# When is an element of a free module over a principal ideal domain contained in a basis?

I'm trying to show the following:

Let $R$ be a principal ideal domain and let $M$ be free $R$-module of rank $n$. Let $Y=\{y_1,\ldots,y_n\}$ be a basis of $M$ and $x\in M$ with $x=y_1a_1+\cdots+y_na_n$. Then $x$ is contained in a base of $M$ if and only if $\gcd(a_1,\ldots,a_n)=1$.

I have shown that if $x$ belongs to a basis of $M$, then $\gcd(a_1,\ldots,a_n)=1$.

This is my idea for the other direction:

If $n=1$, then $a_1\in R^*$, and we are done. Now suppose $n\geq 2$. Since $\gcd(a_1,\ldots,a_n)=1$, there are $s_1,\ldots,s_n\in R$ such that $a_1s_1+\cdots+a_ns_n=1$, then using this equality I tried to get some non-zero $y\in M$ such that $\langle x\rangle\cap \langle y\rangle=0$. Supposing the existence of such $y$, using Zorn's lemma you can get a maximal nonzero submodule $N$ such that $\langle x\rangle\cap N=0$, then I showed that $\langle x\rangle\oplus N=M$, and since $M$ is free and $R$ a principal ideal domain $N$ is free and has a basis $\{x_1,\ldots,x_m\}$, and thus $\{x,x_1,\ldots,x_m\}$ would be a basis of $M$.

My question is: is there such an $y$? Also, I would like to know whether this idea is too complicated, and if so, I want to know alternative approaches to it.

Thanks

• Dear Camilo, There is certainly such a $y$, e.g. $y = 0$, so there is no problem applying your Zorn's lemma argument to get a maximal $N$. The hard part will be to show that $\langle x \range \oplus N$ is actually equal to $M$; this is where you have to use the assumption on the $a_i$. If you have done this (as you say), then you are in good shape! (And there are other ways to prove this result, but your approach is as good as any.) Regards, – Matt E May 14 '13 at 5:14
• Hello Matt I just edited the question to clarify that $y\neq 0$, I do need this assumption to make the sum direct. Thanks for your advice! – Camilo Arosemena-Serrato May 14 '13 at 5:26
• Dear Camilo, What if $n = 1$, so that $x = a_1 y_1$ for some unit $a_1$? Then necessarily $N = 0$ (and there is no non-zero $y$). So (while admittedly knowing little about it) I'm a bit suspicious of your argument if it can't allow for the possibility that $N = 0$. At the very least, I would check it carefully. Regards, – Matt E May 14 '13 at 6:59
• again, sorry for the mistake, I wanna get such $y\neq 0$, if the dimension of $M$ is $\geq 2$, I corrected the question. – Camilo Arosemena-Serrato May 14 '13 at 12:47
• I suppose that the $Y$ you start with is a basis of $M$. Since $M$ is finitely generated, I think that what you are trying to prove follows more or less immediately from the following (which should be true :) ): an element $x\in M$ belongs to some basis of $M$ if and only if the quotient module $M/Rx$ is torsion-free. – Andrea Mori May 14 '13 at 13:11

Let $R$ be a PID $M$ a free $R$-module of finite rank $n$. Let $\{x_1,\dots,x_k\}$ be $k\leq n$ elements in $M$, and denote $N$ their $R$-span in $M$. Then the claim is that the $k$-ple $\{x_1,\dots,x_k\}$ can be completed to a basis of $M$ over $R$ if and only if $$rk(N)=k\qquad\text{and}\qquad\text{M/N is torsion-free.}$$ Indeed, if $\{x_1,\dots,x_k,x_{k+1},\dots,x_n\}$ is an $R$-basis for $M$, then $M=N\oplus N^\prime$ where $N^\prime$ is the submodule of $M$ generated by$\{x_{k+1},\dots,x_n\}$ and $M/N\simeq N^\prime$. Thus $rk(N)=k$ because $rk(N^\prime)\leq n-k$ and $rk(N)+rk(N^\prime)=n$, and $M/N$ is free because it is isomorphic to a submodule of a free module.

