# Show that $(A[X])^\times=A^\times+nil(A[X])$

I am currently reading through the book Basic Commutative Algebra, by Balwant Singh, wherein the exercise I.XVI reads like:

Show that $(A[X])^\times=A^\times+nil(A[X])$.

Here, for a ring $A$, $A^\times$ means the set of all units, and $nil(A)$ is the intersection of all prime ideals, or the nilpotent radical, the radical of the zero ideal $(0)$.

P.S. I have shown that $A^\times=A^\times+nil A$, but cannot conquer this exercise.
Thanks for any help.

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As you say, the sum of a unit and a nilpotent element is a unit. This immediately gives one inclusion.

Here's a sketch for a proof of the reverse inclusion. Let $f(X) = a_nX^n + \cdots + a_0$ be invertible. Let $g(X) = b_mX^m + \cdots + b_0$ be the inverse of $f$. Clearly $b_0 \in A^*$. Let's show that $a_n$ is nilpotent. To begin with, $a_nb_m = 0$. Looking at the coefficient of $X^{n + m - 1}$ in the $fg$, we see that $a_nb_{m - 1} + a_{n - 1}b_m = 0$. Multiplying this by $a_n$, we get $a_n^2b_{m - 1} + a_{n - 1}a_nb_m = a_n^2b_{m - 1} = 0.$ The idea is to continue doing this until we reach $a_n^{m + 1}b_0 = 0$.

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But, after we show that $a_n$ is nilpotent, why does this tell us that $f$ is a sum of a unit and a nilpotent? – awllower Jan 20 '12 at 4:20
@awllower The point is then that $a_nX^n$ is nilpotent. So $f - a_nX^n$ is a unit. And you keep going from there. – Dylan Moreland Jan 20 '12 at 4:21
Oh, I see. Thus it is to say that, as the sum of unit and nilpotent is a unit, it is itself such a sum. Well, thanks a lot for providing such an interesting answer. – awllower Jan 20 '12 at 4:24

Let me show without calculation the difficult part:

If $f(X)=a_0+a_1X+...+a_n X^n\in (A[X])^\times$, then for all $i\gt0$, the coefficient $a_i$ is nilpotent.

Proof
For an arbitrary prime ideal $\mathfrak p\subset A$, the class $\bar f(X)\in A/\mathfrak p [X]$ is invertible in $A/\mathfrak p [X]$.
Since $A/\mathfrak p [X]$ is a domain, this implies $\bar a_i=0$ (because degree considerations show that an invertible polynomial with coefficients in a domain is a constant).
But this says that $a_i\in \mathfrak p$ for all prime ideals in $\mathfrak p \subset A$. Hence $a_i$ is nilpotent.

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"An egg of Columbus or Columbus's egg refers to a brilliant idea or discovery that seems simple or easy after the fact." From the Wikipedia entry Egg of Columbus. (I try to vary my comments...) – Pierre-Yves Gaillard Jan 20 '12 at 8:38
Dear @Pierre-Yves: thank you for this unexpected and amusing comment ! (By the way, in the Wikipedia entry you link to, there is an engraving by Hogarth, an artist I really like : I am grateful to you for that too.) – Georges Elencwajg Jan 20 '12 at 11:19
@Georges This is great! The situation reminds me of the two common proofs of Gauss's lemma: you can either fiddle around with coefficients, or reduce modulo a prime and obtain a much cleaner proof. – Dylan Moreland Jan 20 '12 at 14:10
Sorry for not reading it after some delay, during which I was not using this site. Hope you do not mind. Also, this is indeed a clever and magnificent proof, which I did not even expect, of the seemingly perplexed statement, whose verisimilitude I once doubted. εστι θαυμασιος.(It is wonderful!) – awllower Jan 29 '12 at 10:48
Ι shall add that this now makes everything appear easy, and thanks very much. – awllower Jan 29 '12 at 10:54