Polynomials on Steroids

The whole abstract algebra endeavour revolves around polynomials and their roots. Eventually, we want to prove the non-existence of a “quintic” formula, but for now we should acquaint ourselves with the basics of polynomials.

Let $R$ be an integral domain with multiplicative unit $1$ .

Lemma 1. For any set $K$ , equip $\mathcal F(K, R)$ with point-wise addition and multiplicative operations: for any $x \in K$ ,

$\begin{aligned} (f+g)(x) &:= f(x) + g(x),\\ (f \cdot g)(x) &:= f(x) \cdot g(x). \end{aligned}$

Then $\mathcal F(K, R)$ forms a ring with multiplicative unit $1$ . Furthermore, for any $u \in R$ , define $u(\cdot) \in \mathcal F(K, R)$ by $u(x) = u$ for any $x \in K$ . Then the map $\iota : R \to \mathcal F(K, R)$ defined by $\iota(u) = u(\cdot)$ is an injective homomorphism, and

$R \cong \iota(R) \subseteq \mathcal F(K, R)$

as a subring. We then abbreviate $u(\cdot) \equiv u$ .

Proof. Book-keeping.

Definition 1. Define the monomials $\mu_n : R \to R$ by $\mu_0 := 1$ ,

$\mu_1(x) := x,\quad x \in R,$

and for $n \in \mathbb N_0$ , $\mu_{n+1} := \mu_n \cdot \mu$ . Note that $\mu_{m+n} = \mu_m \cdot \mu_n$ inductively. Define a polynomial as a function $f \in \mathcal F(R, R)$ by

$\displaystyle f = c_0 + c_1 \mu + \cdots + c_n \mu^n.$

for some constants $c_0,\dots, c_n$ with $c_n \neq 0$ . To explicitly indicate the argument being plugged into the polynomial $f \equiv f(x)$ , we write

$\displaystyle f(x) = c_0 + c_1 x + \cdots + c_n x^n \equiv \sum_{k=0}^n c_k x^k.$

In this case, define $\deg(f) = n$ , and if $f = 0$ , define $\deg(f) = -\infty$ for convenience. We note that $\deg(f) = 0$ if and only if $f$ is a nonzero constant, i.e. $f = c$ for some $c \neq 0$ . We say that $f$ is monic if $c_n = 1$ . Define

$R[x] := \{ f \in \mathcal F(R,R): f\ \text{is a polynomial}\} \subseteq \mathcal F(R,R).$

The ring of polynomials are not just convenient—they form an integral domain. Furthermore, if the coefficients of the polynomial come from a field $\mathbb F$ , the ring of polynomials $\mathbb F[x]$ is the most convenient ring that we could have hoped for.

Theorem 1. $R[x]$ is an integral domain. Furthermore, for any field $\mathbb F$ , $\mathbb F[x]$ is a Euclidean domain.

Proof. Let $f,g \in R[x]$ be non-zero polynomials defined by

$\displaystyle f(x) = \sum_{i=0}^m c_i x^i, \quad g(x) = \sum_{j = 0}^n d_j x^j,$

so that $c_m \neq 0$ and $d_n \neq 0$ . Then

$\begin{aligned} (f \cdot g)(x) &= \left( \sum_{i=0}^m c_i x^i \right) \left( \sum_{j = 0}^n d_j x^j \right) \\ &= \sum_{i=0}^m \sum_{j = 0}^n c_id_j x^{i+j} \\ &= \sum_{k=0}^{m+n} \left( \sum_{i = 0}^k c_id_{k-i} \right) x^{k}. \end{aligned}$

Since $R$ is an integral domain, the coefficient of $x^{m+n}$ is nonzero:

$\displaystyle \sum_{i = 0}^{m+n} c_id_{k-i} = c_m d_n \neq 0.$

Therefore, $\deg(f \cdot g) = \deg(f) + \deg(g) \geq 0$ , so that $f \cdot g \neq 0$ . In fact, we claim that $\deg : \mathbb F[x] \to \mathbb N$ is the required norm function such that $\mathbb F[x]$ forms a Euclidean domain.

