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On this page:

• Introduction
• Polynomial Rings of Integral Domains
• Gauss’s Lemma
• Irreducibility Criteria
• Reducing Coefficients Modulo an Ideal
• Eisenstein’s Criterion
• Worked Exercises

Introduction

TODO

Polynomial Rings of Integral Domains

All rings on this page will be assumed to be commutative.

[theorem] [thmtitle]Proposition %counter% (Basic properties of $R[x]$ for $R$ an ID)[/thmtitle]

Let $R$ be an integral domain and $f, g \in R[x]$. Then

1. $\deg(fg) = \deg f + \deg g$.
2. The units of $R[x]$ are the units in $R$.
3. $R[x]$ is an integral domain. [/theorem]

[proof] (1) Let $a_nx^n$ and $b_m x^m$ be the leading terms of $f$ and $g$, respectively. Then the leading term of $fg$ is $a_nb_mx^{n+m}$. Indeed, if $a_n b_m = 0$, then either $a_n = 0$ or $b_m = 0$ due to $R$ being an integral domain. Thus, the claim follows.

(2) Clearly any unit of $R$ is a unit of $R[x]$. For the converse, suppose we make use of Part 1. An inverse of a non-constant polynomial would have negative degree which is obviously not possible.

(3) Guaranteed by Part (1). [/proof]

[theorem] [thmtitle]Proposition %counter% ($I[x] \unlhd R[x]$ for $I \unlhd R$)[/thmtitle]

Let $I \subseteq R$ be an ideal and define $I[x]$ to be the polynomials in $R[x]$ with coefficients in $I$. Then

1. $I[x]$ is an idael of $R[x]$.
2. $(R/I)[x] \cong R[x]/I[x]$.
3. $I \in \Spec R$ implies $I[x] \in \Spec R[x]$. [/theorem]

[proof] (1) Obvious.

(2) Define the map $\phi:R \to R/I$ to be the canonical projection map. Extend $\phi$ as map $\Phi:R[x] \to (R/I)[x]$ by sending

This map is obviously surjective and it is immediate that this map is an ring homomorphism. Furthermore, we see that $\ker\Phi = {I[x]}$ so the First Isomorphism theorem implies the desired result.

(3) By the properties from the previous proposition, $(R/I)[x]$ is an integral domain and part (2) of this proposition says that $R[x]/I[x]$ is isomorphic to an integral domain. So it follows that $I[x]$ is prime in $R[x]$. [/proof]

A well-known result of algebra is polynomial rings over fields are Euclidean domains which amounts to the polynoial division algorithm.

[theorem] [thmtitle]Theorem %counter% (Polynomial rings over fields are Euclidean domains)[/thmtitle]

Let $F$ be a field. Then $F[x]$ is a Euclidean domain where norm is given by degree. In particular, for any $g, f \in F[x]$ with $f \neq 0$, there exist unique polynomials $q, r\in F[x]$ such that

or $r = 0$. [/theorem]

[proof] (Existence) We first dispose of some trivial cases. If $f = 0$, then we just set $q = r = 0$. Furthermore, if $\deg g < \deg f$, then we just set $q = 0$ and $r = g$. Now assume that $\deg f \le \deg g$ and write both as

We will proceed by induction on $n$. It’s clear that the division algorithm holds if $g$ has degree $0$. Now define

where

Then it follows that $G$ is a polynomial of degree of at most $n-1$ and so there are $Q$ and $R$ for which

and $\deg R < \deg f$ or $R = 0$. Rearranging terms gives

and setting $q \coloneqq Q + u$ and $r \coloneqq R$ gives the desired form in the statement.

