Example 5.11.
The polynomial \(x^2 - 2 \in {\mathbb Q}[x]\) is irreducible since it cannot be factored any further over the rational numbers. Similarly, \(x^2 + 1\) is irreducible over the real numbers.
Section 5.3 Irreducible Polynomials
A nonconstant polynomial \(f(x) \in F[x]\) is irreducible over a field \(F\) if \(f(x)\) cannot be expressed as a product of two polynomials \(g(x)\) and \(h(x)\) in \(F[x]\text{,}\) where the degrees of \(g(x)\) and \(h(x)\) are both smaller than the degree of \(f(x)\text{.}\) Irreducible polynomials function as the “prime numbers” of polynomial rings.
Example 5.12.
The polynomial \(p(x) = x^3 + x^2 + 2\) is irreducible over \({\mathbb Z}_3[x]\text{.}\) Suppose that this polynomial was reducible over \({\mathbb Z}_3[x]\text{.}\) By the division algorithm there would have to be a factor of the form \(x - a\text{,}\) where \(a\) is some element in \({\mathbb Z}_3[x]\text{.}\) Hence, it would have to be true that \(p(a) = 0\text{.}\) However,
\begin{align*}
p(0) & = 2\\
p(1) & = 1\\
p(2) & = 2.
\end{align*}
Therefore, \(p(x)\) has no zeros in \({\mathbb Z}_3\) and must be irreducible.
Not only is the definition of a irreducible polynomial similar to that of a prime integer, but it has similar properties as well. The following lemma is similar to Lemma 2.13, both in its statement and in its proof, and we also have unique factorization of polynomials into irreducibles, similar to Theorem 2.15, the Fundamental Theorem of Arithmetic.
Lemma 5.13.
Let \(p(x)\) be an irreducible polynomial in \(F[x]\text{,}\) where \(F\) is a field. If \(p(x)\) divides a product \(f(x)g(x)\text{,}\) then \(p(x)\) divides either \(f(x)\) or \(g(x)\text{.}\)
Proof.
Suppose that \(p(x)\) does not divide \(f(x)\text{.}\) Since the only polynomials that divide \(p(x)\) are non-zero multiples of \(p(x)\) and constants, but \(p(x)\) does not divide \(f(x)\text{,}\) then the greatest common divisor of \(f(x)\) and \(p(x)\) must be \(1\text{.}\) Therefore, by Proposition 5.10, there exist polynomials \(r(x)\) and \(s(x)\) such that \(1 = r(x)p(x) + s(x) f(x)\text{.}\) Multiplying on both sides by \(q(x)\text{,}\) we get:
\begin{equation*}
g(x) = r(x) p(x) g(x) + s(x) f(x) g(x)
\end{equation*}
By assumption, \(p(x)\) divides \(f(x) g(x)\) and so \(p(x)\) divides both of the summands, and so \(p(x)\) divides \(g(x)\text{.}\)
Theorem 5.14.
Let \(f(x)\) be a non-zero polynomial in \(F[x]\text{,}\) where \(F\) is a field. Then \(f(x)\) can be written \(c p_1(x) \cdots p_n(x)\text{,}\) where \(c \in F\) is a constant, and \(p_1(x), \ldots, p_n(x)\) are irreducible, monic polynomials. Moreover, this factorization is unique, up to reordering, in the sense that if \(f(x)\) also factors as \(d q_1(x) \cdots q_m(x)\text{,}\) where \(d \in F\) and \(q_1(x),\cdots q_m(x)\) are irreducible, then \(c = d\text{,}\) \(n = m\text{,}\) and there is a reordering of \(q_1(x), \cdots, q_n(x)\) that is equal to \(p_1(x),\cdots,p_n(x)\)
Proof.
We prove the existence of a factorization by induction on the degree of \(f(x)\text{.}\) If \(f(x)\) has degree \(0\text{,}\) then it is equal to \(c\) for some \(c \in F\text{.}\) Now suppose that the degree of \(f(x)\) is greater than \(0\text{.}\) If \(f(x)\) is irreducible, then let \(c\) be the leading coefficient of \(f(x)\) and \(p_1(x) = f(x)/c\) is a monic, irreducible polynomial such that \(f(x) = c p_1(x)\text{.}\) If \(f(x)\) is not irreducible, then \(f(x) = g(x) h(x)\) for some polynomials \(g(x)\) and \(h(x)\) of smaller degree. By induction, both \(g(x)\) and \(h(x)\) can be written as the product of a constant an irreducible, monic polynomials, and so \(f(x)\) can as well.
