I'm taking a class in finite fields and have not been able to conceptualize how modulo + finite fields works in polynomial space. I understand the basic premises of modular arithmetic, but can't work out how to actually generate a finite field of polynomials.

For example:

Find all $f(x)$ and $g(x)$ in $\mathbb Z_3[x]$: $$(x^3 + x +1) f(x) + (x^2 + x +1)g(x) = 1$$

I know conceptually how to solve this sort of equation when the coefficients are integers and $f(x), g(x)$ are simple variables, but I don't know how to generate fields in $\mathbb Z_3[x]$ and then how exactly to use them to solve this sort of equation for polynomials once I have their $\mathrm{gcd}$ in $\mathbb Z_3[x]$.

  • $\begingroup$ Do you know how to do long division on $$\frac{x^3 + x +1}{x^2 + x +1}$$ in ${\mathbb Z}_3[x]$? i.e. compute quotient and remainder? That's what you need to compute the Extended Euclidean algorithm for polynomials. For example, these pdf notes $\endgroup$ – user2468 Mar 29 '12 at 0:33
  • $\begingroup$ @J.D. I understand how to do polynomial long division, but am kind of confused on how it differs when in $\mathbb Z_3[x]$ $\endgroup$ – Stephen Young Mar 29 '12 at 0:41
  • $\begingroup$ @Stephen Young: Almost like the usual, except that the arithmetic on the coefficients goes modulo $3$. So $x^2+x+1$ "goes into $x^3+x+1$" $x$ times. Multiply, you get $x^3+x^2+x$. Subtract from $x^3+x+1$. We get (x^3+x+1)-(x^3+x^2+x)=-x^2+1=2x^2+1$. Continue. $\endgroup$ – André Nicolas Mar 29 '12 at 0:46
  • $\begingroup$ Thanks @AndréNicolas - so is the $\mathbb Z_3[x]$ just referring to the modulo of the coefficients then? I'm trying to wrap my head around how to generate a finite field from a statement like $\mathbb Z_3[x]$ / m(x) $\endgroup$ – Stephen Young Mar 29 '12 at 0:50
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    $\begingroup$ Thanks everyone, really helpful stuff - learned more tonight than in the last two months of lectures. $\endgroup$ – Stephen Young Mar 29 '12 at 6:18

There aren't any, since the left side is $0$ at $x=1$.

But one can get a general description of $f(x)$ and $g(x)$ if you replace the right-hand side by, say, $x-1$ (or equivalently $x+2$).


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