Viceversa, consider the quotient map $$\pi:M\longrightarrow M/N.$$ Let $\{z_1,\dots,z_r\}$ be a basis of $M/N$. Choose $y_i\in M$ such that $\pi(y_i)=z_i$ and let $N^\prime\subset M$ the submodule generated by the $y_i$'s. Since the $z_i$ are $R$-linearly independent so are the $y_i$, i.e. they form a basis of $N^\prime$. For any $m\in M$ write $$\pi(m)=a_1z_1+\cdots+a_rz_r,\qquad a_i\in R.$$ Then $m-\sum_{i=1}^ra_iy_i\in\ker(\pi)=N$, from which follows readily that $M=N\oplus N^\prime$ and thus $\{x_1,\dots,x_k,y_1,\dots,y_r\}$ is a basis for $M$ containing $\{x_1,\dots,x_k\}$. The claim is proved.

Now, let $x\in M$, $x\neq0$, and let $\{y_1,..,y_n\}$ be a basis of $M$. Then $x=a_1y_1+\dots+a_ny_n$ for some $a_i\in R$. Let $d=\operatorname{gcd}(a_1,\dots,a_n)$.

If $d\neq1$, the element $y=\frac1dx\in M$ gives a torsion element in $M/Rx$. On the other hand, suppose that $d=1$ and $z\in M/Rx$ is such that $rz=0$ for some non-zero $r\in R$. Choose $y\in M$ such that $\pi(y)=z$ and write $y=\sum_{i=1}^nb_iy_i$. Then $ry=sx$ for some $s\in R$, i.e. $$rb_i=sa_i,\qquad\text{for all i=1,\dots,n}$$ Thus $r\operatorname{gcd}(b_i)=s$, i.e. $r$ divides $s$ in $R$. Therefore $y=\frac sr x\in Rx$ and $\pi(y)=z=0$, thus proving that $M/Rx$ is torsion-free.

Let $R$ be a PID and $M$ a free $R$-module of any rank, finite or infinite. Then $x\in M$ belongs to some basis of $M$ iff there exists a linear functional $g\colon M\to R$ such that $g(x)=1$. Moreover, let $(e_t\mid t\in T)$ be a basis of $M$, and $x=\sum_{t\in T}\xi_t e_t$ (where $\xi_t\neq 0$ for only finitely many $t\in T$); then $x$ belongs to some basis of $M$ iff $\gcd\{\xi_t\mid t\in T\}=1$.

Proof. Necessity. Suppose there is a basis $(e_t\mid t\in T)$ of $M$ such that $x=e_s$ for some $s\in T$. For each $t\in T$ choose an $\alpha_t\in R$, taking care to choose $\alpha_s=1$. There exists a unique linear functional $g$ on $M$ such that $g(e_t)=\alpha_t$ for every $t\in T$. Since $g(x)=g(e_s)=1$, we have a linear functional $g$ as required.

Sufficiency. Let $g$ be a linear functional on $M$ so that $g(x)=1$. We claim that $M=Rx\oplus\ker(g)$. If $y\in M$, then $g(y-g(y)x)=0$, thus $y-g(y)x\in\ker(g)$; this shows that $M=Rx+\ker(g)$. If $\xi x+z=0$ for some $\xi\in R$ and $z\in\ker(g)$, then $\xi=g(\xi x+z)=0$, thus also $z=0$; this shows that the sum $Rx+\ker(g)$ is direct. Now $\ker(g)$, as a submodule of a free module over a PID, is free, so we can extend $x$ to a basis of $M$ by any basis of $\ker(g)$.

The reformulation in terms of the coordinates $\xi_t$ of $x$ with respect to a basis $(e_t\mid t\in T)$.

Every linear functional $g$ on $M$ is uniquely determined by the values $\alpha_t=g(e_t)\in R$, $t\in T$, which can be chosen arbitrarily. The value of $g$ on $x$ is $g(x)=\sum_{t\in T}\alpha_t\xi_t$ (a finite linear combination). Since every linear combination of $\xi_t$'s is divisible by $d=\gcd\{\xi_t\mid t\in T\}$, which is itself a linear combination of $\xi_t$'s (because $R$ is a PID), it is clear that there exists a linear functional $g$ such that $g(x)=1$ iff $d=1$. Done.

You may enjoy proving the following generalization.