(Uniqueness) Now suppose that

where $\deg r, \deg R < \deg f$ or $r = 0$ or $R = 0$. Then

Now, $(Q - q)f$ has degree $\deg f + \deg(Q - q)$ which has degree larger than $R - r$. It follows that $Q - q$ must be $0$ which implies that $R = r$. [/proof]

[example] [extitle]Remark %counter%[/extitle]

It is well-known that this result can be considerably generalized to rings that aren’t even integral domains so long as the leading coefficient of $b$ is a unit, but we lose uniqueness in this case. [/example]

An immediate result that follows from $F[x]$ being a Euclidean domain is the following:

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

$F[x]$ is a PID and UFD. [/theorem]

[theorem] [thmtitle]Theorem %counter% (Factor theorem)[/thmtitle]

Let $f \in F[x]$ and $a\in F$. Then $f$ has $x - a$ as a factor if and only if $f(a) = 0$. [/theorem]

[proof] The $\implies$ direction is obvious. For $\impliedby$, we apply the division algorithm to obtain

where $\deg r < 1$ or $r = 0$. Plugging in $x = a$ gives

Since $r$ is a constant, it follows that $r = 0$. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $f \in F[x]$ be of degree $n$. Then $f$ has at most $n$ roots. [/theorem]

[proof] Apply the theorem inductively until there are no roots left. This leaves $f$ in the form

Since $n = \deg f = \deg p + k$, it follows that $k \le n$. [/proof]

A well-known result from elementary number theory is that $(\ZZ/p\ZZ)^\times$ is a cycle group when $p$ is a prime number. This is straightforward to prove using the results we just proved.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $p \in \ZZ$ be prime. Then the group $(\ZZ/p\ZZ)^\times$ is cyclic. [/theorem]

[proof] Notice that $(\ZZ/p\ZZ)^\times$ is a finitely-generated $\ZZ$-module so the invariant factor decomposition implies that

where $d_n \mid d_{n-1} \mid \cdots \mid d_2 \mid d_1$. In particular, this implies that

in $(\ZZ/p\ZZ)^\times$. Equivalently, $a$ is a root of $x^{d_1} - 1$ in $(\ZZ/p\ZZ)[x]$. By the corollary we just proved, $\ZZ/p\ZZ$ has at most $d_1$ elements which implies that $(\ZZ/p\ZZ)^\times \cong \ZZ/d_1\ZZ$, implying the cyclicity as desired. [/proof]

Gauss’s Lemma

[theorem] [thmtitle]Theorem %counter% (Gauss’ Lemma)[/thmtitle]

Let $R$ be a UFD and $F = \Frac R$. If $p \in R[x]$ is reducible in $F[x]$, then $p(x)$ is reducible in $R[x]$.

More precisely, if $p(x) = A(x)B(x)$ for some nonconstant polynomials $A(x), B(x) \in F[x]$, then there are nonzero elements $r,s\in F$ such that $a(x) \coloneqq rA(x)$ and $b(x) \coloneqq sB(x)$ are elements of $R[x]$. and $p(x) = a(x)b(x)$ is a factorization in $R[x]$. [/theorem]

[proof] Suppose that $p(x)$ factors as

in $F[x]$ where $A(x)$ and $B(x)$ are both nonzero. As $F$ is the field of fractions of $R$, all coefficients of $A(x)$ and $B(x)$ have denominators that are elements of $R$. Thus, we can clear denominators by multiplying by a common multiple $d$ of all the denominators to obtain

where $a_0(x), b_0(x) \in R[x]$ and $d$ is nonzero. In the particular case where $d$ is a unit, we can just take $a(x) = d^{-1}(x)$ and $b(x)$ and call it a day. If not, then as $R$ is a UFD, we can factor $d$ into irreducibles

in $R$. Our goal is to redistribute each $1/p_i$ to $a_0(x)$ and $b_0(x)$ so that we do not “knock them out of” $F[x]$. Passing to the quotient ring $(R/\generator{p_1})[x]$, we see that

We showed earlier that $(R/\generator{p_1})[x]$ is an integral domain so either $\overline{a_0} = 0$ or $\overline{b_0}=0$. Whichever of the two it is, this tells us that $a_0/p_1\in R[x]$ or $b_0/p_1 = 0$. Thus, this gives us the new equation

where both $a_1(x), b_1\in R[x]$. Iterating this process, this obtain

where $a_r(x), b_r(x) \in R[x]$. Thus, we can let $a(x) = a_r(x)$ and $b(x) = b_r(x)$ and we are done. [/proof]

A version of the converse of Gauss’ Lemma holds so long as you cannot factor out a common factor from the coefficients of your $p(x)$. The following statement is much more commonly referred to as Gauss’ Lemma, in my experience.

