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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.

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1 Answer 1

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 $c=1-b$.

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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

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