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I have a problem here that asks: "Express the multiplicative inverse of $1+2^{1/3}-3\cdot2^{2/3}$ as $a_0+a_1\cdot2^{1/3}+a_2\cdot2^{2/3}$."

I believe they are asking us to find it by utilizing the Euclidean algorithm. I am pretty confused about how to do this. I am also confused about the whole concept and how this relates to fields...

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    $\begingroup$ Convince yourself that $\mathbb{Q}[x]/(x^3-2)$ is same as $\mathbb{Q}(\theta)$ where $\theta =2^{\frac{1}{3}}$.. NOw, Inverse of $1+2^{\frac{1}{3}}-32^{\frac{2}{3}}$ is same as inverse of $1+\theta -3 \theta ^2$... $\endgroup$
    – user87543
    Dec 3 '13 at 8:17
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    $\begingroup$ Hint: Run Euclid's algorithm to express the gcd of $x^3-2$ and $1+2x-3x^2$ as their polynomial linear combination. $\endgroup$ Dec 3 '13 at 8:26
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Convince yourself that $\mathbb{Q}[x]/(x^3-2)$ is same as $\mathbb{Q}(\theta)$ where $\theta =2^{\frac{1}{3}}$..

Now, Inverse of $1+2^{\frac{1}{3}}-32^{\frac{2}{3}}$ is same as inverse of $1+\theta -3 \theta ^2$...

and then...

as $\mathbb{Q}(\theta)$ is a field, inverse of $1+\theta -3 \theta ^2$ exists

and as $\mathbb{Q}(\theta)$ is of degree $3$ every element is expressed as $a_0 +a_1\theta +a_2 \theta^2$..

In your language, every element can be written as $a_0+a_1 2^{\frac{1}{3}} +a_2 2^{\frac{2}{3}}$.

In particular, inverse of $1+2^{\frac{1}{3}} -3 2^{\frac{2}{3}}$ can also be written as $a_0+a_1 2^{\frac{1}{3}} +a_2 2^{\frac{2}{3}}$..

Can you proceed further?

Slight Extension:

As $x^3-2$ is irreducible, g.c.d of this with any other polynomial will be $1$

in particular, g.c.d of $x^3-2, 1+2x-3x^2$ is $1$

i.e., I have two polynomials $m_x$ and $n_x$ in $\mathbb{Q}[x]$ such that :

$m_x.(x^3-2)+ n_x (1+2x-3x^2)=1$

As this is an identity, we would have $m_{\theta}.(\theta^3-2)+ n_{\theta} (1+2\theta-3\theta^2)=1$

But then, $\theta^3-2=0$

So, we would have $n_{\theta} (1+2\theta-3\theta^2)=1$

so, $n_{\theta}$ will then be inverse of $(1+2\theta-3\theta^2)$.

Now, What is $n_{\theta}$??

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  • $\begingroup$ I'm not quite sure what to do next. $\endgroup$
    – Stacey
    Dec 3 '13 at 8:23
  • $\begingroup$ you have $\theta^3-2=0$ and then, you should be able to write some relation between $x^3-2$ and $1+2x-3x^2$ using euclidean algorithm.. I am sure in the very next step you will be able to see the answer.. $\endgroup$
    – user87543
    Dec 3 '13 at 8:28
  • $\begingroup$ Okay I'll try that out. Thank you! $\endgroup$
    – Stacey
    Dec 3 '13 at 8:29
  • $\begingroup$ may be you should first calculate what is g.c.d of $x^3-2$ and $1+2x-3x^2$.. $\endgroup$
    – user87543
    Dec 3 '13 at 8:33

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