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $R$ be a UFD and $F = \Frac R$. Suppose the greatest common divisor of the coefficients of $p(x)$ is $1$. Then $p(x)$ is irreducible in $R[x]$ if and only if it is irreducible in $F[x]$.

In particular, if $p(x)$ is a monic polynomial that is irreducible in $R[x]$, then $p(x)$ is irreducible in $F[x]$. [/theorem]

[proof] ($\implies$) This is just the contrapositive of Gauss’ Lemma.

($\impliedby$) Now suppose that $p(x)$ is reducible in $R[x]$, then $p(x)$ we claim that $p(x)$ is reducible in $F[x]$. The assumption with the greatest common divisor implies non of the coefficients have a common factor. Thus,

implies that neither $a(x)$ nor $b(x)$ are constant. So $p(x) = a(x)b(x)$ is a (nontrivial) factorization in $F[x]$. [/proof]

The last statement of the Corollary needs care. It is NOT true if we drop the UFD assumption and just make $R$ and integral domain.

[example] [extitle]Example %counter%[/extitle]

It is not difficult to show that $\ZZ[2i]$ is an integral domain. Let $p(x) = x^2 + 1 \in \ZZ[2i][x]$. The fraction field of $\ZZ[2i]$ is just $\QQ[i]$ and so $p(x) = (x - i)(x + i) \in Q[i][x]$. So it follows that $p(x)$ is reducible in $F[x]$. [/example]

Irreducibility Criteria

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $F$ be a field and $p(x) \in F[x]$ be of degree 2 or 3. Then $p(x)$ is reducible if and only if it has a root in $F$. [/theorem]

[proof] ($\implies$) If $p(x)$ is reducible, one of its factors is necessary a linear factor. So the claim follows by the Factor Theorem.

($\impliedby$) Same argument but in reverse. [/proof]

[theorem] [thmtitle]Theorem %counter% (Rational roots theorem)[/thmtitle]

Let $R$ be a UFD with $F = \Frac R$. Let

Suppose that $r/s \in F$ is in lowest terms and is a root of $p(x)$. Then

• $r \mid a_0$
• $s \mid a_n$.

In particular, if $p(x)$ is a monic polynomial in $R[x]$ and $p(d) \neq 0$ for every $d \in R$ such that $d\mid a_0$, then $p(x)$ has no roots in $F$. [/theorem]

[proof] Since $r/s$ is a root of $p(x)$, it follows that

Multiplying through by $s^n$ gives

Thus,

So $s$ divides $a_n r^n$. As $s$ does not divide $r$, it follows that $s \mid a_n$. A similar argumet shows that $r\mid a_0$. The last claim follows immediaately by the result we just proved. [/proof]

[example] [extitle]Example %counter%[/extitle]

We claim that $x^3 - 3x - 1$ is irreducible in $\ZZ[x]$. Indeed, Gauss’ Lemma implies that $x^3 - 3x - 1$ is irreducible in $\ZZ[x]$ if and only if it is irreducible in $\QQ[x]$ as the gcd of the coefficients is $1$.

Now since $x^3 - 3x - 1$ is a polynomial of degree three, it is reducible if and only if it has a root in $\QQ[x]$. The rational roots theorem tells us that $\pm 1$ are the only possibilities for roots and neither is a root. So $x^3 - 3x - 1$ is irreducible in $\QQ[x]$ which implies that it is irreducible in $\ZZ[x]$. [/example]