We prove the existence requirement by induction on $n$ . For $n = 0$ , $f,g$ are nonzero constants, and hence we have $g = (f/g) \cdot f + 0$ .

Suppose the division algorithm holds whenever $\deg(g) = k$ . We need to show that it holds for $n = \deg(g) = k +1$ . If $m > n$ , then

$g = 0 \cdot f + g, \quad \deg(g) < \deg(f).$

Hence, suppose $m \leq n$ . Define the polynomial $h \in \mathbb F[x]$ by

$\displaystyle h(x) = \frac {d_n}{c_m} \cdot x^{n-m} \cdot f(x) = \sum_{i=0}^m \frac{c_i \cdot d_n}{c_m} \cdot x^{n-m+i}.$

Note that $h$ has degree $n$ and the coefficient of $x^n$ is $d_n$ . Therefore, $g-h$ has degree $\leq n-1$ . By the induction hypothesis, there exist unique polynomials $q, r \in \mathbb F[x]$ such that

$g - h = q \cdot f + r,\quad \deg(r) < \deg(f).$

By algebruh,

$\displaystyle g(x) = \left(q(x) + \frac {d_n}{c_m} \cdot x^{n-m}\right) \cdot f(x) + r(x),\quad \deg(r) < \deg(f),$

as required.

For uniqueness, consider two outcomes from the division algorithm:

$g = q_1 \cdot f + r_1 = q_2 \cdot f + r_2, \quad \deg(r_k) < \deg(f).$

Then $(q_1 - q_2) \cdot f = r_2 - r_1$ . Then $f \mid (r_2 - r_1)$ . If $r_2 - r_1 \neq 0$ , then

$\deg(r_2 - r_1) \leq \max \{r_1, r_2\} < \deg(f) < \deg(r_2 - r_1),$

a contradiction. Therefore, $r_1 = r_2$ , so that $q_1 \cdot f = q_2 \cdot f$ . Since $\mathbb F[x]$ is an integral domain, $q_1 = q_2$ , establishing uniqueness, as required.

Since $\mathbb F[x]$ is a Euclidean domain, it is a UFD. In particular, unique factorisation holds in $\mathbb F[x]$ . We will use $\mathbb F[x]$ heavily to construct field extensions.

Now suppose $R$ is a UFD with fraction field $\mathbb F_R:= Q(R)$ .

Lemma 1. For any $r$ , $r \in R$ is irreducible if and only if $r(\cdot) \in R[x]$ is irreducible.

Proof. We prove $(\Rightarrow)$ and leave $(\Leftarrow)$ as an exercise. Suppose $r \in R$ is irreducible and $r(\cdot) = f(\cdot) g(\cdot)$ . Since $r \neq 0$ and

$0 \leq \deg(f) + \deg(g) = \deg(f \cdot g) = \deg(r) = 0,$

we have $\deg(f) = \deg(g) = 0$ , so that $f(\cdot) \equiv f,g(\cdot) \equiv g$ are also nonzero constants. Since $r$ is irreducible, either $f \equiv f(\cdot)$ or $g \equiv g(\cdot)$ is a unit. Hence, $r(\cdot)$ is irreducible, as required.

Given $c_0,c_1,\dots$ , inductively define

$\gcd(c_0,c_1,\dots, c_{k+1}):= \gcd(\gcd(c_0,c_1,\dots, c_k), c_{k+1}),$

where $\gcd(c_0, c_1)$ is defined uniquely up to multiplication by a unit.

Definition 2. Define the content of a polynomial $f = \sum_{i=0}^n c_i \mu_i \in R[x]$ by

$c(f):= \gcd(c_0,c_1,\dots, c_n),$

which can be shown to be unique up to multiplication by a unit. We call $f$ primitive if there exists a unit $u \in R$ such that $c(f) = u$ .

