# R is a PID , M a free R module, R-basis

We are a group of Finnish second semester undergrad students who are trying to solve old exercises from this linear algebra class for the benefit of our whole semester and even all prospective students (aka TeXing the solutions for future generations and putting them on our Dropbox). So far so good, this one exercise really leaves us clueless (probably due to non of us quite understanding how to handle modules) and we hope somebody who knows "non linear" Algebra can teach us by solving this with explainations so we don't lose more time just on this one exercise ( we spent heartfelt 2-3 hours already trying to figure something, but we haven't got anything substantial which isn't just a mere guess) :

This is the "evil" exercise:

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a) Let $R$ be a PID, $M$ a free $R$ module and let $S$ be a $R$-submodule with $\{0\} \ne S \subset M$. Let $(x_{1},\dots,x_{n})$ be a $R$-basis of $M$ and let $t_{1},\ldots,t_{m}\in R$, with $t_{1}|\cdots|t_{m} \ne 0$ so that $(t_{1}x_{1},\cdots,t_{m}x_{m})$ is a $R$-basis of $S$. If we now let $$F=\bigl\{f_{1}(y_{1})+\cdots+f_{k}(y_{k});k\ge 1, f_{1},...,f_{k}\in \mathrm{Hom}(M,R), \ y_{1},\ldots,y_{k}\in S\bigr\}.$$ Then we can show that $F=t_{1}R$

b) Let more concretely be $R=\mathbf{Z}$, $M= \mathbf{Z}^{3}$, $S= \mathbf{Z}a_{1}+\mathbf{Z}a_{2}+\mathbf{Z}a_{3}$. Putting $a_{1}=(1,2,3)$, $a_{2}=(4,5,6)$, $a_{3}=(7,8,9)$. Compute $t_{1}>0, \ldots , t_{m} > 0$ in (a)

c) Show that $M/S$ is isomorphic to $(\mathbf{Z}/3\mathbf{Z}) \times \mathbf{Z}$

When we asked our prof where he takes the exercises from, he told us he takes them from a book called "Algebra" by Artin. We couldn't find it in the book itself, so excuse us for not giving a direct reference. Our professor puts $30$% of pure Algebra stuff in the exercises and it is obvious that they are his favourite exercises (so if you want a good grade you need to know this material). We sincerely hope that a seasoned algebraist takes pity/empathy on us and excuse ourselves for the English mistakes and our mathematical incapability.

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Is there supposed to be a question in a)? –  Brandon Carter Dec 15 '11 at 17:24
Yes. You were 30 seconds too fast for us :-). –  user21302 Dec 15 '11 at 17:33
(a) Doesn't make sense as written. If $f_1(y_1)+\cdots+f_k(y_k)$ is an element of the set of homomorphisms from $M$ to $R$, then what are $f_i$? What does that mean? I suspect that you mean that $F$ consists of all elements of $R$ of the form $f_1(y_1)+\cdots+f_k(y_k)$ with $f_i\in\mathrm{Hom}(M,R)$ and $y_1,\ldots,y_k\in S$. Is that correct? –  Arturo Magidin Dec 15 '11 at 17:57
Yes, the exercise says exactly what you thought. Thanks also for cleaning up our thread. –  user21302 Dec 15 '11 at 18:23

(a) I'm assuming that $F$ is supposed to be: $$F=\Bigl\{ f_1(y_1)+\cdots + f_k(y_k)\mid f_i\in \mathrm{Hom}(M,R),y_j\in S\Bigr\}.$$

I claim that if $f\in\mathrm{Hom}(M,R)$ and $y\in S$, then $f(y)\in t_1R$.

Indeed, if $y\in Y$, then there exist $a_1,\ldots,a_m\in R$ such that $y = a_1(t_1x_1)+\cdots+a_m(t_mx_m)$. Therefore, $$f(y) = f\bigl( a_1(t_1x_1)+\cdots + a_m(t_mx_m)\bigr) = t_1 f(a_1x_1)+\cdots + t_mf(a_mx_m).$$ Since $t_1|t_k$ for all $k\geq 1$, $t_jf(a_mx_m)\in t_1R$ for all $j$. Thus, $F\subseteq t_1R$.

Conversely, let $t_1a\in t_1R$. Since $M$ is free, there is a homomorphism $f\colon M\to R$ such that $f(x_1)=a$, $f(x_j)=0$ for $j\gt 1$. Then $r(t_1x_1) = t_1a\in F$, so $t_1R\subseteq F$. This proves the equality.

(b) Is an exercise in Smith Normal Form, or on something like Theorem II.1.6 in Hungerford's Algebra. I'm going to do it that way, though one can simplify things in this case (see below).

Let $F$ be the set corresponding to $S$ from part (a); in our case, we consider all maps $f\colon\mathbb{Z}^3\to\mathbb{Z}$, evaluated at all linear combinations of $(1,2,3)$, $(4,5,6)$, and $(7,8,9)$. It is not hard to check that $F$ is always a subgroup of $R$, and so is a subgroup of $\mathbb{Z}$, hence is of the form $m\mathbb{Z}$ for some $m\geq 0$. In this case, $F$ will be $\mathbb{Z}$, since we can show $1\in F$ by a suitable choice of morphisms and elements of $S$: take $f_1\colon\mathbb{Z}^3\to\mathbb{Z}$ be given by $f_1(x,y,z) = x$, and consider $f_1(1,2,3)\in F$. So $x_1 = (1,2,3)$, $t_1=1$.