Reducing Coefficients Modulo an Ideal

The technique that is described in the above example is limited because it requires us to look at polynomials of degreee $\le 3$. One fairly useful trick for checking irreducibility involves passing to a quotient ring. Why? Well, suppose we have

in $R[x]$. If $I$ is an ideal of $R$, then modding $R[x]$ out by $I[x]$ and passing to the quotient gives

as the quotient map $R[x] \to R[x]/I[x] \cong (R/I)[x]$ is a ring homomorphism. This implies any factorization in $R[x]$ is also a factorization in $(R/I)[x]$. So, if we can prove irreducibility in $(R/I)[x]$, then we immediately get irreducibility in $R[x]$. Choosing $I$ appropriately can let us obtain irreducibility with ease.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

Let $I$ be a proper ideal of the integral domain $R$ and $p(x)$ be a nonconstant monic polynomial in $R[x]$. If $\overline{p(x)}$ is irreducible in $(R/I)[x]$, then $p(x)$ is irreducible in $R[x]$. [/theorem]

[example] [extitle]Example %counter%[/extitle]

Consider the $p(x) = x^3 + x + 1 \in \ZZ[x]$. We claim that $p(x)$ is irreducible in $\ZZ[x]$.

Indeed, consider $\overline{p(x) \in (\ZZ/2\ZZ)[x]}$. Then

and

Thus, $\overline{p(x)}$ is irreducible in $(\ZZ/2\ZZ)[x]$ so $p(x)$ is irreducible in $\ZZ[x]$. [/example]

[example] [extitle]Example %counter%[/extitle]

Consider the $p(x) = x^5 + x^2 + 1 \in \ZZ[x]$. We claim that $p(x)$ is irreducible in $\ZZ[x]$.

Indeed, consider $\overline{p(x) \in (\ZZ/2\ZZ)[x]}$. Then

and

Thus, $\overline{p(x)}$ is irreducible in $(\ZZ/2\ZZ)[x]$ so $p(x)$ is irreducible in $\ZZ[x]$. [/example]

Eisenstein’s Criterion

Given a polynomial

such that $a_{n-1}, a_{n-2}, \ldots, a_1, a_0 \in P$ where $P \in \Spec R$ and $a_0 \notin P^2$, then $f(x)$ is irreducible in $R[x]$. This idea originates from the fact that if $f(x)$ did factor into $g(x)h(x)$. Writing $g(x)$ and $h(x)$ as

then

Now, if we assume that $c_0 \notin P$, then $b_0$ MUST be in $P$. Similarly, $b_1, \ldots, b_r$ are all in $P$. So then $b_r$ cannot be a unit and so $b_rs_t \neq 1$, contradicting the claim that $f(x)$ factors nontrivially. So if we assume that $a_0 \notin P^2$, then $f(x)$ cannot be irreducible. This is known as Eisenstein’s criterion.

[theorem] [thmtitle]Theorem %counter% (Eisenstein’s Criterion)[/thmtitle]

Let $R$ be an integral domain and $P \in \Spec R$. If

is a polynomial in $R[x]$ of degree $\le 1$ such that $a_{n-1}, a_{n-2}, \ldots, a_1, a_0 \in P$ but $a_0 \notin P^2$, then $f(x)$ is irreducible in $R[x]$. [/theorem]

There is a special case of this theorem to which Eisenstein is most commonly applied.

[theorem] [thmtitle]Corollary %counter% (Eisenstein’s Criterion)[/thmtitle]

Let $p$ be a prime integer. If

is a polynomial in $\ZZ[x]$ of degree $\le 1$ such that $p\mid a_{n-1}, a_{n-2}, \ldots, a_1, a_0$ but $p^2\not\mid a_0$, then $f(x)$ is irreducible in both $\ZZ[x]$ and $\QQ[x]$. [/theorem]

[proof] Since $\ZZ[x]$ is a PID, its prime ideals are generated by prime elements (as PIDs are UFDs). Thus, we obtain the statement of Eisenstein specialized to $\ZZ[x]$. Furthermore, we know that irreducibility of a monic polynomial in $\ZZ[x]$ is equivalent to irreducibility in $\QQ[x]$. [/proof]

[example] [extitle]Example %counter%[/extitle]

The polynomial $x^4 + 10x + 5 \in \ZZ[x]$ is irreducible by Eisenstein by choosing $p = 5$. [/example]

[example] [extitle]Example %counter%[/extitle]

The polynomial $x^n - a \in \ZZ[x]$ for $n \ge 1$ is irreducible by Eisenstein whenever $a$ is divisible by $p$ but not $p^2$.
[/example]

[example] [extitle]Example %counter% (the $x \mapsto x + 1$ trick)[/extitle]

The polynomial $f(x) = x^4 + 1 \in \ZZ[x]$ is irreducible. It is obvious that $x^4 + 1$ can’t have any linear factors, but it’s not nearly as obvious if it has quadratic factors.