Let $R$ and $M$ be as above, and $x_1,\ldots,x_n\in M$. Then $(x_1,\ldots,x_n)$ can be extended to a basis of $M$ iff there exists a linear map $h\colon M\to R^n$ such that $h(x_1)=e_1$, $\ldots$, $h(x_n)=e_n$, where $(e_1,\ldots,e_n)$ is the standard basis of $R^n$. Moreover, let $(u_t\mid t\in T)$ be a basis of $M$, and $x_i=\sum_{t\in T}\xi_{it} u_t$ for $1\leq i\leq n$ and $t\in T$. Then $(x_1,\ldots,x_n)$ can be extended to a basis of $M$ iff the $\gcd$ of the determinants of all $n\!\times\! n$ submatrices of the $\{1,\ldots,n\}\!\times\! T$-matrix $[\xi_{it}]$ is $1$. (Since there exists a finite subset $S$ of $T$ such that $\xi_{it}=0$ for $1\leq i\leq n$ and $t\in T\setminus S$, the condition in fact involves only finitely many $n\!\times\! n$ matrices.)

I like to prove things like this as algorithmically as possible, using matrices and row and column operations that are as elementary as possible.

I construct an invertible matrix $A$ such that $A \matrix(a_1,\ldots,a_n)^T$ = $(1,0,\ldots,0)^T$. Let $B$ be the linear map corresponding to $A$ relative to the basis $\{y_1,\ldots,y_n\}$. Then the desired basis is $\{x = B^{-1}y_1,\ldots,B^{-1}y_n\}$.

If $n = 1$, take $A = \matrix(1/a_1)$.

If $n = 2$, let $c = \gcd(a_1,a_2) = 1$ and choose $s_1$ and $s_2$ such that $s_1 a_1 + s_2 a_2 = c$. Take $$A = \begin{bmatrix} s_1 \ \ \ \ s_2 \\ -a_2/c \ \ a_1/c \end{bmatrix}.$$ Then $A \matrix(a_1,a_2)^T = \matrix(c,0)^T$. In this step, $c = 1$, but we write it as $c$ for use in the induction step. $A$ is invertible because its determinant is $1$.

If $n > 2$, use induction on the number of nonzero entries in $\matrix(a_1,\ldots,a_n)$.

If there is only $1$ nonzero entry, then it is a $\gcd$ of the entries and in fact $1$ by the choice made in the previous step. Apply the most elementary row operation of interchanging rows to move the nonzero entry to the first entry. Row interchange operations expressed in matrix form correspond to multiplication on the left of the matrix $A$ constructed in previous stages of the induction. The product remains invertible.

If there are $2$ or more nonzero entries, first interchange rows so that $a_1$ and $a_2$ are nonzero (this will actually make the row interchange in the above step never occur). Apply the $n = 2$ step to the first $2$ coordinates (more precisely, to the first $2$ rows) (augment the $2 \times 2$ matrix with a $(n-2) \times (n-2)$ identity matrix). Now $c = \gcd(a_1,a_2)$ is not necessarily a unit, but $A$ is invertible because we divided its second row by $c$ to make its determinant $1$. Most importantly, reducing $\matrix(a_1,a_2,a_3,\ldots)^T$ to $\matrix(c,0,a_3,\ldots)^T$ leaves the $gcd$ of the entries unchanged, so the induction works.

All the matrices constructed correspond to not-always-elementary row operations when they are used to multiply column vectors on the left. Some are non-elementary since they use a linear combination of rows while for pure elementary row operations only the special linear combination $unit * row_1 + r_2 * row_2$ is permitted. If $R$ is Euclidean, then the Euclidean algorithm essentially gives a construction of the $2 \times 2$ $A$ as a product of matrices for elementary row operations, so the final $A$ is also such a product. I don't know of any practical algorithms for constructing the $2 \times 2$ $A$ for PIDs that are not Euclidean.

This is the dual of a special case of the algorithm for reduction to Hermite normal form. In the general case of it, we start with an arbitrary matrix instead of a column vector, and do reversible (left) row operations on it to get a triangular matrix (with some complications for the matrix not being square). Hermite normal form is what you can naturally get by doing reversible (right) column operations instead. There is also Smith normal form. It is what you can naturally get by doing both reversible row operations and reversible column operations -- a diagonal matrix. All of the elementary theory of modules over PIDs can be read off from these matrices.