Question:

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

\[ \left. \mathbb{Z}_3[x] \middle/ \langle x^3 - x + 1 \rangle \right. \]

of \(\mathbb{Z}_3\) is equal to ______. (answer in integer)

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Check if x^3-x+1 has a root in Z3; if not it is irreducible, and the extension degree equals its degree.
Updated On: Jul 21, 2026
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Correct Answer: 3

Solution and Explanation

Step 1: Understand the setup.
We are working with \(\mathbb{Z}_3 = \{0, 1, 2\}\), the integers modulo 3, and looking at the polynomial ring \(\mathbb{Z}_3[x]\) over it.
We take the quotient of this ring by the ideal generated by \(f(x) = x^3 - x + 1\), written \(\mathbb{Z}_3[x] / \langle f(x) \rangle\).
A standard fact from field theory says: if \(f(x)\) is irreducible over \(\mathbb{Z}_3\), then this quotient is a field, and its degree as an extension of \(\mathbb{Z}_3\) equals the degree of \(f(x)\).

Step 2: Check whether \(f(x)\) is irreducible over \(\mathbb{Z}_3\).
Since \(f(x)\) has degree 3, it fails to be irreducible only if it has a root in \(\mathbb{Z}_3\) (any factorization of a cubic must include at least one linear factor).
So we test all three elements of \(\mathbb{Z}_3\):
\(f(0) = 0^3 - 0 + 1 = 1\), which is not \(0\) in \(\mathbb{Z}_3\).
\(f(1) = 1^3 - 1 + 1 = 1\), which is not \(0\) in \(\mathbb{Z}_3\).
\(f(2) = 2^3 - 2 + 1 = 8 - 2 + 1 = 7 \equiv 1 \pmod{3}\), which is not \(0\) in \(\mathbb{Z}_3\).

Step 3: Conclude irreducibility.
Since none of \(0, 1, 2\) is a root of \(f(x)\) in \(\mathbb{Z}_3\), \(f(x)\) has no linear factor, so it cannot be factored into lower degree polynomials over \(\mathbb{Z}_3\). This means \(f(x)\) is irreducible over \(\mathbb{Z}_3\).

Step 4: Find the degree of the extension.
Because \(f(x)\) is an irreducible polynomial of degree 3, the quotient ring \(\mathbb{Z}_3[x] / \langle f(x) \rangle\) is a field, and as a vector space over \(\mathbb{Z}_3\) it has dimension equal to the degree of \(f(x)\), which is 3. This dimension is exactly the degree of the field extension.

Final Answer:
The degree of the field extension is 3. \[ \boxed{3} \]
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