# Units in the rational polynomial ring

• Jan 25th 2013, 06:29 AM
mathmansam
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.
• Jan 25th 2013, 07:58 AM
Deveno
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:

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

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

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

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

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

we have, for the product:

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

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:

$\displaystyle b_0 = \frac{1}{a_0}$.

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

verify that this definition makes r an inverse for u.
• Jan 25th 2013, 08:00 AM
mathmansam
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!