Results 1 to 2 of 2

Math Help - Irreducible polynomial

  1. #1
    Member
    Joined
    Nov 2008
    Posts
    76

    Irreducible polynomial

    Why is the polynomial XY is irreducible in R=k[X,XY, XY^2,XY^3,...]? Assume XY=pq for some p,q \in R. I want to show that either p or q are units. Would anyone help me on this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17

    Re: Irreducible polynomial

    It has been 5 days now so you probably have figured it out by now, but still, here is my take. First, we find the units R^{\times}. Since R is a subring of k[x,y] any unit of R is also a unit of k[x,y]. But we know (from a theorem) that the units of k[x,y] are the non-zero elements k^*. Thus we see that R^{\times} = k^*.

    Thus, to show that xy is irreducible in R, we need to show that if xy = pq for p,q\in R, then p or q is in k^*. So, let xy = pq for some p,q \in R. We can view any polynomial of R as a polynomial in k[x,y] and thus we'll work over k[x,y] instead. There, we have that

    p(x,xy,\dots) = p_1(x) + xy p_2(x,y)

    and

    q(x,xy,\dots) = q_1(x) +xy q_2(x,y)

    for some p_1,p_2,q_1,q_2. Multiplying these out gives us

    xy = pq = (p1+xyp_2)(q_1+xyq_2) = p_1q_1 + xy(p_2q_1+p_1q_2+xyp_2q_2)

    Since p_1q_1 \in k[x], we see that it must be zero for the LHS and RHS to agree. Since k[x,y] is an integral domain, p_1 = 0 or q_1 = 0. We choose p_1 = 0, that is

    xy = xy p_2(x,y) (q_1(x) + xy q_2(x,y))

    Since k[x,y] is integral, we can cancel out the xy. We are left with

    p_2(x,y)(q_1(x)+xy q_2(x,y)) = 1

    and thus p_2(x,y),q_1(x)+xy q_2(x,y) \in (k[x,y])^{\times} = k^*. Thus we see that p = c xy and q = d for some c,d \in k^*. Therefore, q is a unit, and we are done.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Irreducible polynomial
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: February 9th 2010, 01:08 AM
  2. Irreducible polynomial
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 22nd 2010, 07:15 AM
  3. Irreducible polynomial
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: February 20th 2009, 02:22 AM
  4. Irreducible polynomial
    Posted in the Advanced Algebra Forum
    Replies: 11
    Last Post: February 15th 2009, 08:59 AM
  5. irreducible polynomial
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: December 2nd 2008, 10:01 AM

Search Tags


/mathhelpforum @mathhelpforum