inverse of polynomial in division ring

May 2014
18
0
Europe
Hello,
I'm supposed to find inverse to the polynomial:
$ x^2+1+(f)$ in $Q[x]/((f),+,*)$
and $f=x^3+5$.
I know the solution somehow involves extended Euclidean algorithm, but generally don't know what to do. I tried to find the coefficients for Bezout identity which worked, but don't know how to continue.

Can somebody help me with this?
 
Jun 2013
1,129
601
Lebanon
yes, use the Euclidean algorithm

\(\displaystyle x^3+5=\left(x^2+1\right)x +(5-x)\)

\(\displaystyle x^2+1=(5-x)(-x-5)+26\)

Therefore

\(\displaystyle \left(x^2+1\right)\left(\frac{1}{26}-\frac{5}{26}x-\frac{1}{26}x^2\right)+\)

\(\displaystyle \left(x^3+5\right)\left(\frac{x}{26}+\frac{5}{26}) =1\)

so the inverse is

\(\displaystyle \frac{1}{26}-\frac{5}{26}x-\frac{1}{26}x^2\)
 
Last edited:
May 2014
18
0
Europe
The textbook states that this is the way we verify it - $(x^2+1)*inverse=1 \mod f$ - and it works... but why?

I don' really understand the notation - We want an inverse to $x^2+1+(f)$ but we verify correctness with formula using multiplication. Would you point me in the right direction? (wiki link is more than sufficient. thanks.)
 
Jun 2013
1,129
601
Lebanon
The textbook states that this is the way we verify it - $(x^2+1)*inverse=1 \mod f$ - and it works... but why?

I don' really understand the notation - We want an inverse to $x^2+1+(f)$ but we verify correctness with formula using multiplication. Would you point me in the right direction? (wiki link is more than sufficient. thanks.)
Q[x] / (f) is a field, (f) is the ideal generated by f

a typical element is written in the form h(x) + (f) where h(x) is in Q[x]

the 0 element is written as 0 +(f) = (f)

the multiplicative identity 1 + (f)

so the inverse of h(x) would be g(x) such that ( h(x) + (f) ) * ( g(x) + (f) ) = 1 + (f)

which is the same as h(x)*g(x) + (f) = 1 + (f)

This is why we use multiplication to verify the answer.

Is this what you are asking?
 
May 2014
18
0
Europe
Yes, thank you.

This would work with other fields too, right? eg. if we used Z_prime instead of Q?
 
Jun 2013
1,129
601
Lebanon
sure.

you need an irreducible polynomial f in \(\displaystyle \mathbb{Z}_p[x]\)

to make sure that the quotient is a field so every nonzero element has an inverse
 
May 2014
18
0
Europe
Great,
thanks a lot for help. I have another question from algebra that I'll post soon, I'd be really grateful if you helped me with that one too.