We begin by considering

Applying Eisenstein with $p = 2$ shows that $f(x + 1)$ is irreducible. This implies that $f(x)$ is irreducible as any factorization $f(x) = g(x)h(x)$ is a factorization of $f(x + 1)$ by noting that $f(x+1) = g(x+1)h(x+1)$. Thus, $f(x)$ is irreducible. [/example]

[example] [extitle]Example %counter% (Cyclotomic polynomials)[/extitle]

The polynomial $\Phi_p(x) = \frac{x^p - 1}{x - 1} \in \ZZ[x]$ is irreducible by applying the shift trick. Indeed,

Clearly, $p$ divides all the coefficients (but the first) yet $p^2 \not\mid p$. So $\Phi_p(x+1)$ is irreducible which implies that $\Phi_p(x)$ is as well. [/example]

Worked Exercises

[example] [extitle]Dummit and Foote Exercise 9.2.1[/extitle]

Let $F$ be a field and let $f \in F[x]$ be a polynomial of degree $n \ge 1$. Prove that $\{\overline{1}, \overline{x}, \ldots, \overline{x^{n-1}}\}$ is a basis for $F[x]/\generator{f}$ as an $F$-vector space. [/example]

[proof] (Spanning) Let $g\in F[x]$. Then

by the division algorithm where $\deg r < \deg f$ or $\deg r = 0$. Thus,

Clearly $r$ is in the span of the proclaimed basis.

(Linear independence) Suppose that

Then $\sum_{i=0}^{n-1} a_i x^{i}$ is an element of $\generator{f}$. In other words, $f$ divides this polynomial. But since $f$ is degree $n$, this polynomial must be the zero polynomial. [/proof]

[example] [extitle]Dummit and Foote Exercise 9.2.5[/extitle]

Let $F$ be a field. Find all the ideals in the ring $F[x]/\generator{p}$ where $F$ is a field and $p$ is a polynomial in $F[x]$. [/example]

[solution] By the ideal correspondence theorem, the ideals of $F[x]/\generator{p}$ are in bijection with the ideals of $F[x]$ containing $\generator{p}$. Since $F[x]$ is a UFD, this means that the ideals containing $\generator{p}$ are generated by the factors of $p$. [/solution]

[example] [extitle]TODO: Cite[/extitle]

Let $F$ be a field. Show that the polynomial ring $F[x]$ contains infinitely many irreducible elements. [/example]

[proof] For the sake of contradiction, suppose that there were finitely many irreducibles

Now we consider the polynomial

Notice that this implies that

Since $F[x]$ is a UFD, $Q(x)$ factors into irreducibles. One of the factors, say $p_i(x)$ pops up in $p_1(x)p_2(x)\cdots p_r(x)$ so $p_i(x)$ divides the left hand side of the expression. On the other hand, $p_i(x)$ certainly fails to divide the right-hand side of the expression so the claim follows. [/proof]

[example] [extitle]Dummit and Foote Exercise 9.3.4[/extitle]

Let $R = \ZZ + x\QQ[x]$ be the set of polynomials in $x$ with rational coefficients whose constant term is an integer.

