2.3 Multiple roots

Even when polynomials in $F[X]$ have no common factor in $F[X]$ , one might expect that they could acquire a common factor in $\Omega[X]$ for some $\Omega\supset F$ . In fact, this doesn’t happen — greatest common divisors don’t change when the field is extended.

Proposition 2.10

Let $f$ and $g$ be polynomials in $F[X]$ , and let $\Omega$ be an extension of $F$ . If $r(X)$ is the gcd of $f$ and $g$ computed in $F[X]$ , then it is also the gcd of $f$ and $g$ in $\Omega[X]$ . In particular, distinct monic irreducible polynomials in $F[X]$ do not acquire a common root in any extension of $F.$

Proof.

Let $r_{F}(X)$ and $r_{\Omega}(X)$ be the greatest common divisors of $f$ and $g$ in $F[X]$ and $\Omega[X]$ respectively. Certainly $r_{F}(X)|r_{\Omega}(X)$ in $\Omega[X]$ , but Euclid’s algorithm (1.8) shows that there are polynomials $a$ and $b$ in $F[X]$ such that

a(X)f(X)+b(X)g(X)=r_{F}(X),

and so $r_{\Omega}(X)$ divides $r_{F}(X)$ in $\Omega[X]$ .

For the second statement, note that the hypotheses imply that $\gcd(f,g)=1$ (in $F[X]$ ), and so $f$ and $g$ can’t acquire a common factor in any extension field. _□

The proposition allows us to speak of the greatest common divisor of $f$ and $g$ without reference to a field.

Let $f\in F[X]$ . Then $f$ splits into linear factors

equation (4) (4)

f(X)=a\prod_{i=1}^{r}(X-\alpha_{i})^{m_{i}},\text{ }\alpha_{i}\text{ distinct, }m_{i}\geq 1\text{, }\sum_{i=1}^{r}m_{i}=\deg(f),

in $E[X]$ for some extension $E$ of $F$ (see 2.4). We say that $\alpha_{i}$ is a root of $f$ of multiplicity $m_{i}$ in $E$ . If $m_{i}>1$ , then $\alpha_{i}$ is said to be a multiple root of $f$ , and otherwise it is a simple root.

I claim that the unordered sequence of integers $m_{1},\ldots,m_{r}$ in (4) is independent of the extension $E$ chosen to split $f$ . Certainly, it is unchanged when $E$ is replaced with its subfield $F[\alpha_{1},\ldots,\alpha_{r}]$ , and so we may suppose that $E$ is a splitting field for $f$ . Let $E$ and $E^{\prime}$ be splitting fields for $F$ , and suppose that $f(X)=a\prod_{i=1}^{r}(X-\alpha_{i})^{m_{i}}$ in $E[X]$ and $f(X)=a^{\prime}\prod_{i=1}^{r^{\prime}}(X-\alpha_{i}^{\prime})^{m_{i}^{\prime}}$ in $E^{\prime}[X]$ . Let $\varphi\colon E\rightarrow E^{\prime}$ be an $F$ -isomorphism, which exists by (2.7b), and extend it to an isomorphism $E[X]\rightarrow E^{\prime}[X]$ by sending $X$ to $X$ . Then $\varphi$ maps the factorization of $f$ in $E[X]$ onto a factorization

f(X)=\varphi(a)\prod_{i=1}^{r}(X-\varphi(\alpha_{i}))^{m_{i}}

in $E^{\prime}[X]$ . By unique factorization, this coincides with the earlier factorization in $E^{\prime}[X]$ up to a renumbering of the $\alpha_{i}$ . Therefore $r=r^{\prime}$ , and

\{m_{1},\ldots,m_{r}\}=\{m_{1}^{\prime},\ldots,m_{r}^{\prime}\}.

We say that $f$ has a multiple root when at least one of the $m_{i}>1$ , and that $f$ has only simple roots when all $m_{i}=1$ . Thus “ $f$ has a multiple root” means “ $f$ has a multiple root in one, hence every, extension of $F$ splitting $f$ ”, and similarly for $``f$ has only simple roots”.

We wish to determine when a polynomial has a multiple root. If $f$ has a multiple factor in $F[X]$ , say $f=\prod f_{i}(X)^{m_{i}}$ with some $m_{i}>1$ , then obviously it will have a multiple root. If $f=\prod f_{i}$ with the $f_{i}$ distinct monic irreducible polynomials, then Proposition 2.10 shows that $f$ has a multiple root if and only if at least one of the $f_{i}$ has a multiple root. Thus, it suffices to determine when an irreducible polynomial has a multiple root.

Example 2.11

Let $F$ be of characteristic $p\neq 0$ , and assume that $F$ contains an element $a$ that is not a $p$ th-power, for example, $a=T$ in the field $\mathbb{F}{}_{p}(T).$ Then $X^{p}-a$ is irreducible in $F[X]$ , but by 1.4 we have $X^{p}-a=(X-\alpha)^{p}$ in its splitting field. Thus an irreducible polynomial can have multiple roots.

The derivative of a polynomial $f(X)=\sum a_{i}X^{i}$ is defined to be $f^{\prime}(X)=\sum ia_{i}X^{i-1}$ . The usual rules for differentiating sums and products still hold, but note that in characteristic $p$ the derivative of $X^{p}$ is zero.

Proposition 2.12

For a nonconstant irreducible polynomial $f$ in $F[X]$ , the following statements are equivalent:

(a)

$f$ has a multiple root;
(b)

$\gcd(f,f^{\prime})\neq 1$ ;
(c)

$F$ has nonzero characteristic $p$ and $f$ is a polynomial in $X^{p}$ ;
(d)

all the roots of $f$ are multiple.

