Irreducibility of the Polynomial ( x^3 - 5 ) over (mathbb{R}[x]) and (mathbb{Q}[x])

In this article, we will explore the irreducibility of the polynomial ( x^3 - 5 ) over two different fields: the field of real numbers (mathbb{R}[x]) and the field of rational numbers (mathbb{Q}[x]). The irreducibility of a polynomial refers to whether it can be factored into the product of polynomials of lower degree with coefficients in the given field.

1. Irreducibility over (mathbb{R}[x])

Over the field of real numbers (mathbb{R}[x]), a polynomial is said to be irreducible if it cannot be factored into the product of two non-constant polynomials with real coefficients.

1.1 Degree and Real Roots

The polynomial ( x^3 - 5 ) is of degree 3, which is an odd degree. A polynomial of odd degree always has at least one real root. We can find the real root by solving:

( x^3 - 5 0 )

This gives us:

( x sqrt[3]{5} )

1.2 Factoring and Irreducibility

Given that we have a real root ( x sqrt[3]{5} ), the polynomial can be factored as:

( x^3 - 5 (x - sqrt[3]{5})(x^2 sqrt[3]{5}x sqrt[3]{25}) )

Here, ( (x - sqrt[3]{5}) ) is a linear polynomial and ( (x^2 sqrt[3]{5}x sqrt[3]{25}) ) is a quadratic polynomial with real coefficients. Since the quadratic polynomial ( x^2 sqrt[3]{5}x sqrt[3]{25} ) can be further factored if it has real roots or remains irreducible if it has complex roots, the presence of a real root ensures that ( x^3 - 5 ) is reducible over (mathbb{R}[x]).

2. Irreducibility over (mathbb{Q}[x])

Over the field of rational numbers (mathbb{Q}[x]), a polynomial is said to be irreducible if it cannot be factored into polynomials of lower degree with rational coefficients.

2.1 Rational Root Theorem

The Rational Root Theorem states that any rational root of the polynomial ( x^3 - 5 ) must be of the form ( pm frac{p}{q} ), where ( p ) divides the constant term (-5) and ( q ) divides the leading coefficient (1). The possible rational roots are ( pm 1, pm 5 ).

2.2 Testing Rational Roots

Let's test the possible rational roots:

For ( x 1 ):( 1^3 - 5 -4 ) (not a root) For ( x -1 ):( (-1)^3 - 5 -6 ) (not a root) For ( x 5 ):( 5^3 - 5 120 ) (not a root) For ( x -5 ):( (-5)^3 - 5 -130 ) (not a root)

Since none of the possible rational roots are actual roots of the polynomial ( x^3 - 5 ), the polynomial cannot be factored into polynomials of lower degree with rational coefficients. Therefore, ( x^3 - 5 ) is irreducible over (mathbb{Q}[x]).

3. Conclusion

Based on our analysis:

Over (mathbb{R}[x]), the polynomial ( x^3 - 5 ) is reducible. Over (mathbb{Q}[x]), the polynomial ( x^3 - 5 ) is irreducible.

This is because the polynomial has a real root over (mathbb{R}[x]) and no rational roots over (mathbb{Q}[x]).

Additional Insights

Since ( x^3 - 5 0 ), we can write:

( x^3 5 )

The other two roots are the complex cube roots of 5, which are:

( 5^{1/3} e^{2pi ki/3} ) for ( k 0, 1, 2 )

Thus, the polynomial ( x^3 - 5 ) is reducible over (mathbb{R}[x]) and not reducible over (mathbb{Q}[x]).