Question

Let \(R\) be a UFD. (a) Show that a monic polynomial of degree two or three in \(R[X]\) is irreducible if and only if it has no roots in \(R\). (b) The polynomial \(X^{2}-a \in R[X]\) is irreducible if and only if \(a\) is not a square in \(R\).

Short answer

(a) Prove that a monic polynomial of degree two or three in the UFD \(R[X]\) is irreducible if and only if it has no roots in \(R\). Proof: 1. If the polynomial is irreducible, it has no roots in \(R\): If \(f(a) = 0\) for an irreducible monic polynomial \(f(X)\), it can be factored as \(f(X) = (X-a)g(X)\), contradicting its irreducibility since neither \((X-a)\) nor \(g(X)\) are units. Thus, \(f(X)\) has no roots in \(R\). 2. If the polynomial has no roots in \(R\), it is irreducible: If a monic polynomial \(f(X)\) has no roots in \(R\), any factorization \(f(X) = g(X)h(X)\) must involve either \(g(X)\) or \(h(X)\) having a root in \(R\), contradicting the root-free property. Thus, \(f(X)\) is irreducible. (b) Prove that a monic polynomial of degree two, \(X^2-a\), is irreducible in \(R[X]\) if and only if \(a\) is not a square in \(R\). Proof: 1. If the polynomial is irreducible, \(a\) is not a square in \(R\): If \(a = b^2\) for some \(b \in R\), then \(f(X) = (X - b)(X + b)\), contradicting its irreducibility. Therefore, \(a\) is not a square in \(R\). 2. If \(a\) is not a square in \(R\), the polynomial is irreducible: If a polynomial \(f(X) = X^2 - a\) is reducible, it can be factored as \(f(X) = (X-b)(X-c)\) with \(a = bc\). However, since \(a\) is not a square in \(R\), this factorization contradicts the reducibility of the polynomial. Thus, \(f(X)\) is irreducible in \(R[X]\).

Step-by-step solution

(a) Proof: If the polynomial is irreducible, it has no roots in \(R\)#

Assume that \(f(X) \in R[X]\) is an irreducible monic polynomial of degree two or three. If \(f(a) = 0\) for some \(a \in R\), then \(f(X)\) can be factored as \(f(X) = (X-a)g(X)\), where \(g(X) \in R[X]\) and degree of \(g(X)\) is one or two depends on the original degree of \(f(X)\). But \(f(X)\) is assumed to be irreducible; therefore, any factorization must involve \(f(X)\) itself or a unit element from \(R[X]\). As \((X-a)\) has a degree of \(1\), and \(g(X)\) has a nonzero degree as well, none of those factors can be a unit element. Thus, we have a contradiction, which means that \(f(X)\) must have no roots in \(R\).

(a) Proof: If the polynomial has no roots in \(R\), it is irreducible#

Assume that \(f(X) \in R[X]\) is a monic polynomial of degree two or three with no roots in \(R\). Suppose that \(f(X)\) is reducible; then there exist non-trivial polynomials \(g(X), h(X) \in R[X]\) such that \(f(X) = g(X)h(X)\). Since the degree of \(f(X)\) is either two or three, one of those factors, say \(g(X)\), must have degree one, and the other factor \(h(X)\) has a degree either one or two. As \(g(X)\) has a degree of one, it must have a root in \(R\) which is also a root of the original polynomial \(f(X)\). This contradicts the fact that \(f(X)\) has no roots in \(R\), so our assumption that \(f(X)\) is reducible must be false, and \(f(X)\) is irreducible.

(b) Proof: If the polynomial is irreducible, \(a\) is not a square in \(R\)#

Assume that \(f(X) = X^2 - a \in R[X]\) is an irreducible polynomial. If there exists a \(b \in R\) such that \(a = b^2\), we could rewrite the polynomial as \(f(X) = (X - b)(X + b)\). In this case, \(f(X)\) would be factored into two non-trivial factors, contradicting the assumption that it is irreducible. Thus, \(a\) must not be a square in \(R\).

