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I am asked to find the inverse of $\widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.

I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the degree of polynomial in I.

In previous question, we have calculated the inverse of $\widehat {X} $ which is $1/2$(X). And i am trying to breaking down $X^3 - 1 $ into $X(X^2-2)$ + $(2X+1)$. Is this bit some hints to approach this question or is it a new idea to find the inverse with a polynomial of a larger degree?

Thanks a lot!!

Thomas
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3 Answers3

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Hint $\ \ \color{#c00}{x^2 = 2}\ \Rightarrow\ \dfrac{1}{1+\color{#c00}{x^2} x} \,=\, \dfrac{1}{1+\color{#c00}2x} = \dfrac{1}{1+2\color{#c00}x}\, \dfrac{1-2x}{1-2\color{#c00}x} = \dfrac{1\,-\,2x\ }{1-4(\color{#c00}2)}$

i.e. $ $ rationalize the denom of $\, \dfrac{1}{1+2\sqrt{2}}.\,$ More generally, use the Extended Euclidean Algorithm.

For much further discussion see here and its links.

Bill Dubuque
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Applying the generalized Euclidean algorithm to $x^3+1$ and to $x^2-2$, you get that$$-\frac74=\left(\frac{x^2}2-\frac x4+1\right)(x^2-2)-\left(\frac x2-\frac14\right)(x^3+1).$$Therefore,$$\left(-\frac{2x^2}7+\frac x7-\frac47\right)(x^2-2)+\left(\frac{2x}7-\frac17\right)(x^3+1)=1$$and so the inverse of $x^3+1$ in $\mathbb{Q}[x]/\langle x^2-2\rangle$ is $\dfrac{2x}7-\dfrac17$.

user26857
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José Carlos Santos
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extended Euc:

$$ \left( x^{3} + 1 \right) $$

$$ \left( x^{2} - 2 \right) $$

$$ \left( x^{3} + 1 \right) = \left( x^{2} - 2 \right) \cdot \color{magenta}{ \left( x \right) } + \left( 2 x + 1 \right) $$ $$ \left( x^{2} - 2 \right) = \left( 2 x + 1 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 1 }{ 4 } \right) } + \left( \frac{ -7}{4 } \right) $$ $$ \left( 2 x + 1 \right) = \left( \frac{ -7}{4 } \right) \cdot \color{magenta}{ \left( \frac{ - 8 x - 4 }{ 7 } \right) } + \left( 0 \right) $$ $$ \frac{ 0}{1} $$ $$ \frac{ 1}{0} $$ $$ \color{magenta}{ \left( x \right) } \Longrightarrow \Longrightarrow \frac{ \left( x \right) }{ \left( 1 \right) } $$ $$ \color{magenta}{ \left( \frac{ 2 x - 1 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x^{2} - x + 4 }{ 4 } \right) }{ \left( \frac{ 2 x - 1 }{ 4 } \right) } $$ $$ \color{magenta}{ \left( \frac{ - 8 x - 4 }{ 7 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ - 4 x^{3} - 4 }{ 7 } \right) }{ \left( \frac{ - 4 x^{2} + 8 }{ 7 } \right) } $$ $$ \left( x^{3} + 1 \right) \left( \frac{ 2 x - 1 }{ 7 } \right) - \left( x^{2} - 2 \right) \left( \frac{ 2 x^{2} - x + 4 }{ 7 } \right) = \left( 1 \right) $$

Will Jagy
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  • If one insists on using the longer way of the extended Euclidean algorithm, then it is much clearer & simpler to use it in the augmented matrix form. – Bill Dubuque Feb 04 '19 at 20:46
  • @BillDubuque I am very fond of continued fractions, and was delighted to learn Bezout for (rational) polynomials could also be done that way. Matter of taste, I guess, or of effort invested. – Will Jagy Feb 04 '19 at 20:53
  • Certainly CFs are useful but - as in this critique of a similar prior answer - it will likely prove incomprehensible to most readers without a link to a careful explanation of the method. I always link my extended Euclidean algorithm answers to a detailed exposition and this seems to work well in this regard (I get very few questions). – Bill Dubuque Feb 11 '19 at 16:24