We now want to prove uniqueness of the factorization. Suppose we have two factorizations
\begin{equation*}
c p_1(x) \cdots p_n(x) = d q_1(x) \cdots q_m(x).
\end{equation*}
Because the \(p_i(x)\) and the \(q_j(x)\) are monic, the leading coefficient on the left is \(c\) and on the right it is \(d\text{.}\) Therefore, \(c = d\text{.}\)
To show the equality of the irreducible factors, up to reordering, we use induction on \(n\text{.}\) If \(n=0\text{,}\) then \(f(x)\) is a constant and so \(m\) must be \(0\text{.}\) Now suppose that \(n > 0\text{.}\) By Lemma 5.13, \(p_1(x)\) divides one of \(q_1(x), \ldots, q_m(x)\text{.}\) By reordering, we can assume that \(p_1(x)\) divides \(q_1(x)\text{,}\) and since \(q_1(x)\) is irreducible and both \(p_1(x)\) and \(q_1(x)\) are monic, then \(p_1(x) = q_1(x)\text{.}\) Cancelling these factors, we have that \(p_2(x) \cdots p_n(x) = q_2(x) \cdots q_m(x)\text{,}\) and by the inductive hypothesis \(n-1 = m-1\text{,}\) and these factors are the same up to reordering, which completes the proof.
Subsection Historical Note
Throughout history, the solution of polynomial equations has been a challenging problem. The Babylonians knew how to solve the equation \(ax^2 + bx + c = 0\text{.}\) Omar Khayyam (1048–1131) devised methods of solving cubic equations through the use of geometric constructions and conic sections. The algebraic solution of the general cubic equation \(ax^3 + bx^2 + cx + d = 0\) was not discovered until the sixteenth century. An Italian mathematician, Luca Pacioli (ca. 1445–1509), wrote in Summa de Arithmetica that the solution of the cubic was impossible. This was taken as a challenge by the rest of the mathematical community.
Scipione del Ferro (1465–1526), of the University of Bologna, solved the “depressed cubic,”
\begin{equation*}
ax^3 + cx + d = 0.
\end{equation*}
He kept his solution an absolute secret. This may seem surprising today, when mathematicians are usually very eager to publish their results, but in the days of the Italian Renaissance secrecy was customary. Academic appointments were not easy to secure and depended on the ability to prevail in public contests. Such challenges could be issued at any time. Consequently, any major new discovery was a valuable weapon in such a contest. If an opponent presented a list of problems to be solved, del Ferro could in turn present a list of depressed cubics. He kept the secret of his discovery throughout his life, passing it on only on his deathbed to his student Antonio Fior (ca. 1506–?).
Although Fior was not the equal of his teacher, he immediately issued a challenge to Niccolo Fontana (1499–1557). Fontana was known as Tartaglia (the Stammerer). As a youth he had suffered a blow from the sword of a French soldier during an attack on his village. He survived the savage wound, but his speech was permanently impaired. Tartaglia sent Fior a list of 30 various mathematical problems; Fior countered by sending Tartaglia a list of 30 depressed cubics. Tartaglia would either solve all 30 of the problems or absolutely fail. After much effort Tartaglia finally succeeded in solving the depressed cubic and defeated Fior, who faded into obscurity.
At this point another mathematician, Gerolamo Cardano (1501–1576), entered the story. Cardano wrote to Tartaglia, begging him for the solution to the depressed cubic. Tartaglia refused several of his requests, then finally revealed the solution to Cardano after the latter swore an oath not to publish the secret or to pass it on to anyone else. Using the knowledge that he had obtained from Tartaglia, Cardano eventually solved the general cubic
\begin{equation*}
a x^3 + bx^2 + cx + d = 0.
\end{equation*}
Cardano shared the secret with his student, Ludovico Ferrari (1522–1565), who solved the general quartic equation,
\begin{equation*}
a x^4 + b x^3 + cx^2 + d x + e = 0.
\end{equation*}
In 1543, Cardano and Ferrari examined del Ferro’s papers and discovered that he had also solved the depressed cubic. Cardano felt that this relieved him of his obligation to Tartaglia, so he proceeded to publish the solutions in Ars Magna (1545), in which he gave credit to del Ferro for solving the special case of the cubic. This resulted in a bitter dispute between Cardano and Tartaglia, who published the story of the oath a year later.
