Chapter 56: Problem 4
Let \(F\) be a field, and let \(f(x)=a x^{2}+b x+c\) be in \(F[x]\), where \(a \neq 0\). Show that if the characteristic of \(F\) is not 2, the splitting field of \(f(x)\) over \(F\) is \(F\left(\sqrt{\left.b^{2}-4 a c\right)}\right.\). [Hint: Complete the square, just as in your high school work, to derive the "quadratic formula"']
Short Answer
Step by step solution
Understand the Exercise
Complete the Square
Define the Discriminant
Derive the Quadratic Formula
Determine the Splitting Field
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Field
- **Closure**: For any two elements in the field, both the sum and the product of these elements remain in the field.
- **Associativity**: Addition and multiplication are associative, meaning the grouping of the operations does not affect the outcome.
- **Commutativity**: Both addition and multiplication are commutative, so the order of elements does not affect the result.
- **Identity Elements**: There exist two distinct elements within the field, 0 and 1, functioning as identity elements for addition and multiplication, respectively.
- **Inverses**: Every element has an additive inverse (or negative), and every non-zero element has a multiplicative inverse.
Characteristic of a Field
- If no such \(n\) exists, the field is said to have characteristic 0. This is typical for fields like \(\), \(\mathbb{Q}\), \(\mathbb{R}\), and \(\mathbb{C}\).
- If there is such an \(n\), the characteristic is that number. For finite fields, typical examples have positive characteristics.
Quadratic Formula
\[ax^2 + bx + c = 0\]where \(a eq 0\). Applying the quadratic formula, the solutions for \(x\) are:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula involves both addition/subtraction and square roots. It efficiently finds the roots by calculating the discriminant, \(b^2 - 4ac\), which decides how many and what type of roots exist:
- If the discriminant is positive, there are two distinct real roots.
- If it is zero, there is one double root.
- If negative, the roots are complex.
Completing the Square
1. Factor out \(a\) from the \(ax^2 + bx\) part:
\ ax^2 + bx = a(x^2 + \frac{b}{a}x) \2. Take half of the coefficient of \(x\) inside the parenthesis, square it, and add/subtract: \ (x + \frac{b}{2a})^2 - (\frac{b}{2a})^2 \3. Rewrite the expression: \ f(x) = a [(x + \frac{b}{2a})^2 - (\frac{b}{2a})^2] + c \4. Simplify to get: \ f(x) = a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a}) \Completing the square is useful because it prepares the polynomial for application of the quadratic formula. It highlights how changing the form of a polynomial can simplify solving it and understanding its behavior.