To show that \(F / \{0\} \cong F\), we need to establish an isomorphism between the quotient ring \(F / \{0\}\) and the field \(F\).
Let \(F\) be a field. \(\{0\}\) is the zero ideal in \(F\).
The quotient ring \(F / \{0\}\) is constructed by partitioning \(F\) into cosets of the ideal \(\{0\}\). Since \(\{0\}\) contains only the zero element, every element \(a \in F\) forms a distinct coset \(a + \{0\}\).
Each coset \(a + \{0\}\) is simply \(a\). Hence, the set of cosets is just the set of elements of \(F\).
The addition and multiplication operations on the cosets are defined as follows: \[(a + \{0\}) + (b + \{0\}) = (a + b) + \{0\}\] \[(a + \{0\}) \cdot (b + \{0\}) = (a \cdot b) + \{0\}\] Since \(\{0\}\) is the zero ideal, these operations reduce to: \[a + b + \{0\} = a + b\] \[a \cdot b + \{0\} = a \cdot b\]
Define the map \(\varphi: F / \{0\} \to F\) by \(\varphi(a + \{0\}) = a\). This map is well-defined because each coset \(a + \{0\}\) corresponds uniquely to \(a\).
Check that \(\varphi\) is a ring homomorphism: \(\varphi\) is injective: If \(\varphi(a + \{0\}) = \varphi(b + \{0\})\), then \(a = b\). \(\varphi\) is surjective: For any \(a \in F\), \(a = \varphi(a + \{0\})\).
Since \(\varphi\) is a bijective homomorphism, it is an isomorphism. Therefore, \(F / \{0\} \cong F\).
Thus, we have shown that \(F / \{0\}\) is isomorphic to \(F\). \(\square\)
Lecture Exercise for Reader
(a) Want to show that \(f = x^3 + 2x + 1\) is irreducible over \(\mathbb{Z}_3\) and that \(\mathbb{Z}_3[x] / I\) is a field of order 27 containing a subfield isomorphic to \(\mathbb{Z}_3\). A polynomial of degree 3 is irreducible over \(\mathbb{Z}_3\) if it has no roots in \(\mathbb{Z}_3\) and cannot be factored into polynomials of lower degrees. To check for roots, evaluate \(f(x) = x^3 + 2x + 1\) at \(x = 0, 1, 2\) in \(\mathbb{Z}_3\). For \(f(0) = 0^3 + 2 \cdot 0 + 1 = 1\), \(f(1) = 1^3 + 2 \cdot 1 + 1 = 1 + 2 + 1 = 4 \equiv 1 \pmod{3}\), and \(f(2) = 2^3 + 2 \cdot 2 + 1 = 8 + 4 + 1 = 13 \equiv 1 \pmod{3}\), none of these are zero, so \(f(x)\) has no roots in \(\mathbb{Z}_3\). Since \(f(x)\) has no roots, it cannot be factored into a product of a linear polynomial and a quadratic polynomial in \(\mathbb{Z}_3[x]\). Thus, \(f(x)\) is irreducible over \(\mathbb{Z}_3\). Since \(f\) is irreducible, the ideal \((f)\) is maximal, and the quotient ring \(\mathbb{Z}_3[x] / (f)\) is a field. The polynomial \(f(x)\) is of degree 3, and the quotient ring \(\mathbb{Z}_3[x] / (f)\) consists of equivalence classes of polynomials of degree less than 3: \(\mathbb{Z}_3[x] / (f) = \{ a + bx + cx^2 + (f) \mid a, b, c \in \mathbb{Z}_3 \}\). There are \(3^3 = 27\) such equivalence classes, so \(\mathbb{Z}_3[x] / (f)\) is a field of order 27. The field \(\mathbb{Z}_3\) is naturally embedded in \(\mathbb{Z}_3[x] / (f)\) as \(\{ a + (f) \mid a \in \mathbb{Z}_3 \}\), forming a subfield isomorphic to \(\mathbb{Z}_3\). \(\square\)
(b) Want to show that \(\mathbb{Z}_3[x] / (f)\) is a field extension of \(\overline{\mathbb{Z}}_3\) and is a 3-dimensional vector space over \(\overline{\mathbb{Z}}_3\) with basis \(\{1 + (f), x + (f), x^2 + (f)\}\). Let \(\overline{\mathbb{Z}}_3 = \{ 0 + (f), 1 + (f), 2 + (f) \}\), which is isomorphic to \(\mathbb{Z}_3\). The field \(\mathbb{Z}_3[x] / (f)\) contains \(\overline{\mathbb{Z}}_3\) as its subfield, making it a field extension. Any element of \(\mathbb{Z}_3[x] / (f)\) can be written as \(a + bx + cx^2 + (f)\) where \(a, b, c \in \mathbb{Z}_3\), and the elements \(\{1 + (f), x + (f), x^2 + (f)\}\) span \(\mathbb{Z}_3[x] / (f)\). To show that \(\{1 + (f), x + (f), x^2 + (f)\}\) is a basis, we need to establish linear independence and spanning. For linear independence, suppose \(\alpha (1 + (f)) + \beta (x + (f)) + \gamma (x^2 + (f)) = 0 + (f)\), which implies \(\alpha + \beta x + \gamma x^2 \equiv 0 \pmod{f}\). Since \(1, x, x^2\) are linearly independent in \(\mathbb{Z}_3[x]\) modulo \(f(x)\), it follows that \(\alpha = \beta = \gamma = 0\). For spanning, any element in \(\mathbb{Z}_3[x] / (f)\) can be expressed as a linear combination of \(1, x, x^2\) over \(\overline{\mathbb{Z}}_3\). Thus, \(\mathbb{Z}_3[x] / (f)\) is a 3-dimensional vector space over \(\overline{\mathbb{Z}}_3\) with basis \(\{1 + (f), x + (f), x^2 + (f)\}\). \(\square\)