Thread: 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 + ....+ a_nxⁿ was in Q[[t]] then its inverse would be (a₀ + a₁x + ....+ a_nxⁿ )^-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.

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:



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



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

with ur = 1.

we have, for the product:

, where

if this is to be equal to 1, then c₀ = 1, c_k = 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 b_k like so:

.



verify that this definition makes r an inverse for u.

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!