Lemma 2. If $f,g \in R[x]$ are primitive, so is $f \cdot g \in R[x]$ . More generally, given non-constant polynomials $f,g \in \mathbb F_R[x]$ , $c(f \cdot g) = c(f) \cdot c(g)$ .

Proof. Given primitive polynomials $f = \sum_{i=0}^m c_i \mu_i$ and $g = \sum_{j=0}^n d_j \mu_j$ , recall that

$\displaystyle f \cdot g = \sum_{k=0}^{m+n} \left( \sum_{i+j = k} c_i d_j \right) \mu_k.$

We proceed by contradiction. Suppose $f \cdot g$ is not primitive. Then

$\gcd\{ c_i d_j : i+j = k\} = c(f \cdot g) \neq u$

for any unit $u$ .

Since $R$ is a UFD, there exists an irreducible element $p \in R$ such that $p \mid c(f \cdot g)$ . If $p \mid c_i$ for all $i$ , then $p \mid c(f)$ and $f$ is not primitive, a contradiction. Therefore, let $r$ denote the smallest $i$ such that $p \nmid c_i$ . Similarly, let $s$ denote the smallest $j$ such that $p \nmid d_j$ . By expanding the coefficient of $\mu_{r+s}$ , we obtain

$\displaystyle c(f \cdot g) = \cdots + c_{r+1}d_{s-1} + c_r d_s + c_{r-1}d_{s+1} + \cdots.$

By definition, $p \mid d_j$ for $j \leq s-1$ and $p \mid c_i$ for $i \leq r-1$ . Therefore,

$p \mid (c(f \cdot g) - c_r d_s).$

Together with $p \mid c(f \cdot g)$ , we must have $p \mid c_r d_s$ , so that $p \mid c_r$ or $p \mid d_s$ , a contradiction in either way.

Theorem 2. Fix $f \in R[x]$ . If $f$ is reducible in $\mathbb F_R[x]$ , then $f$ is reducible in $R[x]$ . The converse holds if $f$ is primitive.

Proof. We first note that for any polynomial $f \in R[x]$ , there exists a unique primitive polynomial $f_0 \in R[x]$ such that $f = c(f) \cdot f_0$ . Hence, if $f = g \cdot h$ for $g,h \in R[x]$ , then

$\begin{aligned} f &= g \cdot h \\ c(f) \cdot f_0 &= ( c(g) \cdot g_0 ) \cdot ( c(h) \cdot h_0 ) \\ c(g \cdot h) \cdot f_0 &= (c(g) \cdot c(h)) \cdot ( g_0 \cdot h_0 ). \end{aligned}$

By Lemma 2, $g_0 \cdot h_0$ is primitive too, and $c(f_0) = c(g_0 \cdot h_0) = 1$ without loss of generality, so that $c(g \cdot h) = c(g) \cdot c(h)$ .

Suppose $f$ is reducible in $\mathbb F_R[x]$ . Write $f = g \cdot h$ for non-units $g, h \in \mathbb F_R[x]$ , where

$\displaystyle g = \sum_{i=0}^m \frac{ c_i }{ u_i } \cdot \mu_i,\quad h = \sum_{j=0}^n \frac{ d_j }{ v_j } \cdot \mu_j,$

and $u_i, v_j \in R$ are irreducible without loss of generality. Define

$\begin{aligned} u &:= u_0 u_1 \cdot \dots \cdot u_m, \\ v&:= v_0 v_1 \cdot \dots \cdot v_n \end{aligned}$

so that $uvf = ug \cdot vh \in R[x]$ . Define the non-units $g_u := ug, h_v := vg \in R[x]$ . Now $f = c(f) \cdot f_0$ , where $f_0$ is primitive. We remark that in a similar manner, $g_{u,0}, h_{v,0}$ are primitive non-units. Therefore,

