Csdrhrt

Let $theta$ be a root of $p(x)=x^3+9x+6$, find the inverse of $1+theta$ in $mathbb{Q(theta)}$.

So problems like this really annoy me but I did crappy on the last homework after making a lot of arithmetic mistakes so I want to run everything by you guys. This one isn't actually for homework it's just a suggested practice problem but okay lets go!!!

So I'm not sure if the method I used to do this was standard or not, I don't remember my professor showing me, but what I did first was used the canonical euclidean algorithm in $mathbb{Q[x]}$ and wrote:

$x^3+9x+6$

$=(1+x)(x^2-x+10-frac{4}{x+1})$

$=(1+x)(x^2-x+10)-4$

the reduced modulo the minimal polynomial of $theta$ i.e. $x^3+9x+6$ and so

$4=(1+x)(x^2-x+10)$

$rightarrow$

$1= frac{1}{4}(1+x)(x^2-x+10)$ and sooo ah $(theta^2-theta+10)frac{1}{4}$ is what I believe is the inverse of $(1+theta)$ in the quotient field... Like i said I hate these problems but ehh is this correct? On a lighter note it's finally getting cool enough for me to go on proper runs here and so that's pretty bomb!

That's correct. It takes only $1$ step (= division) in the extended Euclidean algorithm to invert a linear polynomial $,g,$ since $ f = g h + c ,Rightarrow, bmod f!: ghequiv -c,Rightarrow, 1/g equiv -h/c,,$ just as you calculated.

Generally it is easier if you use the augmented-matrix form of the extended Euclidean algorithm, e.g. below we compute $,1/g pmod{!f} = 1/(x^2!+!1) pmod{!x^3!+!2x!+!1},$ over $,Bbb Z_3,,$ from this answer.

$begin{eqnarray}
[![1]!]&& &&f = x^3!+2x+1 &!!=&, left<,color{#c00}1,,color{#0a0}0,right>quad , {rm i.e.} qquad f, = color{#c00}1cdot f, +, color{#0a0}0cdot g\
[![2]!]&& &&qquad , g =x^2!+1 &!!=&, left<,color{#c00}0,,color{#0a0}1,right>quad ,{rm i.e.} qquad g, = color{#c00}0cdot f, +, color{#0a0}1cdot g\
[![3]!]&=&[![1]!]-x[![2]!]!!!!!!!!!!!!!!! &&qquadqquad x+1 ,&!!=&, left<,color{#c00}1,,color{#0a0}{-x},right> {rm i.e.}quad x!+!1, =, color{#c00}1cdot f,color{#0c0}{-,x}cdot g\
[![4]!]&=&[![2]!]+(1!-!x)[![3]!]!!!!!!!!!!!!!!!!!!!!!!!! &&qquadqquadqquad 2 ,&!!=&, left<,color{#c00}{1!-!x},, color{#0a0}{1!-!x+x^2},right>\
end{eqnarray}$

Hence the prior line implies: $ 2 = (color{#c00}{1!-!x})f + (color{#0a0}{1!-!x!+!x^2})g $

Thus in $,Bbb Z_3[x] bmod f!:, {-}1equiv 2 equiv (color{#0a0}{1!-!x!+!x^2})g Rightarrow bbox[6px,border:1px solid red]{g^{-1}equiv, {-}(color{#0a0}{1!-!x!+!x^2})}$

$$ left( frac{ x^{2} - x + 10 }{ 4 } right) $$

==================================================

$$ left( x^{3} + 9 x + 6 right) $$

$$ left( x + 1 right) $$

$$ left( x^{3} + 9 x + 6 right) = left( x + 1 right) cdot color{magenta}{ left( x^{2} - x + 10 right) } + left( -4 right) $$
$$ left( x + 1 right) = left( -4 right) cdot color{magenta}{ left( frac{ - x - 1 }{ 4 } right) } + left( 0 right) $$
$$ frac{ 0}{1} $$
$$ frac{ 1}{0} $$
$$ color{magenta}{ left( x^{2} - x + 10 right) } Longrightarrow Longrightarrow frac{ left( x^{2} - x + 10 right) }{ left( 1 right) } $$
$$ color{magenta}{ left( frac{ - x - 1 }{ 4 } right) } Longrightarrow Longrightarrow frac{ left( frac{ - x^{3} - 9 x - 6 }{ 4 } right) }{ left( frac{ - x - 1 }{ 4 } right) } $$
$$ left( x^{3} + 9 x + 6 right) left( frac{ 1}{4 } right) - left( x + 1 right) left( frac{ x^{2} - x + 10 }{ 4 } right) = left( -1 right) $$

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To display the notation used above: here is a gcd for integers, with the continued fraction displayed in the traditional sideways manner:

$$ gcd( 54321, 12345 ) = ??? $$

$$ frac{ 54321 }{ 12345 } = 4 + frac{ 4941 }{ 12345 } $$
$$ frac{ 12345 }{ 4941 } = 2 + frac{ 2463 }{ 4941 } $$
$$ frac{ 4941 }{ 2463 } = 2 + frac{ 15 }{ 2463 } $$
$$ frac{ 2463 }{ 15 } = 164 + frac{ 3 }{ 15 } $$
$$ frac{ 15 }{ 3 } = 5 + frac{ 0 }{ 3 } $$
Simple continued fraction tableau:
$$
begin{array}{cccccccccccc}
& & 4 & & 2 & & 2 & & 164 & & 5 & \
frac{ 0 }{ 1 } & frac{ 1 }{ 0 } & & frac{ 4 }{ 1 } & & frac{ 9 }{ 2 } & & frac{ 22 }{ 5 } & & frac{ 3617 }{ 822 } & & frac{ 18107 }{ 4115 }
end{array}
$$
$$ $$
$$ 18107 cdot 822 - 4115 cdot 3617 = -1 $$

$$ gcd( 54321, 12345 ) = 3 $$
$$ 54321 cdot 822 - 12345 cdot 3617 = -3 $$