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Find the multiplicative inverse of $1+2x$ in $\mathbb{Z_8}[x]$

My work:

I know that since $1$ is a unit in $\mathbb{Z_8}[x]$ and $2$ is nilpotent of index $3$, $1 + 2x$ has a multiplicative inverse. Now $(1+2x)(a+bx)=1$ and from here I arrive at a system with the following conditions: $a=1$, $b+2a=0$ and $2b=0$. Cómo se pronuncia But with that I do not reach anything concrete and I think there is an error in my calculation because I am not taking into account what happens with the zero dividers. Any suggestion will be welcome.

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    $\begingroup$ You don't need to solve linear systems. Just recall how you proved that $1 + $ nilpotent is invertible. It's done through an explicit formula, and all you need to do is to plug $2x$ into that formula. $\endgroup$ Commented Oct 16, 2019 at 18:34
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    $\begingroup$ This is one of those times calculus can inform algebra. What is $$\frac{1}{1+2x}$$ expanded as a geometric series? $\endgroup$ Commented Oct 16, 2019 at 18:36
  • $\begingroup$ @darijgrinberg 1+2=3 is a unit. in which formula do I have to put 2x? and because? $\endgroup$ Commented Oct 16, 2019 at 18:43
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    $\begingroup$ Not my area of expertise but $\frac 1{1-r}= 1+r+r^2+...$ so wouldn't $\frac 1{1+2x} = 1-2x + 4x^2 - 8x^3+....\equiv 1-2x+4x^2$. And.... lessee.... $(1-2x+4x^2)(1+2x)=1+8x^3\equiv 1$. seems to work.... $\endgroup$ Commented Oct 16, 2019 at 18:52
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    $\begingroup$ How could I not? Brian asked how to express $\frac 1{1+2x}$ as a geometric series. I didn't necessarily know why he was asking but it's well known $\frac 1{1-r}=1+r+r^2+....$ so just needed to solve for $r$ knowing wanting $\frac 1{1-r} = \frac 1{1+2x}$. How could I not know that I'd replace $r$ with $-2x$? Even if I didn't know formula I could have done long division. $1+2x$ into $1$ is $1$. That leaves $-2x$. $1+2x$ to $-2x$ is $-2x$. Leaves $4x^2$. $1+2x$ into $4x^2$ is $4x^2$ and that leave $8x^3\equiv 0\pmod 8$. The hard part is knowing ahead of time that it is invertable. $\endgroup$ Commented Oct 16, 2019 at 19:17

3 Answers 3

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Hint: If $R$ is a ring and $t\in R$ is nilpotent, then $1+t$ is a unit and its inverse is given by a finite geometric series.

In your case, $t=2x$ and so $t^3=0$. Thus, the inverse of $1+2x$ is $1-2x+(2x)^2$, which is easily checked.

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  • $\begingroup$ In this case how to calculate the expression in series of $\frac{1}{1+2x}$? In which bibliography can I find a demonstration of that result? Was it unknown to me until now... $\endgroup$ Commented Oct 16, 2019 at 18:46
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    $\begingroup$ $1/(1-t)=1+t+t^2+t^3+\cdots$. Put $t=-2x$ into that. $\endgroup$ Commented Oct 16, 2019 at 18:47
  • $\begingroup$ @LordSharktheUnknown I already got it! That hint was very important. Thank you. Do you know where I can read any proof of this result with geometric series? $\endgroup$ Commented Oct 16, 2019 at 18:53
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    $\begingroup$ @HendrikMatamoros The proof is elementary by just multiplying it out. $\endgroup$ Commented Oct 16, 2019 at 19:38
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This is just what the other answers say, but I think this is how one can remember it: If you have $1+t$ with $t \in R$ nilpotent, then you can make use of the third binomial formula: $$(1 + t)(1 - t) = 1 - t^2$$ You see that you now have $t^2$ instead of $t$. So multiplying by $(1 + t^2)$ yields $$ (1+t)(1-t)(1+t^2) = 1 - t^4 $$ In your example, $t = 2x$ has order $3$, so $t^4 = 0$ and you are done.

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$$(1+2x)(a+bx+cx^2) =1$$

Solve for $a,b,c$

I have $$ a=1, b=6,c=4$$

$$(1+2x)(1+6x+4x^2) = 1+ (8)x+(16)x^2+(8)x^3 \equiv 1 (\mod 8)$$

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