$\begin{aligned} uv \cdot c(f) \cdot f_0 &= uv f \\ &= g_u \cdot h_v \\ &= c(g_u) \cdot g_{u,0} \cdot c(h_v) \cdot h_{v,0} \\ &= c(g_u) \cdot c(h_v) \cdot g_{u,0}\cdot h_{v,0} \\ &= c(g_u \cdot h_v) \cdot g_{u,0}\cdot h_{v,0} \\ &= c(uvf) \cdot g_{u,0}\cdot h_{v,0} \\ &= uv \cdot c(f) \cdot g_{u,0}\cdot h_{v,0}. \end{aligned}$

Therefore, $f_0 = g_{u,0} \cdot h_{v_0}$ . Hence,

$f = c(f) \cdot f_0 = c(f) \cdot g_{u,0} \cdot h_{v,0},$

so that $f \in R[x]$ is reducible.

For the converse claim, suppose $f \in R[x]$ is reducible. Write $f = g \cdot h$ for non-units $g, h \in R[x]$ . If $f$ is primitive, then $c(g) \cdot c(h) = c(g \cdot h) = c(f)$ is a unit, so that $c(g), c(h)$ are units, and hence, $g,h$ are primitive non-units. Therefore, $g,h$ are primitive non-units in $\mathbb F_R[x]$ , and so $f$ is reducible in $\mathbb F_R[x]$ .

Why do we care about irreducible polynomials?

Example 1. The polynomial $x^2 + 1 \in \mathbb R[x]$ is irreducible since it has no real roots. By Theorem 1, $\mathbb R[x]$ is a PID. Therefore, the prime ideal $\langle x^2 + 1 \rangle \subseteq \mathbb R[x]$ is maximal, so that $\mathbb R[x]/\langle x^2 + 1 \rangle$ forms a field. In fact,

$\mathbb R[x]/\langle x^2 + 1 \rangle \cong \mathbb C.$

Hence, we can represent $i := x + \langle x^2 + 1 \rangle$ .

Proof. Define the ring homomorphism $\phi : \mathbb R[x] \to \mathbb C$ by

$\displaystyle \phi\left( \sum_{k=0}^n c_k x^k\right) := \sum_{k=0}^n c_k i^k.$

The ring homomorphism is surjective because

$\begin{aligned} \phi(a + bx) &= a + bi. \end{aligned}$

To evaluate $\ker(\phi)$ , fix $f \in \ker(\phi)$ . Use the division algorithm to obtain polynomials $q, r \in \mathbb R[x]$ such that

$f(x) = q(x) \cdot (x^2 + 1) + r(x),$

where $r(x) = a +bx$ . Applying $\phi$ on both sides, we leave it as an exercise to check that

$\begin{aligned} 0 = \phi(f) &= a + bi. \end{aligned}$

Hence, $a = b = 0$ implies that $r = 0$ , so that

$f(x) = q(x) \cdot (x^2 +1) \in \langle x^2 + 1 \rangle.$

Therefore, $\ker(\phi) = \langle x^2 + 1 \rangle$ . By the first isomorphism for rings,

$\mathbb R[x]/\langle x^2 + 1 \rangle = \mathbb R[x]/{\ker(\phi)} \cong \phi(\mathbb R[x]) = \mathbb C.$

Hence, irreducible polynomials help us construct field extensions. Here, $\mathbb R \subseteq \mathbb C$ as fields, so we say that $\mathbb C$ is a field extension of $\mathbb R$ .

Remark 1. In principle, then, we could have defined $\mathbb C := \mathbb R[x]/\langle x^2 + 1 \rangle$ and then recover the matrix representation of $\mathbb C$ as a theorem. However, since we needed to use $\mathbb C$ for more elementary purposes,

Let $\mathbb F$ be a field.

Corollary 1. If $f \in \mathbb F[x]$ is irreducible, then $\mathbb F[x]/\langle f(x) \rangle$ forms a field extension of $\mathbb F$ .

Proof. Follow the discussion in Example 1.

