Results 1 to 2 of 2

Math Help - Direct Summand of Free Module over a P.I.D.

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    11

    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 R be a Principal Ideal Domain (P.I.D.) and M be a free module of rank 2 with some basis set \{b_1,b_2\}. Then \forall m \in M define Rm to be the submodule of M generated by m (this is also a free module since R is a P.I.D.). Let m=xb_1+yb_2 \in M for some x,y \in R. For x=y, show that Rm is a direct summand of M (i.e there exists some submodule N of M such that M = Rm \bigoplus N) iff x is invertible in R. Furthermore for any x,y, Rm is a direct summand of M iff the ideal (a,b)=R. If R=\mathbb{Z}, x=3, and y=5, find some v \in M such that \mathbb{Z}m \bigoplus \mathbb{Z}v = M. Any help here would be appreciated thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Basis of dual space involving free Z-module
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 1st 2011, 01:30 PM
  2. Replies: 6
    Last Post: November 30th 2011, 03:50 AM
  3. free abelian group as semi-direct products?
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: July 25th 2009, 10:55 PM
  4. free R-module, subset basis
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 14th 2008, 03:12 AM
  5. Direct sum of free modules
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 25th 2008, 11:06 PM

Search Tags


/mathhelpforum @mathhelpforum