(In fact, in a situation such as these, where $M$ is given as $R^n$ and $S$ is given by generators in terms of the basis of $M$, it is not hard to verify that $F$ is the ideal of $R$ generated by the coordinates of the generators of $S$; in this case, we would want the ideal generated by $1,2,3,4,5,6,7,8,9$, which is nothing more than $\mathbb{Z}$ itself).

So $F=1\mathbb{Z}$, and $t_1$ will be $1$. Now find an element of $S$ that has a coordinate equal to $t_1$: $(1,2,3)$ will do. Here, $t_1=1$ so there is nothing to do, but if $t_i\gt 0$, then we will want to show that every coordinate of the vector must in fact be divisible by $t_1$ (consider the projection on the $i$th coordinate, which must be an element of $F$ as well).

Now let $S'=\{(x,y,z)\in S\mid x=0\}$ ($x$ because it was the first coordinate that was equal to $t_1$ in our choice above). Then $\langle(1,2,3)\rangle\cap S' = \{(0,0,0)\}$, and $\langle(1,2,3)\rangle+S' = S$. Indeed, if $(x,y,z)\in S$, then $(x,y,z)=x(1,2,3) + \bigl((x,y,z)-x(1,2,3)\bigr)$, and $(x,y,z)-x(1,2,3)\in S'$. So $S=\langle(1,2,3)\rangle\oplus S'$.

Our first basis element will then be $t_1(1,2,3)$, and now we proceed to work recursively on $S'$.

Note that since $S$ is generated by $(1,2,3)$, $(4,5,6)$, and $(7,8,9)$, it is also generated by $(1,2,3)$, $(4,5,6)-4(1,2,3) = (0,-3,-6)$, and $(7,8,9)-7(1,2,3)=(0,-6,-12)$. So $S'$ is generated by $(0,3,6)$ and $(0,6,12)$; in fact, $S'$ is generated by $(0,3,6)$.

Consider now $F(S')$. It is easy to check that $F(S') = 3\mathbb{Z}$ (again, $F(S')$ is the ideal generated by the coordinates of the generators of $S'$, in this case $0$, $3$, $6$, and $12$, so $F(S') = (3)=3\mathbb{Z}$); find an element of $S'$ that has some coordinate equal to $3$: $(0,3,6)$ will do. Note that both coordinates are divisible by $3$, so write $(0,3,6) = 3(0,1,2)$. Then $t_2=3$, and $x_2=(0,1,2)$.

Now let $S'' = \{(0,y,z)\in S'\mid y=0\}$. As above, $\langle (0,3,6)\rangle\cap S''=\{(0,0,0)\}$, $\langle (0,3,6)\rangle+S''=S'$. So we can now work in $S''$.

In fact, $S''=\{(0,0,0)\}$ so we are done. We have $t_1=1$, $t_2=3$, $m=2$, $x_1=(1,2,3)$, $x_2=(0,1,2)$. Now we just need to complete the list of $x_i$ to a basis for $M$; take $x_3 = (0,0,1)$, for example.

(c) Now it is easy: \begin{align*} \frac{M}{S} &\cong \frac{(1,2,3)\mathbb{Z} \oplus (0,1,2)\mathbb{Z}\oplus (0,0,1)\mathbb{Z}}{(1,2,3)\mathbb{Z}\oplus 3(0,1,2)\mathbb{Z}\oplus 0(0,0,1)\mathbb{Z}}\\ &\cong \frac{(1,2,3)\mathbb{Z}}{(1,2,3)\mathbb{Z}} \oplus \frac{(0,1,2)\mathbb{Z}}{3(0,1,2)\mathbb{Z}} \oplus \frac{(0,0,1)\mathbb{Z}}{0(0,0,1)\mathbb{Z}}\\ &\cong \frac{\mathbb{Z}}{\mathbb{Z}} \oplus \frac{\mathbb{Z}}{3\mathbb{Z}} \oplus \frac{\mathbb{Z}}{0\mathbb{Z}}\\ &\cong \frac{\mathbb{Z}}{3\mathbb{Z}} \oplus \mathbb{Z}\cong \frac{\mathbb{Z}}{3\mathbb{Z}}\times\mathbb{Z}. \end{align*}

Now, I did (b) using the algorithmic proof that appears, for example, in Hungerford, and using part (a). In fact, we can do it simpler in this instance by doing some preliminary considerations:

First, simplify the generating set by doing row-reduction on the matrix whose rows are the generators of $S$: $$\left(\begin{array}{ccc} 1 & 2& 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array}\right) \to \left(\begin{array}{rrr} 1 & 2 & 3\\ 0 & -3 & -6\\ 0 & -6 & -12 \end{array}\right) \to \left(\begin{array}{rrr} 1 & 2 & 3\\ 0 & 3 & 6\\ 0 & 0 & 0 \end{array}\right).$$

So $S$ is generated by $(1,2,3)$ and $(0,3,6) = 3(0,1,2)$.

Then notice that we have $(1,2,3)$, $(0,1,2)$, and $(0,0,1)$, which are a basis for $\mathbb{Z}^3$; that $(1,2,3)$ and $3(0,1,2)$ are a basis for $S$; and that setting $t_1=1$ and $t_2=3$ will solve the problem.

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