Remark 2. Fix $g := \sum_{k=0}^n c_k x^k \in \mathbb F[x]$ . Since each $c_k$ belongs to $\mathbb F \subseteq \mathbb F[x]/\langle f(x) \rangle$ , we can regard $g \in (\mathbb F[x]/\langle f(x) \rangle)(x)$ . Then for each

$u(x) + \langle f(x)\rangle \in \mathbb F[x]/\langle f(x) \rangle,$

we can evaluate

$\begin{aligned} g(u(x) + \langle f(x)\rangle) &= \sum_{k=0}^n c_k (u(x) + \langle f(x)\rangle)^k \\ &= \sum_{k=0}^n c_k u(x)^k + \langle f(x)\rangle \\ &= g(u(x)) + \langle f(x)\rangle. \end{aligned}$

Therefore, $g(u(x) + \langle f(x) \rangle) = \langle f(x)\rangle$ if and only if $g(u(x)) \in \langle f(x) \rangle$ . Denoting the additive identity of $\mathbb F[x]/\langle f(x) \rangle$ by $0:= \langle f(x) \rangle$ , we have

$g(u(x)) \in \langle f(x) \rangle \quad \iff \quad g(u(x) + \langle f(x)\rangle) = 0.$

Hence, $u(x) + \langle f(x) \rangle$ is a root of $f$ if and only if $f(u(x)) \in \langle f(x) \rangle$ . In particular, by setting $u(x) = x$ ,

$\alpha := x + \langle f(x) \rangle = u(x) + \langle f(x) \rangle$

will always be a root of $f$ , so that $f(\alpha) = \langle f(x) \rangle$ . Since $\langle f(x) \rangle$ is the additive identity $0 \in \mathbb F[x]/\langle f(x) \rangle$ , we can write $f(\alpha) = 0$ i.e. $\alpha$ is a root of $f$ .

Corollary 2. For any non-constant polynomial $g \in \mathbb F[x]$ , there exists an irreducible polynomial $f \in \mathbb F[x]$ and some root $\alpha \in \mathbb F[x]/\langle f(x) \rangle$ of $g$ .

Proof. If $g$ is reducible, compute an irreducible factor $f$ . Therefore, suppose the case $g=f$ is irreducible. Since $f$ is non-constant, $\deg(f) \geq 1$ . If $\deg(f) = 1$ , write $f(x) = a_0 + a_1x$ for $a_1 \neq 0$ . Then set

$\alpha := -a_0/a_1 \in \mathbb F \subseteq \mathbb F[x]/\langle f(x) \rangle.$

If $\deg(f) \geq 2$ , set

$\alpha := x + \langle f(x) \rangle \in \mathbb F[x]/\langle f(x) \rangle$

and follow the discussion in Remark 1.

It remains for us to determine the various conditions for which a polynomial $f \in \mathbb F[x]$ is irreducible.

Lemma 3. Fix $f \in \mathbb F[x]$ . If $\deg(f) \in \{2, 3\}$ , then $f$ is reducible in $\mathbb F[x]$ if and only if $f$ has a root in $\mathbb F$ .

Proof. $(\Rightarrow)$ Write $f = g \cdot h$ for non-units $g, h$ , so that

$\min\{ \deg(g), \deg(h) \} \geq 1.$

Since $\deg(g) + \deg(h) =\deg(f) \in \{2, 3\}$ , $1 \in \{ \deg(g), \deg(h) \}$ . Suppose $\deg(g) = 1$ without loss of generality. Then $g(x) = x-\alpha$ for some $\alpha \in \mathbb F$ , and

$f(x) = g(x) \cdot h(x) = (x-\alpha) \cdot h(x).$

Hence, $f$ has a root $\alpha$ in $\mathbb F$ . The converse holds by the factor theorem and a similar case-by-case argument.

What about polynomials with higher degrees?

Theorem 3 (Eisenstein’s Criterion). Let $p$ be a prime number and fix a polynomial

$\displaystyle f(x) = \sum_{k=0}^n c_k x^k \in \mathbb Z[x] \subseteq \mathbb Q[x],\quad n \geq 2.$

Suppose all of the following hold:

$p \mid c_i$ for $0 \leq i < n$ .
$p \nmid c_n$ .
$p^2 \nmid c_0$ .

