. The module Jordan theorem gives the two one-dimensional blocks $i$ and $-i$, whose dimensions sum to two.

For $e_1=(1,0)$, one has $T{e}_1=(0,1)=e_2$, so $e_1,T{e}_1$ is a basis of ${\mathbb{Q}}^2$.

Direct multiplication gives $T^2{e}_1=-2e_1$, and no nonzero polynomial of degree below two kills $e_1$ because the vectors in step 1.1 are independent. Thus the vector annihilator is $x^2+2$, and [L1] gives its displayed companion matrix.

The cyclic map $\mathbb{Q}[x]\to V_T$, $p\mapsto p(T)e_1$, has kernel $(x^2+2)$ and is surjective, so it gives the displayed module isomorphism. Consequently the sole invariant factor is $x^2+2$; it is the minimal polynomial, and the companion determinant gives the same characteristic polynomial.

Let $A=J_2(0)\oplus J_2(0)$ and $B=J_2(0)\oplus J_1(0)\oplus J_1(0)$. Both are four-dimensional nilpotent matrices with characteristic polynomial $x^4$ and largest block size two, hence minimal polynomial $x^2$.

The invariant factors of $A$ are $x^2,x^2$, while those of $B$ are $x,x,x^2$; equivalently, their ranks are $2$ and $1$. By [L1] they are not similar, despite the equal characteristic and minimal polynomials.

Let $R=\mathbb{Z}[x]$ and $I=(2,x)$. The ring is a domain, and $I$ is generated by the displayed elements, so it is a two-generated torsion-free $R$-module.

Any nonzero cyclic submodule of the torsion-free ideal is isomorphic to $R$. If a direct-sum decomposition contained two nonzero cyclic summands with nonzero generators $a,b\in I$, the relation $b\cdot a-a\cdot b=0$ would be a nontrivial relation between them, contradicting directness. Thus a cyclic decomposition could have at most one nonzero summand.

A single nonzero cyclic summand would make $I$ principal by [F1]. But a generator would divide both $2$ and $x$ in $\mathbb{Z}[x]$, hence would be a unit; that would give $I=R$, while reduction modulo $(2,x)$ shows $1\mathrm{\notin }I$. Thus $I$ is not cyclic and has no direct-sum decomposition into cyclic modules, refuting the claim and isolating the PID hypothesis.

Under the usual integer action, $\mathbb{Q}$ is a $\mathbb{Z}$-module. If $nq=0$ with $n\mathrm{\ne }0$, field cancellation gives $q=0$, so it is torsion-free by [F1].

Suppose, for contradiction, that $\mathbb{Q}$ has a $\mathbb{Z}$-basis $B$. It cannot be empty because $\mathbb{Q}\mathrm{\ne }0$, so choose $b\in B$. Express $b\mathrm{/}2$ as a finite integer linear combination of basis elements and multiply by $2$. Uniqueness of basis coordinates would make the coefficient of $b$ simultaneously $1$ and an even integer, which is impossible.

Step 2.1 rules out every nonempty basis, and the empty basis cannot span the nonzero module. Thus $\mathbb{Q}$ is torsion-free over the PID $\mathbb{Z}$ but is not free; finite generation is essential.

Let $R$ be any PID. The free modules $R$ and $R^2$ are torsion-free, so both have empty elementary-divisor lists. The zero module $R^0$ has the same empty torsion data as well.

By [L1], their free ranks are respectively $1$, $2$, and $0$ and are genuinely distinct. Thus identical elementary-divisor data does not determine the free rank.