Ring

**slevvio** · November 25th 2009, 08:24 AM

Show that the only units of $\mathbb{C} [ X]$ are the non-zero complex number, i.e. The non zero complex polynomials.

Hint: suppose p = $\sum_{r=o}^{n} a_r X^r$ is a unit and let $p^{-1} = \sum_{s=o}^{m} b_s X^s$ be its inverse.
Consider the highest power of X in the product $p p^{-1}$ .

Can anybody give me a hand with this, any help would be much appreciated. I can't think why considering the highest power of x would help me

**tonio** · November 25th 2009, 08:54 AM

Originally Posted by slevvio

Show that the only units of $\mathbb{C} [ X]$ are the non-zero complex number, i.e. The non zero complex polynomials.

Hint: suppose p = $\sum_{r=o}^{n} a_r X^r$ is a unit and let $p^{-1} = \sum_{s=o}^{m} b_s X^s$ be its inverse.
Consider the highest power of X in the product $p p^{-1}$ .

Can anybody give me a hand with this, any help would be much appreciated. I can't think why considering the highest power of x would help me

Suppose $deg(p)=n\,\Longrightarrow\,deg(pq)\geq n\,\,\,\forall\,q\in\mathbb{C}[x]\,\Longrightarrow\,p$ is a unit iff $n=0=deg(1)$

Tonio

**slevvio** · November 25th 2009, 11:25 AM

Thanks Tonio I see this now

clearly $pp^{-1} = X^{n+m} + l.o.t. = 1$

so degree (1) = 0 = degree ( $X^{n+m}$ ) = n+m

so n = - m but n, m both greater than or equal to 0 , hence n = m = 0 and the power series of p has only one term, a complex number. Thanks

Thread: Ring

LinkBack

Thread Tools

Display

Ring

Similar Math Help Forum Discussions

Quotient of polynomial ring is a localization of a polynomial ring

[SOLVED] Prove the Artinian ring R is a division ring

example of prime ring and semiprime ring

Ideals of ring and isomorphic ring :)

ring

Search Tags