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Math Help - Units in the rational polynomial ring

  1. #1
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    Units in the rational polynomial ring

    Hello, I have two problems that relate, and I have been tossing them over for a couple of days but they seem to contradict each other.

    The first: Show that 1-t is a unit in Q[[t]].

    The second: Classify all units of Q[[t]].

    Now the second seems to be that all units are all non-zero zero degree polynomials, aka rationals. As if f(x) = a0 + a1x + ....+ anxn was in Q[[t]] then its inverse would be (a0 + a1x + ....+ anxn )-1 which is not a polynomial.
    Thus f(x) does not have an inverse and is not a unit.

    However, I also stumbled across a theorem/problem that states if a+bx is a unit if a and b are units in the ring. Thanks.
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  2. #2
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    Re: Units in the rational polynomial ring

    you need to be clearer about what Q[[t]] is. usually, this notation means all formal power series, not polynomials (a polynomial is a "finite power series").

    it is true that all units in Q[t] (the ring of rational polynomials) are elements of Q*.

    but let's consider what we have in power series.

    consider the power series:

    1 + t + t2 + t3+...+ tn +.....

    what happens when we multiply this by 1 - t?

    (1 - t)(1 + t + t2 + t3+...+ tn +.....)

    = (1 + t + t2 + t3+...+ tn +.....) - (t + t2 + t3+...+ tn +.....) = 1 + 0t + 0t2 + 0t3 +.... = 1.

    more succintly:

    \frac{1}{1 - t} = \sum_{k = 0}^{\infty} t^k

    this is not a statement of convergence, we are not considering "values" for t, it is a statement about FORMAL power series.

    the general problem of determining the units in Q[[t]] is a bit more complicated. if

    u = \sum_{k=0}^{\infty} a_kt^k

    is a unit of Q[[t]], there is some element:

    r = \sum_{k=0}^{\infty} b_kt^k with ur = 1.

    we have, for the product:

    ur = \sum_{k=0}^{\infty} c_kt^k, where c_k = \sum_{j=0}^{k} a_jb_{k-j}

    if this is to be equal to 1, then c0 = 1, ck = 0, for k > 0.

    now c0 = a0b0, so we see a0 cannot be 0, if u is to be a unit. if a0 ≠ 0, define the bk like so:

    b_0 = \frac{1}{a_0}.

    b_k = -\left(\frac{1}{a_0}\right)\left(\sum_{j=1}^{k} a_jb_{k-j}\right)

    verify that this definition makes r an inverse for u.
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  3. #3
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    Re: Units in the rational polynomial ring

    Thank you so much. I actually went back after posting this and realized that double brackets indicates the power series, which obviously changes the problem completely. I proved what all units are for the simple polynomial ring. Thanks for your help!
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