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Can someone explain how to find inverses in truncated polynomial rings manually (i.e. on pen and paper)? As an example from the tutorial:

Example. Take $N=7$, $q=11$, $a=3+2X^2-3X^4+X^6$. The inverse of $a \bmod 11$ is $a^{-1}=-2+4X+2X^2+4X^3-4X^4+2X^5-2X^6$, since

$(3+2X^2-3X^4+X^6)*(-2+4^X+2X^2+4X^3-4X^4+2X^5-2X^6) \equiv -10+22X+22X^3-22X^6 \equiv 1 \mod 11$.

How to find $a^{-1} = -2+4X+2X^2+4X^3-4X^4+2X^5-2X^6$, manually, not using the pseudocode?

Gilles 'SO- stop being evil'
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Sunia Raharja
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    I don't get your "not pseudo-code". To do it manually, you should execute the pseudo-code manually. – CodesInChaos Mar 11 '13 at 10:13
  • I want to know all step of the process, so i can solve it with handwriting, if its posible – Sunia Raharja Mar 11 '13 at 10:22
  • idon't understand the code on the Techncl report – Sunia Raharja Mar 11 '13 at 10:25
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    To be honest, there is typically a fair amount of work required to do this by hand even for such a small example. It requires long division of polynomials as you are basically doing the extended euclidean algorithm, then substituting back up until you get something of the form $p_1a+p_2b=1$ where $p_1,p_2$ are polynomials and $b=x^7-1$. Since $x^7-1\equiv 0$ in your ring, you have $p_1a=1$ so $a^{-1}=p_1$. Not sure if someone here will be willing to work everything out by hand. We'll see. Just being realistic though. – mikeazo Mar 11 '13 at 13:02
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    A fairly simple algorithm (extended euclidean algorithm) tells you exactly what you need to do to solve the problem. To better understand the algorithm, start with simple integers as in this question – mikeazo Mar 11 '13 at 13:11
  • ok that's great, i just understand about using extended euclidean algorithm, i can find invers p1 (mod p2) if both p1 and p2 are polynomial, but if p1 is polynomial(a) and p2 integer(11) i can't figure it out.

    Can anyone tell me how to divide polynomial a by integer 11?

     – Sunia Raharja Mar 11 '13 at 17:22
  • @SuniaRaharja Take the coefficients (which are integers) modulo 11. – Henno Brandsma Mar 11 '13 at 20:14
  • @SuniaRaharja typo alert: $4^X$ should be $4X$; do you the result of the multiplication as correct, using $X^7 \equiv 1, X^8 \equiv X$, etc? This we do in $\mathbb{Z}[X]$ first, then reduce the coeffficients modulo $11$ (so $22 \equiv 0$, $-10 \equiv 1$). – Henno Brandsma Mar 11 '13 at 21:32
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    Basically, using Extended Euclidean, do the inverse computation of $a$ in $\mathbb{Z}[X]$, doing the gcd with $X^7 - 1$, and reduce the coefficients modulo 11 afterwards. – Henno Brandsma Mar 11 '13 at 21:39
  • @HennoBrandsma do you mean like this?; gcd(x6 - 3x4 +2x2 + 3, X7 - 1), in this time(using Ext Euclid) i must find that; x(x6 - 3x4 +2x2 + 3) + y(x7 - 1) = 1 so i can get the inverse is x, that's how it works right? – Sunia Raharja Mar 12 '13 at 07:11
  • @SuniaRaharja yes, the polynomial $x$ (bad choice of notation) is indeed the inverse, and then you need to reduce mod 11. If all is well, you shouls get $a^{-1}$ or an equivalent (mod 11) polynomial. – Henno Brandsma Mar 12 '13 at 08:56
  • @HennoBrandsma are you sure that if i'm doing it right i can get gcd(x6 - 3x4 +2x2 + 3, X7 - 1) is 1?, because im doing it now and i'm realize that's need alot of paper, and i'm doubt it that i can't get the answer – Sunia Raharja Mar 12 '13 at 10:28
  • Pretty sure, if I have more time, I'll try to compute it too, by hand, of course. – Henno Brandsma Mar 12 '13 at 12:11

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