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I shall prove some properties on a polynomial ring $R[x]$ over a commutative ring $R$, and there are two with which I struggle:

For some $f=a_0+a_1x+...+a_nx^n$,

  1. $f$ is invertible in $R[x] \Leftrightarrow a_0$ is invertible, and $a_1,...,a_n$ are nilpotent.

  2. $fg$ primitive $\Leftrightarrow f$ and $g$ primitive, where primitive means: $(a_0,...,a_n)=(1)$.

The first one is quite intuitive and $a_0$ must be invertible since the first coefficient of the unit-polynomial shall be $1$, but I do not follow the nilpotency of the other coefficients.

Can anyone help?

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Let $g=b_0+b_1x+...+b_mx^m$ be the inverse of $f$. Then $fg=1$ means...... –  N. S. Nov 16 '11 at 2:46
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N.S.: This isn't even the easiest step of the solution. This is a completely obvious beginning. –  darij grinberg Nov 16 '11 at 19:42
    
In artofproblemsolving.com/Forum/viewtopic.php?f=61&t=89417 , olorin has proven 1 constructively. This proof can give rise to explicit lower bounds. –  darij grinberg Nov 16 '11 at 19:44
    
For 2, see artofproblemsolving.com/Forum/viewtopic.php?f=64&t=16593 . (This is secretly a constructive proof, and you will see how to get an algorithm out of it if you know Lombardi's method of trees. If you don't care about constructivity, it's just a quick 2-line proof.) –  darij grinberg Nov 16 '11 at 19:48
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2 Answers

up vote 1 down vote accepted

For 2: write $f=a_0+\cdots+a_nx^n$, $g=b_0+\cdots+b_mx^m$, $fg=c_0+\cdots+c_d x^d$, with $d\leq n+m$, $a_n\neq 0$, $b_m\neq 0$, $c_d\neq 0$.

Let $\mathfrak{m}$ be a maximal ideal of $R$; if $f$ and $g$ are each primitive, then there exists $i,j$ such that $a_i\notin\mathfrak{m}$ but $a_{0},\ldots,a_{i-1}\in\mathfrak{m}$; and $b_j\notin\mathfrak{m}$, but $b_0,\ldots,b_{j-1}\in\mathfrak{m}$. Then $$\begin{align*} c_{i+j} &= a_0b_{i+j} + \cdots + a_{i-1}b_{j+1}\\ &\quad\mathop{+} a_ib_j\\ &\quad\mathop{+}a_{i+1}b_{j-1}+\cdots + a_{i+j}b_0. \end{align*}$$ Every term on the first line of the right hand side is in $\mathfrak{m}$, and every term on the last line of the right hand side is in $\mathfrak{m}$; so $c_{i+j}\in\mathfrak{m}$ if and only if $a_ib_j\in\mathfrak{m}$. Hence...

Conversely, note that $(c_0,c_1,\ldots,c_d)\subseteq (a_0,\ldots,a_n)(b_0,\ldots,b_m)$. So if $fg$ is primitive, then...

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Thanks, that makes sense! –  Marie. P. Nov 17 '11 at 10:04
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Here are indications for 1.

1) Suppose that $f$ is invertible
i) As you already know $a_0$ is invertible.
ii) Consider a prime ideal ${\mathfrak p } \subset R$. If $fg=1$ then $ \bar f\bar g=\bar 1$ in $(R/{\mathfrak p }) [x]$.
But in a domain a polynomial ( like here $\bar f$) can only be invertible if it is constant ( what is the degree of a product of polynomials in a domain? )
So every $\bar a_i, \; i\geq1$ satisfies $\bar a_i=0 $ and thus $ a_i\in {\mathfrak p }$ for all prime ideals in $R$. This shows that $a_i$ is nilpotent.

ii) Suppose $a_o$ invertible and the other $a_i$ nilpotent
Use that if $a$ is invertible and $n$ nilpotent then $a+n$ is invertible (and to prove this statement try $a=1$ first).

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This is a wonderful solution. It's certainly cleaner than the hint in Atiyah-Macdonald (not that having two proofs is bad). –  Dylan Moreland Nov 16 '11 at 18:05
    
Thank you, @Dylan. –  Georges Elencwajg Nov 16 '11 at 18:22
    
This is awesome! –  Keenan Kidwell Nov 16 '11 at 21:06
    
Thank you, @Keenan. –  Georges Elencwajg Nov 16 '11 at 22:08
    
Thanks for that, I got it now! –  Marie. P. Nov 17 '11 at 10:04
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