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 $F[x]$, where the degrees of $g(x)$ and $h(x)$ are both smaller than the degree of $f(x)$. Irreducible polynomials function as the “prime numbers” of polynomial rings.

##### Example17.11

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.

##### Example17.12

The polynomial $p(x) = x^3 + x^2 + 2$ is irreducible over ${\mathbb Z}_3[x]$. Suppose that this polynomial was reducible over ${\mathbb Z}_3[x]$. By the division algorithm there would have to be a factor of the form $x - a$, where $a$ is some element in ${\mathbb Z}_3[x]$. Hence, it would have to be true that $p(a) = 0$. 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.

##### Example17.16

Let $p(x) = x^4 - 2 x^3 + x + 1$. We shall show that $p(x)$ is irreducible over ${\mathbb Q}[x]$. Assume that $p(x)$ is reducible. Then either $p(x)$ has a linear factor, say $p(x) = (x - \alpha) q(x)$, where $q(x)$ is a polynomial of degree three, or $p(x)$ has two quadratic factors.

If $p(x)$ has a linear factor in ${\mathbb Q}[x]$, then it has a zero in ${\mathbb Z}$. By Corollary 17.15, any zero must divide 1 and therefore must be $\pm 1$; however, $p(1) = 1$ and $p(-1)= 3$. Consequently, we have eliminated the possibility that $p(x)$ has any linear factors.

Therefore, if $p(x)$ is reducible it must factor into two quadratic polynomials, say \begin{align*} p(x) & = (x^2 + ax + b )( x^2 + cx + d )\\ & = x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd, \end{align*} where each factor is in ${\mathbb Z}[x]$ by Gauss's Lemma. Hence, \begin{align*} a + c & = - 2\\ ac + b + d & = 0\\ ad + bc & = 1\\ bd & = 1. \end{align*} Since $bd = 1$, either $b = d = 1$ or $b = d = -1$. In either case $b = d$ and so \begin{equation*}ad + bc = b( a + c ) = 1.\end{equation*} Since $a + c = -2$, we know that $-2b = 1$. This is impossible since $b$ is an integer. Therefore, $p(x)$ must be irreducible over ${\mathbb Q}$.

##### Example17.18

The polynomial \begin{equation*}f(x) = 16 x^5 - 9 x^4 + 3x^2 + 6 x - 21\end{equation*} is easily seen to be irreducible over ${\mathbb Q}$ by Eisenstein's Criterion if we let $p = 3$.

Eisenstein's Criterion is more useful in constructing irreducible polynomials of a certain degree over ${\mathbb Q}$ than in determining the irreducibility of an arbitrary polynomial in ${\mathbb Q}[x]$: given an arbitrary polynomial, it is not very likely that we can apply Eisenstein's Criterion. The real value of Theorem 17.17 is that we now have an easy method of generating irreducible polynomials of any degree.

# SubsectionIdeals in $F\lbrack x \rbrack$¶ permalink

Let $F$ be a field. Recall that a principal ideal in $F[x]$ is an ideal $\langle p(x) \rangle$ generated by some polynomial $p(x)$; that is, \begin{equation*}\langle p(x) \rangle = \{ p(x) q(x) : q(x) \in F[x] \}.\end{equation*}

##### Example17.19

The polynomial $x^2$ in $F[x]$ generates the ideal $\langle x^2 \rangle$ consisting of all polynomials with no constant term or term of degree 1.

##### Example17.21

It is not the case that every ideal in the ring $F[x,y]$ is a principal ideal. Consider the ideal of $F[x, y]$ generated by the polynomials $x$ and $y$. This is the ideal of $F[x, y]$ consisting of all polynomials with no constant term. Since both $x$ and $y$ are in the ideal, no single polynomial can generate the entire ideal.

##### Proof

Throughout history, the solution of polynomial equations has been a challenging problem. The Babylonians knew how to solve the equation $ax^2 + bx + c = 0$. 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.