2) Let \( F = \mathbb{Z} / 2 \mathbb{Z} \) denote the field with just two elements. Let \( x \) denote an
indeterminate, and consider the ring of polynomials \( R = F[x]. \) Consider the ideal \( J \) of \( R \) generated
by \( f(x) = x^4 + x^3 + x^2 + x + 1. \) Viewing the quotient ring \( R / J \) as a vector space over \( F, \) find
the dimension and prove that your answer is correct.
Proof. First show that \( f(x) \) is irreducible over \( F = {\mathbb{Z} / 2 \mathbb{Z}} = \mathbb{Z}_2. \)
\( f(0) = f(1) = 1, \) so there are no linear factors and the only possibility for reducing \( f(x) \) is
\( f(x) = (x^2 + ax + 1)(x^2 + bx + 1), \) where \( a, b \in \mathbb{Z}_2. \) Multiplying those quadratics and
comparing coefficients results in \( a + b = 1 \) and \( ab = 1 \). The second of these forces \( a = b = 1, \)
which is inconsistent with the first. Therefore \( f(x) \) is irreducible in \( \mathbb{Z}_2. \)
Therefore \( K = {\mathbb{Z}_2 [x] / (f(x))} = {R / J} \) is a field: \( K = \{a + bx + cx^2 + dx^3 \; | \; a, b, c, d \in \mathbb{Z}_2 \)
and \( x^4 = -x^3 - x^2 - x - 1 \}. \) The relation provided by \( f(x) \) enables reduction of any product of these
cubics to another cubic in \( K, \) so \( K \) indeed has exactly \( 2^4 \) elements from assigning both values in
\( \mathbb{Z}_2 \) to \( a, b, c \) and \( d \).
Viewing \( K \) as a vector space over \( \mathbb{Z}_2 \), it's evident that \( \{ 1, x, x^2, x^3 \} \) forms
a basis, so the dimension of \( K = {R / J} \) is \( 4. \)
QED.