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Can the equality $r = e + u$ hold for all elements $r \in R[x]$, where $R$ - a commutative ring, $e$ - an idempotent ($e^2 = e$) and $u$ has a multiplicative inverse $u^{-1}$?

I think not. My reasoning is as follows. Let's assume the opposite Then for any $r = \sum_{i=0}^{n} r_i x^i \in R[x]$ we have $r = e + u$. It's well known, that $e$ - an idempotent in $R[x]$ $\Leftrightarrow$ $e$ - an idempotent in $R$. It's also well known, that $u$ has a multiplicative inverse in $R[x]$ $\Leftrightarrow$ $u = u_0 \in R$, $u_0$ has a multiplicative inverse in $R$. Then we get that for any $r \in R$ must be $r = e + u$.

Multiply the equality $r = e + u$ by $u^{-1}$. We get $r u^{-1} = e u^{-1} + 1$. Multiply the equality $r = e + u$ again by $(1 - e) u^{-1}$ and get $r (1 - e) u^{-1} = 1 - e$. Express unity from both equalities and equate them. We get $r u^{-1} - e u^{-1} = r (1 - e) u^{-1} + e$. After simplifications we get $(r - 1) u^{-1} e = e$. Then either $e = 0$ or $(r - 1) u^{-1} = 1$. If $e = 0$ then any element in $R$ has a multiplicative inverse, but this is not necessarily true in $R$. If $(r - 1) u^{-1} = 1$, then $u = r - 1$ and $e = 1$. Is this a contradiction or not?

Nickeil
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    Why the specification "on the right" if $R$ is commutative? – jjagmath Apr 20 '24 at 17:04
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    You're right. I'll fix it in the message. – Nickeil Apr 20 '24 at 17:06
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    From $(r - 1) u^{-1} e = e$ you can't conclude in general that $e=0$ or $(r−1)u^{−1}=1$. For that you will need $R$ to be an integer domain. – jjagmath Apr 20 '24 at 17:12
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    I've corrected the question. It was originally for the ring $R[x]$ – Nickeil Apr 20 '24 at 17:35
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    The claim follows from the second paragraph already: if $e$ has to be from $R$ and so has $u$, then their sum must also be from $R$, and say $x$ can't be expressed this way. – Amateur_Algebraist Apr 20 '24 at 17:40
  • Is it because $x$ has the coefficient $r_0 = 0$? – Nickeil Apr 20 '24 at 19:15
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    Actually your claim for units is incorrect, but the one for idempotents is fine, so the sum of an idempotent and a unit in $R[x]$ should be equal to $r + \sum\limits_{i=1}^n a_i x^i$, where $r, a_i \in R$ and $a_i$'s are nilpotent. $x$ is not of this form. – Amateur_Algebraist Apr 20 '24 at 20:25
  • The question leaks like a sieve. There is no quantification anywhere. Is this asking if a polynomial ring is necessarily a clean ring? There are elements of $\mathbb Z$ and $\mathbb Z[x]$ that are not expressible as a sum of a unit an idempotent because, well, there are only two idempotents and two units... – rschwieb Apr 21 '24 at 03:54

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