Let A be a commutative ring with 1.

1) Prove that a sum of a nilpotent element and an invertible element is invertible.

2) Prove that if $f=a_0+a_1x+\dots+a_nx^n \in A[x]$

a) $\exists f^{-1}\in A[x] \Leftrightarrow a_0$ is invertible and the other coefficients are nilpotent.

b) f is nilpotent $\Leftrightarrow $ all its coefficients are nilpotent.

p.s. Those are the first two in a series of problems. The rest easily follow from each other. I'm only struggling with the first two.

  • $\begingroup$ Please search before asking. All have been asked and answered at least a few times prior. $\endgroup$ – Math Gems Apr 2 '13 at 18:49
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    $\begingroup$ And if the question does somehow survive, please try to pick a title that is more descriptive of your question. $\endgroup$ – rschwieb Apr 2 '13 at 19:01

Here is an abstract argument, which I really like. All you have to know is, that the intersection of all prime ideals of $A$ (called "nilradical") consists exactly of the nilpotent elements of $A$.

  1. Let $u,n \in A$, $u$ invertible, $n$ nilpotent. Let further $p$ be any prime ideal of $A$. Since $n \in p$ we have $u+n \equiv u \not\equiv 0\mod p$. So $u+n$ lies in no prime ideal of $A$ and must therefore be a unit.
  2. a) "$\Leftarrow$" follows from 1. For "$\Rightarrow$" let $p$ be any prime ideal of $A$. Since $A/p$ is an integral domain and the reduction $\overline{f} \in (A/p)[x]$ remains invertible, it follows $\deg(\overline{f}) = 0$, which means $a_0 \notin p$ and $a_i \in p$ for $i = 1, \dots, n$. Since this holds for any prime ideal $p$ of $A$ we conclude that $a_0$ is invertible while $a_i$ is nilpotent for $i = 1, \dots, n$.

    b) Proceed similar to a) and use the fact that the only nilpotent element of an integral domain is $0$.
  • $\begingroup$ How does it follow that $\deg(\overline{f}) = 0$? $\endgroup$ – JoeyBF Jun 21 '16 at 8:58
  • $\begingroup$ @JoeyBF For any integral domain $A$ we have $A[X]^\times = A^\times$. $\endgroup$ – Dune Jun 21 '16 at 9:52
  • $\begingroup$ Thanks! But doesn't that statement require the proposition we are proving? (since "being nilpotent" becomes "being 0") Or is that special case simpler to prove? $\endgroup$ – JoeyBF Jun 22 '16 at 6:14
  • $\begingroup$ It is certainly a special case of the proposition, but way easier to prove (degree formula for polynomials). $\endgroup$ – Dune Jun 22 '16 at 11:05

Hints: Supposing $a^n=0$, in order to show that $u+a$ is a unit for some unit $u$, it suffices to show that $1+au^{-1}$ is a unit. (Note that $au^{-1}=b$ has the same index of nilpotency that $a$ has.

To show that $1+b$ is a unit, use the identity $(1+b)(1-b+b^2-b^3+\dots)=1$

For the second part, to show that the polynomial is invertible, if $a_0$ is a unit, then $a_1x$ is clearly nilpotent, and obviously (by what you have been working on!) $a_0+a_1x$ must be a unit. Proceed by induction.

(Trying to think of a good hint for the converse...)


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