# Thread: Direct Summand of Free Module over a P.I.D.

1. ## Direct Summand of Free Module over a P.I.D.

Came across this question while reviewing for an exam and I'm pretty stumped. We let $\displaystyle R$ be a Principal Ideal Domain (P.I.D.) and $\displaystyle M$ be a free module of rank 2 with some basis set $\displaystyle \{b_1,b_2\}$. Then $\displaystyle \forall m \in M$ define $\displaystyle Rm$ to be the submodule of $\displaystyle M$ generated by $\displaystyle m$ (this is also a free module since $\displaystyle R$ is a P.I.D.). Let $\displaystyle m=xb_1+yb_2 \in M$ for some $\displaystyle x,y \in R$. For $\displaystyle x=y$, show that $\displaystyle Rm$ is a direct summand of $\displaystyle M$ (i.e there exists some submodule $\displaystyle N$ of $\displaystyle M$ such that $\displaystyle M = Rm \bigoplus N$) iff $\displaystyle x$ is invertible in $\displaystyle R$. Furthermore for any $\displaystyle x,y$, $\displaystyle Rm$ is a direct summand of $\displaystyle M$ iff the ideal $\displaystyle (a,b)=R$. If $\displaystyle R=\mathbb{Z}$, $\displaystyle x=3$, and $\displaystyle y=5$, find some $\displaystyle v \in M$ such that $\displaystyle \mathbb{Z}m \bigoplus \mathbb{Z}v = M$. Any help here would be appreciated thanks.

2. ## Re: Direct Summand of Free Module over a P.I.D.

Let's do the easy part first:

Suppose $x$ is a unit in $R$, so that there is $u \in R$ with $ux = 1$. Let $N = Rb_2$. I claim $M = Rm \oplus Rb_2$.

To see this, first we show that $M = Rm + Rb_2$. So let $a = c_1b_1 + c_2b_2$ be any element of $M$.

Then we have:

$(c_1u)m + (c_2 - c_1)b_2 = c_1u(xb_1 + xb_2) + c_2b_2 - c_1b_1 = c_1(ux)b_1 + c_1(ux)b_2 + c_2b_2 - c_1b_1 = c_1b_1 + c_1b_2 + c_2b_2 - c_1b_2 = c_1b_1 + c_2b_2 = a$

Next, we show that $Rm \cap Rb_2 = \{0\}$. For if $rm = sb_2$ (for $r,s \in R$) then since $\{b_1,b_2\}$ is a basis for $M$, this means: $rxb_1 + rxb_2 = 0b_1 + sb_2$ so that:

$rx = 0$, and since $x$ is a unit, and $R$ is a domain, $r = 0$.

Now suppose we have $M = Rm \oplus N$. Then $N = Rn$, for some $n \in N$.

If $n = yb_1 + zb_2$, then certainly we have:

$c_1b_1 + c_1b_2 \in M$ with: $c_1b_1 + c_1b_2 = a_1m + a_2n = (a_1x + a_2y)b_1 + (a_1x + a_2z)b_2$ so that: $a_2(y - z) = 0$.

Since $Rm \cap Rn = \{0\}$, if $c_1 \neq 0$, then we must have $a_2 = 0$ (or else $y = z$, and thus $Rm \cap Rn \neq \{0\}$).

In particular, $b_1 + b_2 = a_1m = a_1xb_1 + a_1xb_2$ so that $a_1x = 1$, that is: $x$ is a unit.