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I am supposed to find the inverse of $1 + x^2$ in the division ring $\mathbb{ Z }_{ 11 }[ x ] / \langle x^3 + 1 \rangle$.

I know that that $x^3 = -1 = 10$. I have tried solving for $a$, $b$, and $c$ in $( ax^2 + bx + c )( x^2 + 1 )=1$, but I have found no obvious answer (I've found $a = b = -1/2$, $c = 1/2$, but that obviously doesn't help).

Is there some simple way of doing this that I am missing?

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The multiplicative inverse of $2$ in $\mathbb{Z}_{11}$ is $6$, doesn't that help? –  Julian Kuelshammer Nov 30 '12 at 13:56
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What is $t$? It is the residue class of $x$ in that ring? –  Julian Kuelshammer Nov 30 '12 at 13:58
    
How do you find the inverse of $m$ in $\mathbb Z/\left<n\right>$ when $(m,n)=1$? –  Thomas Andrews Nov 30 '12 at 14:04
    
Notice that $x^3+1$ is actually reducible, and hence the quotient is actually not an integral domain. However, $1+x^2$ is relatively prime to $x^3+1$ and hence has an inverse in the quotient, by the euclidean algorithm. –  Seth Nov 30 '12 at 14:12

2 Answers 2

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use the euclidean algorithm to see that the gcd of $1+x^2$ and $x^3+1$ is 1 in $\mathbb{Z}_{11}[x]$. Now apply the extended version to determine $p,q\in\mathbb{Z}_{11}[x]$ such that $$1=p(x)(1+x^2)+q(x)(x^3+1).$$ Now you see that the second summand is mapped to $0$ under the quotient map $\mathbb{Z}_{11}[x]\to \mathbb{Z}_{11}[x]/\langle x^3+1\rangle$. So your inverse is just the image of $p$ under the quotient map.

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The method here is to use the euclidean algorithm to write $1$ as a polynomial linear combination of $1+x^2$ and $1+x^3$. Then in the quotient, you have found the inverse of $1+x^2$

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