Then $f$ is irreducible in both $\mathbb Z[x]$ and $\mathbb Q[x]$ .

Proof. By the first part of Theorem 2, since $\mathbb Q = \mathbb F_{\mathbb Z}$ , it suffices to check that $f$ is irreducible in $\mathbb Z[x]$ . Suppose for a contradiction that $f$ is reducible in $\mathbb Z[x]$ . Then there exist non-units $g, h \in \mathbb Z[x]$ such that $f = g \cdot h$ . Write

$\displaystyle g(x) = \sum_{i=0}^r u_i x^i,\quad h(x) = \sum_{j=0}^s v_j x^j.$

Since $p \nmid c_n = u_r \cdot v_s$ , $p \nmid u_r$ and $p \nmid v_s$ . Recall that

$\displaystyle f(x) = \sum_{k=0}^n \left( \sum_{i+j = k} u_i v_j \right) x^k.$

In particular, since $c_0 = u_0 \cdot v_0$ , $p \mid u_0 \cdot v_0$ . Suppose $p \mid u_0$ without loss of generality. Since $p^2 \nmid c_0$ , we have $p \nmid v_0$ . Then

$p \mid (c_1 - u_0 v_1) = u_1 v_0,$

so that $p \mid u_1$ . Inductively, we can show that $p \mid u_i$ for $0 \leq i \leq r < n$ . Therefore, $p \mid u_r$ , a contradiction.

Lemma 4. For any polynomial $f$ and $v \in R$ , $f \in R[x]$ is reducible if and only if $f(\cdot + v) \in R[x]$ is reducible.

Proof. $(\Rightarrow)$ Writing $f = g \cdot h$ , we have $f(\cdot + v) = g(\cdot + v ) \cdot h( \cdot + v)$ .

$(\Leftarrow)$ Writing $f_0 := f(\cdot + v) = g \cdot h$ ,

$f = f_0(\cdot - v) = g(\cdot - v) \cdot h( \cdot - v).$

Example 2. For any prime number $p > 2$ , it is obvious that $1$ is a root of $x^p - 1$ , so that $(x-1) \mid (x^p - 1)$ by the factor theorem. Define the $p$ -th cyclotomic polynomial by

$\displaystyle \Phi_p(x) := \frac{x^p - 1}{x-1} \in \mathbb Z[x].$

Then $\Phi_p$ is irreducible in $\mathbb Z[x]$ .

Proof. By Lemma 4, $\Phi_p$ is irreducible in $\mathbb Z[x]$ if and only if $\Phi_p(\cdot +1)$ is irreducible in $\mathbb Z[x]$ . By the binomial theorem,

$\begin{aligned} \Phi_p(x + 1) &= \frac{(x+1)^p - 1}{(x+1)-1} = \sum_{i=0}^{p-1} {p \choose i+1} x^{i}.\end{aligned}$

Now $\displaystyle p \mid {p \choose i}$ for each $0 \leq i \leq p-2$ , the coefficient $c_{p-1}$ of $x^{p-1}$ is $1$ , so that $p \nmid c_{p-1}$ . Furthermore, $\displaystyle c_0 = {p \choose 1} = p$ so that $p^2 \nmid p$ . By Eisenstein’s criterion, $\Phi_p(\cdot + 1)$ is irreducible, as required.

Next time, we shall upgrade our discussion one more time: from rings to fields. A field $\mathbb F$ is basically a ring such that $(\mathbb F, +)$ and $(\mathbb F^*, \cdot)$ form Abelian groups. Fields are, in a sense, the star of the show, and rings afforded us the elementary vocabulary to discourse fields.

Therefore, we’re going on a field trip!

—Joel Kindiak, 3 Feb 26, 2053H

KindiakMath

Polynomials on Steroids

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Polynomials on Steroids

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