5
$\begingroup$

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 $f=a_0+a_1x+\cdots+a_nx^n\in R[x]$,

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

  2. $fg$ is primitive $\Leftrightarrow f$ and $g$ are 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?

$\endgroup$
  • 1
    $\begingroup$ Let $g=b_0+b_1x+...+b_mx^m$ be the inverse of $f$. Then $fg=1$ means...... $\endgroup$ – N. S. Nov 16 '11 at 2:46
  • 3
    $\begingroup$ N.S.: This isn't even the easiest step of the solution. This is a completely obvious beginning. $\endgroup$ – darij grinberg Nov 16 '11 at 19:42
  • $\begingroup$ In artofproblemsolving.com/Forum/viewtopic.php?f=61&t=89417 , olorin has proven 1 constructively. This proof can give rise to explicit lower bounds. $\endgroup$ – darij grinberg Nov 16 '11 at 19:44
  • 1
    $\begingroup$ 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.) $\endgroup$ – darij grinberg Nov 16 '11 at 19:48
3
$\begingroup$

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...

| cite | improve this answer | |
$\endgroup$
9
$\begingroup$

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.

2) Suppose $a_0$ 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).

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This is a wonderful solution. It's certainly cleaner than the hint in Atiyah-Macdonald (not that having two proofs is bad). $\endgroup$ – Dylan Moreland Nov 16 '11 at 18:05
  • 1
    $\begingroup$ This is awesome! $\endgroup$ – Keenan Kidwell Nov 16 '11 at 21:06
  • $\begingroup$ Thanks for that, I got it now! $\endgroup$ – Marie. P. Nov 17 '11 at 10:04
  • $\begingroup$ Thanks for your wise proof! $\endgroup$ – user110706 Dec 24 '14 at 18:34
  • $\begingroup$ You are welcome, dear user110706 $\endgroup$ – Georges Elencwajg Dec 25 '14 at 16:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.