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Calculate Inverses

asked 9 years ago

bruno171092 gravatar image

updated 9 years ago

Hello Guys, i have the following problem.

I got the Ring R=Z/(x71) and the polynom f(x) = x6x4+x3+x21R and want to compute the inverse of f(x), f(x)1. How do I do that?

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answered 9 years ago

tmonteil gravatar image

updated 9 years ago

You will find your inverse if you work over the rationals, that is if R=Q[x]/(x71). There, you can do:

sage: R.<x> = QQ[]
sage: RR = R.quotient(x^7 - 1)
sage: f = RR(x^6-x^4+x^3+x^2-1)
sage: g = f^(-1)
sage: g
16/43*xbar^6 + 11/43*xbar^5 + 21/43*xbar^4 + 1/43*xbar^3 - 2/43*xbar^2 + 4/43*xbar - 8/43
sage: g*f
1
sage: f*g

EDIT answers to the further questions asked in comment:

If you want to work on R=(Z/7Z)/(x71), this is very similar:

sage: Zmod(7)
Ring of integers modulo 7
sage: R.<x> = Zmod(7)[]
sage: R
Univariate Polynomial Ring in x over Ring of integers modulo 7
sage: RR = R.quotient(x^7 - 1)
sage: RR
Univariate Quotient Polynomial Ring in xbar over Ring of integers modulo 7 with modulus x^7 + 6
sage: f = RR(x^6-x^4+x^3+x^2-1)
sage: f
xbar^6 + 6*xbar^4 + xbar^3 + xbar^2 + 6
sage: g = f^(-1)
sage: g
2*xbar^6 + 4*xbar^5 + xbar^3 + 5*xbar^2 + 4*xbar + 6
sage: g.parent()
Univariate Quotient Polynomial Ring in xbar over Ring of integers modulo 7 with modulus x^7 + 6
sage: g*f
1
sage: f*g
1

As for your second question, when i write R.<x>=... i define both the polynomial ring R and its undeterminate x, so x^6-x^4+x^3+x^2-1 is an element of R. when i write RR(x^6-x^4+x^3+x^2-1), i make a conversion of x^6-x^4+x^3+x^2-1 into the quotient ring RR, you can have a look at : http://doc.sagemath.org/html/en/tutor... to learn more.

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Thanks a lot. Sorry to bother you more, but... when i use R=(Z/7Z)/(x71) instead, does it work the same? And in you line 3 you say: "f = RR(x^6-x^4+x^3+x^2-1)"... i dont know exactely what this is good for (like i said... i am totally new to sage). Does that mean you are are looking at x^6-x^4+x^3+x^2-1 as an element in RR?

bruno171092 gravatar imagebruno171092 ( 9 years ago )

I updated my answer to answer your questions.

tmonteil gravatar imagetmonteil ( 9 years ago )

Thanks once more.

bruno171092 gravatar imagebruno171092 ( 9 years ago )

Note that you can also use GF(7) instead of Zmod(7), to see it as a field instead of a ring.

slelievre gravatar imageslelievre ( 9 years ago )
0

answered 9 years ago

FrédéricC gravatar image

updated 9 years ago

Over QQ you can do that

sage: R = PolynomialRing(QQ, 'x')
sage: x = R.gen()
sage: quot = R.quotient_by_principal_ideal(x**7-1)
sage: y = quot.gen()
sage: 1/(y**6-y**4+y**3-y**2+1)
2/13*xbar^6 - 1/13*xbar^5 + 3/13*xbar^4 - 1/13*xbar^3 + 2/13*xbar^2 + 4/13*xbar + 4/13
sage: 1/(y**6-y**4+y**3+y**2-1)
16/43*xbar^6 + 11/43*xbar^5 + 21/43*xbar^4 + 1/43*xbar^3 - 2/43*xbar^2 + 4/43*xbar - 8/43

and you can see that the inverses do not exist over ZZ, as 13 or 43 must be invertible

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Asked: 9 years ago

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Last updated: Nov 22 '15