Chapter 27: Problem 7
Find all \(c \in \mathbb{Z}_{3}\) such that \(\mathbb{Z}_{3}[x] /\left\langle x^{3}+c x^{2}+1\right)\) is a field.
Short Answer
Expert verified
The values of \(c\) are \(0\) and \(2\).
Step by step solution
01
Understanding \\(\mathbb{Z}_3[x]/\langle x^3 + cx^2 + 1 \rangle\\) as a field
A quotient ring \(\mathbb{Z}_3[x]/\langle x^3 + cx^2 + 1 \rangle\) is a field if and only if the polynomial \(x^3 + cx^2 + 1\) is irreducible over \(\mathbb{Z}_3\). This means it cannot be factored into polynomials of lower degree with coefficients in \(\mathbb{Z}_3\).
02
Checking Irreducibility
To check irreducibility, test if \(x^3 + cx^2 + 1\) has any roots in \(\mathbb{Z}_3\), which consists of elements \(0, 1, 2\). Substitute each element as \(x\) and check for roots.
03
Testing Root \\(x = 0\\)
Substitute \(x = 0\) into the polynomial \(x^3 + cx^2 + 1 \) to get \(0^3 + c \cdot 0^2 + 1 = 1\). Since 1 is not zero, \(x = 0\) is not a root for any \(c\).
04
Testing Root \\(x = 1\\)
Substitute \(x = 1\) into the polynomial: \(1^3 + c \cdot 1^2 + 1 = 1 + c + 1 = c + 2\). For \(x = 1\) to be a root, \(c + 2 = 0\) in \(\mathbb{Z}_3\). Solving this, we find \(c = 1\). When \(c = 1\), \(x = 1\) is a root, so the polynomial is reducible.
05
Testing Root \\(x = 2\\)
Substitute \(x = 2\) into the polynomial: \(2^3 + c \cdot 2^2 + 1 = 8 + 4c + 1\). Simplifying modulo 3, \(8 \equiv 2\), \(4c \equiv c\) (since \(4 \equiv 1\)), hence we have \(2 + c + 1 = c + 3 \equiv 0\). Thus, no value of \(c\) makes this zero.
06
Conclusion of Irreducibility
The polynomial \(x^3 + cx^2 + 1\) has no roots in \(\mathbb{Z}_3\) for \(c = 0\) and \(c = 2\), making it irreducible in these cases. Hence, the quotient ring is a field for these values of \(c\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quotient Ring
A quotient ring is a fascinating concept within abstract algebra. To understand a quotient ring like \(\mathbb{Z}_3[x] / \langle x^3 + cx^2 + 1 \rangle\), we first need to understand how it is constructed. This notation refers to a ring of polynomials, \(\mathbb{Z}_3[x]\), divided by an ideal, \(\langle x^3 + cx^2 + 1 \rangle\).
The polynomial \(x^3 + cx^2 + 1\) in the denominator forms what's known as an 'ideal', a special subset of the polynomial ring. By factoring out this ideal, we essentially 'cut down' the ring to a more manageable structure.
The polynomial \(x^3 + cx^2 + 1\) in the denominator forms what's known as an 'ideal', a special subset of the polynomial ring. By factoring out this ideal, we essentially 'cut down' the ring to a more manageable structure.
- When we divide a polynomial ring by this ideal, each element of the new ring can be represented by a remainder from division by the polynomial.
- The result is a new algebraic structure, which contains elements from \(\mathbb{Z}_3\) itself, but within the framework imposed by our `ideal`.
Field in Algebra
In algebra, fields are central because they provide a robust framework where all elements apart from zero have an inverse. This structure makes fields very useful for various algebraic operations and theories.
A field must satisfy two main properties:
A field must satisfy two main properties:
- It must be a commutative group under addition, with an additive identity (0), and every element must have an additive inverse.
- It must be a commutative group under multiplication, except for zero, with a multiplicative identity (1), and every non-zero element must have a multiplicative inverse.
Roots Modulo Arithmetic
Roots in modulo arithmetic are analogous to solving for zeros in standard algebraic equations, but with a twist due to the modular nature. In simple terms, a root of a polynomial in modulo arithmetic is a number that, when plugged into the polynomial, gives a result divisible by the modulo.
For the polynomial \(x^3 + cx^2 + 1\), to find roots in \(\mathbb{Z}_3\), we substitute each element of the field \(\mathbb{Z}_3 = \{0, 1, 2\}\) into \(x\) and check if the result is equivalent to zero:
For the polynomial \(x^3 + cx^2 + 1\), to find roots in \(\mathbb{Z}_3\), we substitute each element of the field \(\mathbb{Z}_3 = \{0, 1, 2\}\) into \(x\) and check if the result is equivalent to zero:
- If \(x = 0\), the polynomial evaluates to 1, which is not a root as it is not zero modulo 3.
- For \(x = 1\), we find \(c + 2 \equiv 0\) modulo 3 for some values of \(c\), indicating a root.
- If \(x = 2\), simplifications show no values of \(c\) yield a zero, thus \(x = 2\) is not a root.