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Chapter 4: Problem 1

Let \(f\) be a polynomial of degree 2 over a field \(K\). Show that either \(f\) is irreducible over \(K\), or \(f\) has a factorization into linear factors over \(K\).

Short Answer

Expert verified
A polynomial of degree 2 over a field K, \(f(x) = ax^2 + bx + c\), is either irreducible or has a factorization into linear factors over K, depending on its roots in K. Specifically: 1. If \(f(x)\) has no roots in K, then it is irreducible over K. 2. If \(f(x)\) has a root k in K, then it has a factorization into linear factors over K, as \(f(x) = (x-k)q(x)\) where q(x) is another linear factor in K[x].

Step by step solution

01

Recall Definitions

Recall the following definitions (we will apply them to the specific case of a polynomial of degree 2 over a field K): Irreducible Polynomial: A polynomial is irreducible over a field K if it cannot be factored into the form p(x) = q(x)r(x), where both q(x) and r(x) are polynomials in K[x] with a degree lower than that of p(x). Linear Factor: A linear factor of a polynomial is a monic (leading coefficient is 1) polynomial of degree 1 that divides the polynomial. Let's consider the polynomial \(f(x) = ax^2 + bx + c\). We want to show that either f(x) is irreducible over K or it has a factorization into linear factors over K.
02

Find Conditions for Irreducible Polynomials

For f(x) to be irreducible over K, it should not be possible to factor it into polynomials of lesser degree in K[x]. So, there should not exist any q(x) and r(x) in K[x] such that f(x) = q(x)r(x), where both q(x) and r(x) have degrees less than 2. Since the only possible degrees of polynomials smaller than 2 are 0 and 1, we are essentially looking for a situation when it is not possible to find two linear factors q(x) and r(x) such that f(x) = q(x)r(x). This means that f(x) has no roots in K. Therefore, f(x) is irreducible over K when it has no roots in K.
03

Find Conditions for Polynomial with Linear Factors

If f(x) has a root in K, say k, then f(k) = 0, which means f(x) can be factored as f(x) = (x-k)q(x), where q(x) is another linear factor in K[x]. In this case, f(x) has a factorization into linear factors over K.
04

Conclusion

We have two cases for the polynomial f(x) = ax^2 + bx + c: 1. f(x) is irreducible over K when it has no roots in K. 2. f(x) has a factorization into linear factors over K when it has a root in K. Thus, a polynomial of degree 2 over a field K is either irreducible or has a factorization into linear factors over K.

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Most popular questions from this chapter

Determine whether \(3-2 \sqrt{2}\) is a square in the ring \(\mathrm{Z}[\sqrt{2}]\).

Given a polynomial \(f \in K\left[t_{1}, \ldots . t_{n}\right]\) prove that a reduced polynomial associated with \(f\) is unique, i.e. there is only one.

Let \(R\) be the set of numbers of the form $$ a+b \sqrt{-5} \quad \text { with } a, b \in \mathbf{Z} $$ (a) Show that \(R\) is a ring. (b) Show that the map \(a+b \sqrt{-5} \mapsto a-b \sqrt{-5}\) of \(R\) into itself is an automorphism. (c) Show that the only units of \(R\) are \(\pm 1\). (d) Show that 3, \(2+\sqrt{-5}\) and \(2-\sqrt{-5}\) are prime elements, and give a non-unique factorization $$ 3^{2}=(2+\sqrt{-5})(2-\sqrt{-5}) $$ (e) Similarly, show that 2, \(1+\sqrt{-5}\) and \(1-\sqrt{-5}\) are prime elements, which give the non-unique factorization $$ 2 \cdot 3=(1+\sqrt{-5})(1-\sqrt{-5}) $$

(a) Let \(f\) be a polynomial of degree 3 over a field \(K\). If \(f\) is not irreducible over \(K\), show that \(f\) has a root in \(K\). (b) Let \(F=\mathbf{Z} / 2 \mathbf{Z}\). Show that the polynomial \(t^{3}+t^{2}+1\) is irreducible in \(F[t] .\)

Show that \(t^{4}+4\) can be factored as a product of polynomials of degree 2 with integer coefficients. [Hint: try \(\left.t^{2} \pm 2 t+2 .\right]\)
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