Say you're given a ordered set of $n$ relatively prime elements, $a_1,\dots,a_n$ in a principal ideal domain $D$. If I relabel these elements $a_{11},\dots,a_{1n}$ in the same order, is it possible to find some remaining $a_{kj}$ in $D$ such that $(a_{kj})$ is an invertible $n\times n$ matrix over $D$?

  • $\begingroup$ In a more general setting, this became known as Serre's problem, and later the Quillen-Suslin Theorem. $\endgroup$ – Gerry Myerson Oct 28 '11 at 22:23
  • $\begingroup$ @Gerry: Can you indicate how this is related to Quillen-Suslin? $\endgroup$ – Soarer Oct 29 '11 at 4:38
  • $\begingroup$ @Soarer, sorry, it's a bit out of my depth. Some statements of the problem/theorem don't look anything like Aria's qiestion, but if you type Quillen Suslin matrix into a search engine you will probably find some references that bring out the relation. $\endgroup$ – Gerry Myerson Oct 29 '11 at 8:17
  • $\begingroup$ @Soarer, Here's a reference: Moshe Roitman, Completing unimodular rows to invertible matrices, J. Algebra 49 (1977), no. 1, 206–211, MR0453779 (56 #12033). From the review: Quillen's solution of Serre's problem implies that if $R$ is a commutative ring, then any unimodular row over $R[x]$ which contains a unitary polynomial is completable to an invertible matrix. The author shows.... Other related results on this completion problem are discussed. $\endgroup$ – Gerry Myerson Oct 31 '11 at 6:11

We can prove this by induction. For $n=1$, the set $\{a_1\}$ is relatively prime if and only if $a_1$ is a unit, and thus $(a_{11})$ is invertible.

So assume that we can extend any set of $n$ relatively prime elements to an invertible matrix, and let $n+1$ relatively prime elements $a_1,\dotsc,a_{n+1}$ be given. Let $g=\gcd(a_1,\dotsc,a_n)$. Then we can write $a_i=gb_i$ for $1\le i\le n$, with the $b_i$ relatively prime, and we can extend the $b_i$ to an invertible $n\times n$ matrix $B$. If we multiply the first row of $B$ by $g$, we can place it into the upper left $n\times n$ block of the matrix $A$ to be constructed, and we'll have $a_i=gb_i$ in the right place for $1\le i\le n$. Now place a multiple of the first row of $B$ into the $n+1$-th row of $A$, with the multiplier $r$ yet to be determined, and fill the $n+1$-th column with zeros except for the first and last elements $a_{n+1}$ and $s$, with $s$ yet to be determined. Now the determinant of the upper left $n\times n$ block is $g\det B$, and the determinant of the lower left $n\times n$ block is $(-1)^{n+1}r\det B$, so by Laplace expansion the determinant of the entire matrix $A$ is

$$\det A=sg\det B+ a_{n+1}r\det B=(sg+ ra_{n+1})\det B\;.$$

Since $a_1,\dotsc,a_{n+1}$ are relatively prime, $g$ and $a_{n+1}$ are relatively prime. Thus we can choose $r$ and $s$ such that the expression in parentheses is $1$ and $\det A = \det B$. But a matrix over a commutative ring is invertible if and only if its determinant is invertible (see Do these matrix rings have non-zero elements that are neither units nor zero divisors? and necessary and sufficient condition for trivial kernel of a matrix over a commutative ring). Thus, since $B$ is invertible, $A$ is invertible.

  • $\begingroup$ In the formula for $\det A$, the two plus signs should be minus signs, which is easily seen for the case where $A$ is a $2 \times 2$ matrix. For the general case, $\det A = (-1)^{n + 1 + n + 1}sg\det B + (-1)^{1 + n + 1}a_{n + 1}(-1)^{n + 1}r \det B = (sg - ra_{n + 1})\det B$. Of course, the signs of $sg$ and $ra_{n + 1}$ do not affect the validity of the proof. $\endgroup$ – Maurice P Dec 16 '16 at 17:20

This follows from the structure theorem of finitely generated modules over a principal ideal domain.

Consider your $n$-tuple $(a_1,\dots,a_n)$ as an element $a$ of the module $D^n$ over $D$, and form the quotient module $Q=D/\langle a\rangle$. It is clearly a finitely generated module over $D$ (since $D^n$ is), and the hypothesis $\gcd(a_1,\dots,a_n)=1$ means that the quotient $Q$ is torsion-free, since the order $d\in D$ (generator of the annihilator) of any torsion element would be is a common divisor of $a_1,\dots,a_n$. By the structure theorem, $Q$ is therefore a free module over $D$. If $[\overline b_1,\ldots,\overline b_k]$ is a basis for $Q$, consisting of images of elements $b_i\in D^n$, then $[a,b_1,\ldots,b_k]$ forms a basis of $D^n$. Since the rank of a free module is well defined one has $k=n-1$, and the $b_i$ are a possible choice for the remaining rows of your matrix.


There is much written on this and related problems, e.g. it is the special case $\rm\:k=1\:$ of below.

MR 82k:15013
Gustafson, William H.; Moore, Marion E.; Reiner, Irving
Matrix completions over Dedekind rings.
Linear and Multilinear Algebra 10 (1981), no. 2, 141--144.

Let d be any element of the ideal generated by the k by k minors of a k by n matrix M over a Dedekind domain R . The authors show that M can be "completed" to form the top k rows of some invertible n by n matrix of determinant d .

In fact, their proof works if R is any Prufer domain whose finitely generated ideals can be generated by $\ 1\ 1/2\ $ elements; merely precede their lemma by the proof of the formula $\rm\ A \oplus B \cong R \oplus AB\ $ given in Theorem 4.1 of R. C. Heitmann and the reviewer [Rocky Mountain J. Math. 5 (1975), 361--373; MR 52 #3141].

Reviewed by Lawrence S. Levy 91k:15028


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.