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Find out if polynomial $x^3-x^2+x-1$ has inverse element in quotient ring ${\displaystyle \mathbb {Z} }_{11}/(x^4+3x^3-3x^2-4x-1)$, if yes find this inverse.

I know that $x^3-x^2+x-1$ has inverse element when $gcd(x^4+3x^3-3x^2-4x-1;x^3-x^2+x-1)$~$1$. I also know that I can use Euclidean algorithm. But I'm not sure about my result, I have got result that the g$cd$~$5$ it means that $x^3-x^2+x-1$ does not have inverse. Am I correct?

Thanks for any help!

José Carlos Santos
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Tomak
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3 Answers3

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I did not check the computations, but if you got $5$, then that means that they are relatively prime and therefore that it has an inverse. That's so because asserting that $5$ is a greatest common divisor of two polynomials is the same thing as asserting that $1$ is a greatest common divisor of them. Note that, in $\mathbb{Z}_{11}$, $5\neq0$.

José Carlos Santos
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  • Thanks very much, I am trying to find inverse, I have found that it should be $5x^3+3x^2+2x+3$, if I haven't made a mistake. If it isn't correct result I can edit contribution, and write down my process. I am little confused, how can I chcek if my inverses is correct, sorry about my simple questions. – Tomak Jun 11 '18 at 14:59
  • @Tomak You check by computing the product of the original polynomial by the one that you got. You should get $1$ modulo $x^4+3x^3-3x^2-4x-1$. – José Carlos Santos Jun 11 '18 at 15:03
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The extended Euclidean algorithm over $\mathbb Q$ gives: $$ 232 = (49 x^3 + 168 x^2 - 75 x - 195)(x^3-x^2+x-1) +(-49 x^2 + 28 x - 37)(x^4+3x^3-3x^2-4x-1) $$ Mod $11$, this becomes $$ 1 \equiv (5 x^3 + 3 x^2 - 9 x - 8)(x^3-x^2+x-1) \bmod (x^4+3x^3-3x^2-4x-1) $$ and so the inverse is $5 x^3 + 3 x^2 - 9 x - 8 \equiv 5x^3+3x^2+2x+3$.

lhf
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Since Jose already covered the fact that your approach is valid, I'll present a cool alternative approach. First, construct the companion matrix of $x^4+3x^3-3x^2-4x-1$, call it $A$.

Here $$A = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 4 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & -3 \end{pmatrix}_.$$

Now consider the matrix, $B = A^3 - A^2 + A - I$. Then $B$ is invertible (recall that the coefficients of the matrix lie in $\mathbb{Z}_{11}$) iff $x^3 - x^2 + x - 1$ is invertible. Even stronger, if you express $B^{-1}$ as a linear combination of $I, A, A^2, A^3$, this tells you the inverse of $(x^3-x^2 + x -1)$!

More explicitly, if $B^{-1} = c_3A^3 + c_2A^2 + c_1A + c_0I$, then $$(x^3-x^2+x-1)^{-1} = c_3x^3+c_2x^2+c_1x+c_0.$$

Ken Duna
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  • Very beautiful, I would never think of that it can be join in this way. But what looks like little difficult for me how to find $B^{-1}$ as a linear combination. Are there some easy way? – Tomak Jun 11 '18 at 18:47
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    Let me preface this by saying that you almost certainly wouldn't want to do this by hand. You can consider $I,A,A^2,A^3,$ and $B^{-1}$ as vectors of length $16$. Put the vectors corresponding to $I, A, A^2, A^3$ as the first 4 columns of a matrix, and the vector corresponding to $B^{-1}$ as the $5^{th}$ column. If you do row reduction to put this matrix in reduced row echelon form, the entries in the $5^{th}$ column will then tell you how to express $B^{-1}$ as a linear combination of the others. Again, this is not something you would want to do by hand, but a computer would be happy to do. – Ken Duna Jun 15 '18 at 16:46