I'm having some difficulties understanding how I'm supposed to construct a field of order $p^r$(where p is a prime number and $r\geq1$) using an irreducible polynomial of degree $r$ in $Z_p[x]$.
(All the $Z$ symbols denote the symbol for the ring of integers.).
The theory states the following
"Let $k(x)$ be an irreducible polynomial of degree $r$ in $Z_p[x]$ and let ~ be the equivalence relation on $Z_p[x]$ defined by
$a(x)$ ~ $b(x) \Leftrightarrow a(x) – b(x)$ is divisible by $k(x)$.
Then the set of equivalence classes of ~ in $Z_p[x]$ is a field of order $p^r$
An example using this is the following:
Construct a field of order 4 by using the irreducible polynomial $x^2 + x + 1$ in $Z_2$.
The solution states that the field consists of the following 4 elements:
{0, 1, x, x+1}
But exactly how have they deduced that from the theory? I understand why the order is 4. But I'm not understanding how they deduced that the elements of the field should be those mentioned above.
Best Answer
The idea here is the following general theorem:
Let $R$ be a commutative ring, and suppose $I\subset R$ is a maximal ideal of $R$, then the quotient ring $R/I$ is a field.
An ideal $I\subset R$ is a set such that $ra\in I$ and $ar\in I$ for each $a\in I$ and $r\in R$, and a maximal ideal is a proper ideal not properly contained in any other proper ideal. Furthermore, the ring $R/I$ is simply the set of all equivalence classes under the relation $r_1\sim r_2$ if and only if $r_1-r_2\in I$.
I'll leave it as an exercise for you to show:
In your example for $\mathbb Z_2[x]/(x^2+x+1)$, notice that $x^2\sim -(x+1)$, so all equivalence classes in your field have a linear representative element, and since $\{0,1,x,x+1\}$ are the only linear elements of $\mathbb Z_2[x]$, you know these are the only ones.