Multiplicative polinomial inverse revisited

Question

As it usually goes with asking questions for a third person... You don't get it right the first time.

Question I asked here is as follows:

Let $A(x)$ be a polynomial with integer coefficients. Is there always a polynomial $B(x)$ for which

$$A(x)\cdot B(x)\equiv 1\pmod n$$

(for a given integer $n$). If the answer isn't yes, an answer "yes if $n$ is ____ (fill in with a characteristic the number has to have" would be interesting as well. Of course, no is also an answer.

This question is on the track of what I'm interested in, it just doesn't have the key ingredient, the mod (that's why it turned out trivial).

The real asker of the question forgot to mention the master key ingredient to me, though. Multiplication in that congruence is not the ordinary one. It is done on $\mathbb{Z}(\mathbb{R})/(x^N-1)$, so it should read

$$A(x)\cdot B(x) \text{ mod } (x^N-1) \text{ mod } n = 1$$

in a bit abused mod notation. I didn't edit the original question, since people already gave the correct answer to that question as it was stated. So, the question is still open - is there a $B$ for every $A$, now in this manner of multiplication?

Ahh! I see. This is a more interesting variant :-) – Jyrki Lahtonen Aug 23 '13 at 19:27 — Jyrki Lahtonen, Aug 23 '13 at 19:27
What's $\mathbb{Z}(\mathbb{R})$ in this case? – Loki Clock Aug 23 '13 at 19:43 — Loki Clock, Aug 23 '13 at 19:43
Gotta be $\mathbb{Z}[x]$, @Loki. – Jyrki Lahtonen Aug 23 '13 at 19:55 — Jyrki Lahtonen, Aug 23 '13 at 19:55

score 1 · Answer 1 · answered Aug 23 '13 at 19:24

1

If $n=4$ and all the coefficients of $A(x)$ are even, then all the coefficients of $A(x)\cdot B(x)$ will be even, hence not 1 modulo $n$.

This remains true if the coefficients of $A(x)$ have any common factor, that divides $n$.

answered Aug 23 '13 at 19:24

vadim123

83,937

score 1 · Answer 2 · answered Aug 23 '13 at 19:39

A necessary and sufficient condition for the existence of such a polynomial $B(x)$ is the following: For each prime divisor $p\mid n$ we must have the condition that in the ring $\mathbb{Z}_p[x]$ we have $\gcd(\overline{A}(x),x^N-1)=1$. Here the notation $\overline{A}(x)$ means the polynomial $A(x)$ with all its coefficient projected down to the field $\mathbb{Z}_p$ (that you may also call reduction modulo $p$). Because $\mathbb{Z}_p$ is a field, this g.c.d. can be computed using the usual Euclidean algorithm of polynomial rings.

The necessity of this condition is clear. If for some $p\mid N$ we have that $\gcd(\overline{A}(x),x^N-1)=d(x)$ for some polynomial $d(x)$ that is not a constant, then that same factor $d(x)$ will be a factor modulo $p$ of $A(x)B(x)$ even if you subtract a multiple of $x^N-1$ (Gosh, it would be so much easier and clearer to say this in the language of quotient rings). So $A(x)B(x)$ reduced modulo $p$ and $x^N-1$ will be a non-constant. A fortiori the same will hold, if we only reduce it modulo $n$.

The sufficiency is a bit trickier, and depends on an easy version of Hensel lifting and the Chinese Remainder Theorem. If you want I can try and elaborate

Now I (or we, to be precise) would like to know how to find $B(x)$, if it exists? — H. M. Šiljak, Aug 24 '13 at 15:01

Multiplicative polinomial inverse revisited

2 Answers2