Let \(x^3 - x + 1 \in \mathbb{Z}_3[x]\), where \(\mathbb{Z}_3[x]\) is the ring of all polynomials with coefficients in \(\mathbb{Z}_3\). Then the degree of the field extension
There is another way to get the same answer, by directly counting the elements of the quotient ring instead of quoting the rule that the degree equals the degree of the irreducible polynomial.
Confirm $f(x) = x^3-x+1$ has no root in $\mathbb{Z}_3$: checking $x=0,1,2$ gives $f(0)=1$, $f(1)=1$, $f(2)=7 \equiv 1 \pmod 3$. None of these is $0$, so $f(x)$ has no linear factor over $\mathbb{Z}_3$, and being a cubic with no root means it cannot be broken into a product of a linear and a quadratic factor, so it is irreducible.
Describe elements of the quotient ring: When we divide any polynomial in $\mathbb{Z}_3[x]$ by $f(x)$, the remainder always has degree less than 3, so it can be written as $a_0 + a_1 x + a_2 x^2$ with each $a_i \in \{0,1,2\}$. Every element of $\mathbb{Z}_3[x]/\langle f(x) \rangle$ is uniquely represented this way.
Count the elements: There are 3 choices for each of $a_0, a_1, a_2$, giving $3 \times 3 \times 3 = 27$ elements in total. Since $f(x)$ is irreducible, this quotient ring is a field with 27 elements.
Relate the size to the degree: A finite field that is an extension of $\mathbb{Z}_3$ with $3^n$ elements has degree $n$ over $\mathbb{Z}_3$, because it is a vector space over $\mathbb{Z}_3$ with $n$ basis elements ($1, x, x^2, \dots, x^{n-1}$). Here the field has $27 = 3^3$ elements, so $n=3$.
Let's summarize:
$f(x)=x^3-x+1$ has no root in $\mathbb{Z}_3$, so it is irreducible.
The quotient field has $3^3=27$ elements, matching a degree-3 extension of $\mathbb{Z}_3$.
