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Hello I am wondering if anyone can help me to understand how I can find the inverse of an element in a quotient ring

for example

I know that $\mathbb{F_{2}}/(<x^3+x+1>)$ is a field as the polynomial is irreducible in $\mathbb{F_{2}}$

Now I want to be able to find inverses ,

for example say the inverse of $(x^2+3)$

My thoughts:

I we wrote it as $x^2+1$ as $3=1$ in this field ( is that valid?)

Then I tried to do $gcd(x^3+x+1,x^2+1)$

to find that $x^{3}+x+1=(x^{2}+1)(x)+1$

and $x^{2}+1=(x+1)(1)$

so then I thought I could maybe do,,

$1=x^{3}+x+1-(x^{2}+1)(x)$

and $1=-1$ in this field

so would this basically tell me that $$(x^{3}+x+1)+(1)=(x)(x{^2}+1)$$ and hence the inverse of $x^{2}+1$ in this field is $x$?

Does any of it make sense to you guys? Any help? It may be very likely that I am wrong, and id like to figure out where.

Further more, Id like to know how I can compute roots of other polynomials in the field.

For example $s^{3}+s^{2}+1$ or $s^{3}+1$

PS: is this also a field with 8 elements? Because I am confused then how to find roots as I thought in the case of a field with prime number of elements we can do something like calculating gcd $x^{p}-x$ , but here we have 8.

If anyone at all can help itd be much appreciated. I have a quiz tomorrow and I have been studying for very long and there are just some things I cant seem to understand.

Thanks

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  • $\begingroup$ Ir is true that the equivalence class of $x$ is the inverse of the equivalence class of $x^2+1$. Your basic Euclidean algorithm argument is sound. The field indeed has $8$ elements. $\endgroup$ – André Nicolas Nov 9 '15 at 3:02
  • $\begingroup$ Thank you, and for finding roots of other elements can I simply plug in s=0 , s=1 and s=2 and see if they come out to zero since we only have not so many cases? for example no s in F2 will satisfy $s^{3}+s^{2}+1=0$ but $s= 1 or -1$ will for $s^{3}+1$ ? Is that make sense or is there a better method? $\endgroup$ – Quality Nov 9 '15 at 3:05
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Your Euclidean algorithm argument is sound. Indeed the equivalence class of $x$ is the inverse of the equivalence class of $x^2+1$.

If you want to solve $t^3+t^2+1=0$ in this field, note that $x^3=x+1$ and therefore the equivalence class of $x^2+x$ is a solution.

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  • $\begingroup$ and it is valid that I went from $x^{2}+3$ to $x^{2}+1$ correct? And yes that is starting to make sense, I defintely understand why $x^{3}=x+1$ in this field, so are you saying I would rewrite the polynomial as $x^2+x=0$ but then do I still need to be the roots of that? $\endgroup$ – Quality Nov 9 '15 at 3:15
  • $\begingroup$ Or is that just the answer as the equivalence classes are elements, and similarly for $t^{3}+1$ the equivalence class of x is a solution? Or would it be like $x+1+<x^{3}+x+1>$ where <x^{3}+x+1> can be written as $ax^{2}+bx+c$ where a b and c are 0 or 1 $\endgroup$ – Quality Nov 9 '15 at 3:20
  • $\begingroup$ $x^2+x$ (or more properly the equivalence class of $x^2+x$) is the solution. No finding of roots of $t^2+t=0$. Remember that here $x$ is not an unknown, it is a polynomial. Similarly, "$x$" is a solution of $t^3+1=0$ in our quotient field. $\endgroup$ – André Nicolas Nov 9 '15 at 3:27

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