understanding roots of polynomials in field extensions I'm running into a conceptual stumbling block understanding the application of the FHT to field extensions and finding roots, if anyone has any pointers on where I might be misunderstanding.
I'm given a polynomial $p(x)\in \mathbb{Z}_3[x]$, $p(x)=x^2-x-1$, and asked to describe a field extension of $\mathbb{Z}_3$ that contains a root of that polynomial.
I know that $\mathbb{Z}_3[x]/p(x)$ $\cong$ $\mathbb{Z}_3[c]$, where $c$ is the root of $p(x)$ and $\mathbb{Z}_3[c]$ is the smallest field extension containing both $\mathbb{Z}_3$ and $c$, but I am very shaky on how to actually find that root.
My textbook says that the coset $x+<p(x)>$ will be a root of the polynomial, but I'm having a real hard time accepting that claim without any other explanation (which Pinter doesn't seem to offer). I don't even know where to start with trying to work through that, like what to set that coset equal to or how to evaluate it.
 A: The hint in the book is correct $c=x+\langle p(x)\rangle$ works. After all
$$
c^2-c-1=x^2-x-1+\langle p(x)\rangle
$$
according to the definitions of arithmetic operations of the ring. But here the r.h.s.
is
$$
x^2-x-1+\langle p(x)\rangle=p(x)+\langle p(x)\rangle=0+\langle p(x)\rangle=0,
$$
because $p(x)$ is in the ideal $\langle p(x)\rangle$, and the ideal itself is the zero element of the quotient ring (actually the quotient ring is a field, as $p(x)$ is irreducible and hence generates a maximal ideal.
It is natural to be a bit reluctant to accept this calculation. In a way we produce the zero out of thin air, and in this case it is difficult (in contrast to extending the field of rational numbers) you really need to do it this way, because we cannot, for example, look for a zero of the polynomial in some bigger field already in our repertoire. Such as the field of complex numbers. Remember that "officially" the complex number $i$ ia also  constructed similarly $i=x+\langle x^2+1\rangle$ in the quotient ring $\Bbb{C}=\Bbb{R}[x]/\langle x^2+1\rangle$ (there are other ways, but this is the sleekest in the sense that you don't have to verify e.g. the associativity law of the product of complex numbers separately.