1. Prove that $R$ is an integral domain and its units are $\pm 1$.
2. Show that the irreducibles in $R$ are $\pm p$ where $p$ is a prime in $\ZZ$ and the polynomials $f(x)$ that are irreducible in $\QQ[x]$ and have constant term $\pm 1$. Prove that these irreducibles are prime in $R$.
3. Show that $x$ cannot be written as the product of irreducibles in $R$ (in particular, $x$ is not irreducible) and conclude that $R$ is not a U.F.D.
4. Show that $x$ is not a prime in $R$ and describe the quotient ring $R/\generator{x}$. [/example]

[proof] (1) Suppose that $p(x), q(x) \in R$ such that

This would imply that the constant terms multiply $0$ and so we would contradict $\ZZ$ being an integral domain. As for the units, clearly $\pm 1$ are units. To see that they are the only units, suppose $f(x), g(x) \in R$ such that

Then the constant terms must multiply to $1$ which means those are both $\pm 1$. Furthermore, since $R$ is an integral domain $\deg fg = \deg f + \deg g$ so $\deg f = \deg g = 0$, proving the desired claim.

(2) For $\pm p$, we note that if $p = a(x)b(x)$ in $R[x]$, we first must have $a$ and $b$ be constants as $R$ being an integral domain implies $\deg p = \deg a + \deg b$. But $\deg p = 0$ so $a$ and $b$ are constants. Since primes are irreducible in $\ZZ$, it follows that either $a$ or $b$ is a unit as desired.

Now suppose that $f(x)$ is irreducible in $\QQ[x]$ and has constant term $\pm 1$. If we factor $f(x)$ nontrivially into

in $R[x]$. Now, since we assume that the constant term of $f(x)$ is $\pm 1$, the constant terms of $a(x)$ and $b(x)$ are forced to be $\pm 1$. We can rule out the cases where $a(x) = \pm 1$ or $b(x) = \pm 1$ because we claimed to have factored nontrivially. Then we see that $f(x)$ is reducible in $\QQ[x]$, which contradicts our assumption that $f(x)$ is irreducible in $\QQ[x]$.

It is straightforward to see that these are the only irreducibles, so we skip the verification of that.

We conclude by showing that these irreducibles are prime. Suppose that

Then it follows that $p$ divides the constant term of $a(x)$ or $b(x)$. So either $p\mid a(x)$ or $p \mid b(x)$. As for $f(x) \mid a(x)b(x)$, we can pass up to factorization in $\QQ[x]$ and then show that the resulting factorization can be changed (similar to Gauss’ Lemma) if necessary. We will omit the proof of this argument.

(3) Follows immediately by Part 2.

(4) To see that note that $x$ is not prime, if we write $x$ as $x = \frac{1}{2}x \cdot 2$, it follows that $x$ does not divide either factor in $R$. So $x$ is not prime. As for $R/\generator{x}$, we claim that $R/\generator{x}$ consists of polynomials of the form $a + bx$ where $b \in [0, 1] \cap \QQ$ and $a \in \ZZ$ which we omit the proof for. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Factor the following polynomials over the indicated rings (some may be irreducible).

1. $f(x) = x^3 + x + 1$ over $(\ZZ/3\ZZ)[x]$.
2. $g(x) = x^3 + x + 1$ over $(\ZZ/5\ZZ)[x]$.
3. $h(x) = x^4 + 1$ over $(\ZZ/2\ZZ)[x]$.
4. $u(x) = x^4 + 1$ over $(\ZZ/3\ZZ)[x]$.
5. $v(x) = \dfrac{(x+2)^p - 2^p}{x}$ over $\QQ[x]$ where $p$ is an odd prime. [/example]

[solution] (1) Since $f(x)$ is a polynomial of degree $3$, it is reducible if and only if it has a root. So we just compute $f(0), f(1), f(2)$ modulo $3$:

So $1$ is a factor. Computing via long division (which is fine as $\ZZ/3\ZZ$ is a field) gives

And then hand-verification shows that $x^2 + x + 2$ is irreducible.

(2) Hand computation shows that $g(x)$ is irreducible.

(3) It is straightforward to see that $x^4 + 1 = (x + 1)^4$.

(4) It is straightforward to show that $u(x)$ has no roots. Solving a system of equations in $\ZZ/3\ZZ$ shows that

(5) Apply Eisenstein with $p$ as the prime. Irreducibility follows immediately. [/solution]