(b) Proof: If \(a\) is not a square in \(R\), the polynomial is irreducible#

Assume that \(a\) is not a square in \(R\). We will prove that \(f(X) = X^2 - a \in R[X]\) is irreducible by contradiction. Suppose that \(f(X)\) is reducible; then there must exist non-trivial polynomials \(g(X), h(X) \in R[X]\) such that \(f(X) = g(X)h(X)\). Since \(f(X)\) has a degree of two, at least one of those factors, say \(g(X)\), must have degree one, and the other factor \(h(X)\) must also have a degree one. Let \(g(X) = X-b\) and \(h(X) = X-c\) for some \(b, c \in R\). It follows that \(f(X) = (X-b)(X-c) = X^2 - (b+c)X + bc\). Comparing the coefficients, we have \(a = bc\). But by assumption, \(a\) is not a square in \(R\), which means that \(b\) and \(c\) are not associates (as unit factors won't affect the square condition), contradicting the reducibility of the polynomial. Therefore, \(f(X)\) must be irreducible in \(R[X]\).

Key concepts

Unique Factorization Domain (UFD)

A Unique Factorization Domain (UFD) is an integral domain in which every non-zero element can be expressed as a product of prime elements, and this factorization is unique, apart from the order of the factors and units. Think of it like the prime factorization of integers, where each number can be uniquely broken down into a product of prime numbers. This uniqueness plays a pivotal role when dealing with polynomials over these domains, especially when trying to determine the irreducibility of a polynomial.

When working with a UFD, like the ring of integers, certain properties of polynomials and their roots become much more manageable to study, due in part to this idea of unique factorization. This fundamental property ensures that if a polynomial is factorable, its factors are uniquely determined by its roots (if any exist) within the domain.

Monic Polynomial

A monic polynomial is simply a type of polynomial where the leading coefficient is one. In other words, the coefficient before the highest degree term of the polynomial is 1. For instance, a quadratic monic polynomial would take the form of \( x^2 + bx + c \), where the coefficient of \( x^2 \) is one. This characteristic of monic polynomials is essential when trying to understand their factorization and roots.

In terms of irreducibility, when you come across a monic polynomial in a UFD, you can employ specific techniques to test whether the polynomial can be broken down into simpler, nontrivial factors, which can, in turn, determine the roots of the polynomial if they exist within the domain.

Roots of Polynomial

The roots of a polynomial are the values for which the polynomial equals zero. In algebra, finding the roots is akin to solving an equation; they're the solutions that satisfy the polynomial equation \( f(x) = 0 \). With monic polynomials in a UFD, if you can find a root within the domain, this root can be used to factor the polynomial. Conversely, if you cannot find such a root, this is a strong indicator that the polynomial is irreducible within the domain.

Understanding the relationship between the roots and factorization in a UFD is crucial since this connection can tell us if the polynomial is irreducible or not. In our exercise, a polynomial's lack of roots in the domain confirms its irreducibility for polynomials of degree two or three.

Algebraic Geometry

While not addressed directly in the given exercise, algebraic geometry is an area of mathematics that bridges together algebra, specifically the study of polynomials, with the geometric representation of these algebraic equations. It allows us to visually interpret the solutions of polynomial equations as geometric objects such as curves, surfaces, and more complex structures.

In the context of our discussion, algebraic geometry provides a geometric viewpoint for understanding the properties of irreducible polynomials. An irreducible polynomial in one variable corresponds to an indecomposable curve in a geometric space, and analyzing its irreducibility is akin to understanding the indivisible nature of that curve.

Degree of Polynomial

The degree of a polynomial is the highest power of the variable that appears in the polynomial with a non-zero coefficient. It's a simple yet crucial measure because it informs us about the behavior and number of possible roots the polynomial can have, according to the Fundamental Theorem of Algebra.

In regards to irreducibility as mentioned in the exercise, the degree becomes notably relevant. For instance, monic polynomials of degree two or three in a UFD have a simplicity that allows for certain shortcuts when checking for irreducibility. The exercise demonstrates that the lack of roots within the domain is equivalent to declaring the polynomial irreducible for those specific degrees, which stems from the fact that any factorization of such polynomials would necessarily lower the degree and produce visible roots.