Proof.

(a) $\Rightarrow$ (b). Let $\alpha$ be a multiple root of $f$ , and write $f=(X-\alpha)^{m}g(X)$ , $m>1$ , in some extension field. Then

equation (5) (5)

f^{\prime}(X)=m(X-\alpha)^{m-1}g(X)+(X-\alpha)^{m}g^{\prime}(X).

Hence $f$ and $f^{\prime}$ have $X-\alpha$ as a common factor.

(b) $\Rightarrow$ (c). As $f$ is irreducible and $\deg(f^{\prime})<\deg(f)$ ,

\gcd(f,f^{\prime})\neq 1\implies f^{\prime}=0.

Let $f=a_{0}+\cdots+a_{d}X^{d}$ , $d\geq 1$ . Then $f^{\prime}=a_{1}+\cdots+ia_{i}X^{i-1}+\cdots+da_{d}X^{d-1}$ , which is the zero polynomial if only if $F$ has characteristic $p\neq 0$ and $a_{i}=0$ for all $i$ not divisible by $p$ .

(c) $\Rightarrow$ (d). By hypothesis, $f(X)=g(X^{p})$ with $g(X)\in F[X]$ . Let $g(X)=\prod_{i}(X-a_{i})^{m_{i}}$ in some extension field. Then each $a_{i}$ becomes a $p$ th power, say, $a_{i}=\alpha_{i}^{p}$ , in some possibly larger extension field. Now

f(X)=g(X^{p})=\prod\nolimits_{i}(X^{p}-a_{i})^{m_{i}}=\prod\nolimits_{i}(X-% \alpha_{i})^{pm_{i}}

which shows that every root of $f(X)$ has multiplicity at least $p$ .

(d) $\Rightarrow$ (a). Obvious. _□

Proposition 2.13

The following conditions on a nonzero polynomial $f\in F[X]$ are equivalent:

(a)

$\gcd(f,f^{\prime})=1$ in $F[X]$ ;
(b)

$f$ has only simple roots.

Proof.

Let $\Omega$ be an extension of $F$ splitting $f$ . From (5), p. 5, we see that a root $\alpha$ of $f$ in $\Omega$ is multiple if and only if it is also a root of $f^{\prime}$ .

If $\gcd(f,f^{\prime})=1$ , then $f$ and $f^{\prime}$ have no common factor in $\Omega[X]$ (see 2.10). In particular, they have no common root, and so $f$ has only simple roots.

If $f$ has only simple roots, then $\gcd(f,f^{\prime})$ must be the constant polynomial, because otherwise it would have a root in $\Omega$ which would then be a common root of $f$ and $f^{\prime}$ . _□

Definition 2.14

A polynomial is separable if it is nonzero and satisfies the equivalent conditions on (2.13).¹¹ 1 This is Bourbaki’s definition. Often (e.g., in the books of Jacobson and in earlier versions of these notes) a polynomial $f$ is said to be separable if each of its irreducible factors has only simple roots.

Thus a nonconstant irreducible polynomial $f$ is not separable if and only if $F$ has characteristic $p\neq 0$ and $f$ is a polynomial in $X^{p}$ (see 2.12). Let $f=\prod f_{i}$ with $f$ and the $f_{i}$ monic and the $f_{i}$ irreducible; then $f$ is separable if and only if the $f_{i}$ are distinct and separable. If $f$ is separable as a polynomial in $F[X]$ , then it is separable as a polynomial in $E[X]$ for every extension $E$ of $F$ .

Definition 2.15

A field $F$ is perfect if it has characteristic zero or it has characteristic $p$ and every every element of $F$ is a $p$ th power.

Thus, $F$ is perfect if and only if $F=F^{q}$ , where $q$ is the characteristic exponent of $F$ .

Proposition 2.16

A field $F$ is perfect if and only if every irreducible polynomial in $F[X]$ is separable.

Proof.

If $F$ has characteristic zero, the statement is obvious, and so we may suppose $F$ has characteristic $p\neq 0$ . If $F$ contains an element $a$ that is not a $p$ th power, then $X^{p}-a$ is irreducible in $F[X]$ but not separable (see 2.11). Conversely, if every element of $F$ is a $p$ th power, then every polynomial in $X^{p}$ with coefficients in $F$ is a $p$ th power in $F[X]$ ,

\textstyle\sum a_{i}X^{ip}=\left(\sum b_{i}X^{i}\right)^{p}\quad\text{if }\quad a_{i}=b_{i}^{p}\text{,}

and so it is not irreducible. _□

Example 2.17

(a)

A finite field $F$ is perfect, because the Frobenius endomorphism $a\mapsto a^{p}\colon F\rightarrow F$ is injective and therefore surjective (by counting).
(b)

A field that can be written as a union of perfect fields is perfect. Therefore, every field algebraic over $\mathbb{F}{}_{p}$ is perfect.
(c)

Every algebraically closed field is perfect.
(d)

If $F_{0}$ has characteristic $p\neq 0$ , then $F=F_{0}(X)$ is not perfect, because $X$ is not a $p$ th power.

Aside 2.18

Let $F$ be a perfect field. We’ll see later (LABEL:ag1) that every finite extension $E/F$ is simple, i.e., $E=F[\alpha]$ with $\alpha$ a root of a (separable) polynomial $f\in F[X]$ of degree $[E\colon F]$ . Thus it follows directly from (2.2b) that, for any extension $\Omega$ of $F$ , the number of $F$ -homomorphisms $E\rightarrow\Omega$ is $\leq[E\colon F]$ , with equality if and only if $f$ splits in $\Omega$ . We can’t use this argument here because it would make the exposition circular.