Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Can we factorize the polynomial $f(a)=1+a$ so that it is a product of 2 polynomials each of which is not a unit in the ring $R[a]$? I don't think it is possible but I am not sure why. The reason I think it is not possible is that the only way I can think of to write $1+a$ as a product is $(1+a)\times 1$. And I think this is the only way to write it.

share|cite|improve this question
up vote 4 down vote accepted

If $R$ is a domain, then you cannot factor $1+X$ any further. Indeed, because $\deg gh = \deg g + \deg h$, if $f=gh$ you'd have $1=\deg f= \deg g+ \deg h$ and so $\deg = 1$ and $\deg h=0$ (or vice-versa). Since the leading coefficient of $f$ is $1$, $h$ is a unit. But it doesn't mean that $h=1$. You could take $h$ to be any unit, say $-1$ for instance.

If $R$ is not a domain, then you may be able to factor $1+X$ non-trivially. For instance, $1+X = (5X+1)(6X+1)$ in $\mathbb Z/(10)[X]$. Note how this works exactly because $\deg gh = \deg g + \deg h$ fails here.

The example above is really about the existence of idempotents in a ring. Indeed, if $b^2=b$ in $R$ then $1+X=(1+bX)(1+cX)$ where $c=1-b$.

share|cite|improve this answer
Thanks, @lhf, but how would I know if $R$ is the domain? All i am told is that $R$ is a ring and $R[a]$ is a polynomial ring over $R$ . What does that mean? I just thought that that means that any element in $R[a]$ has coefficients from $R$? Is that right? So is that the second case? – ringo Feb 15 '12 at 10:53
I think there is something about integral domains (if $R$ is an integral domain then the 1st case is true), but how would I know if $R$ is one? – ringo Feb 15 '12 at 10:56
If you're not told that $R$ is a domain and it's not implicit from the context, then there is not much you can say. However, note my edited answer. – lhf Feb 15 '12 at 10:57
Thank you again, lhf! – ringo Feb 15 '12 at 11:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.