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root

This entry is about the notion of root in algebra. For the notion in representation theory see at root (in representation theory).

Definitions

A square root of an element $a$ in a monoid $M$ is a solution to the equation $x^2=a$ within $M$. If $M$ is the multiplicative monoid of non-zero elements of a (commutative) field or integral domain $K$ (or a submonoid of this, such as the group of units), then there are exactly two square roots, denoted $\pm \sqrt{a}$, if there are any; the element $0$ has only one square root, $\pm \sqrt{0}=0$. But if $K$ is even a non-integral domain or a non-commutative skew field, then there may be more; in the skew field $ℍ$ of quaternions, there are continuumly many square roots of $-1$.

More generally, for $n\in \mathbb{N}$ a solution to the equation $x^n=a$ is called an $n$th root of $a$. Specifically for $a=1$ and working in the field of complex numbers, one speaks of an $n$th root of unity. This terminology can be applied to other fields as well; for example, the field of 7-adic numbers contains non-trivial cube (or $3^{\mathrm{rd}}$) roots of unity.

More generally, for any polynomial $P(x)$ of $x$ with coefficients in a field $K$, a solution to $P(x)=0$ in $K$ is called a root of $P$. When $P(x)\in K[x]$ has no solution in $k$, one can speak of a splitting field? obtained by “adjoining roots” of $P$ to $K$, meaning that one considers roots in an extension field? $i:K\hookrightarrow E$ of the corresponding polynomial $Q=(i{\otimes }_{K}K[x])(P)\in E[x]$, i.e., applying the evident composite map

$$K[x]\cong K{\otimes }_{K}K[x]\stackrel{i{\otimes }_{K}1}{\to }E{\otimes }_{K}K[x]\cong E[x]$$K[x] \cong K \otimes_K K[x] \stackrel{i \otimes_K 1}{\to} E \otimes_K K[x] \cong E[x]

to $P$ to get $Q$, and passing the smallest intermediate subfield? between $K$ and $E$ that contains the designated roots of $Q$ (often writing $P$ for $Q$ by abuse of language).

More generally still, one may refer to roots even of non-polynomial functions $f$ defined on a field, for example of meromorphic functions $f:ℂ\to ℂ$, although it is much more usual to speak of zeroes of $f$ instead of roots of $f$ (e.g., zeroes of the Riemann zeta function); see zero set? and intermediate value theorem.

Roots of unity in fields

In a field $k$, a torsion element of the multiplicative group $k^{*}$ is a root of unity by definition. Moreover we have the following useful result.

Theorem

Let $G$ be a finite subgroup of the multiplicative group $k^{*}$ of a field $k$. Then $G$ is cyclic.

Proof

Let $e$ be the exponent of $G$, i.e., the smallest $n>0$ such that $g^n=1$ for all $g\in G$, and let $m=\mathrm{order}(G)$. Then each element of $G$ is a root of $x^e-1$, so that ${\prod }_{g\in G}(x-g)$ divides $x^e-1$, i.e., $m\le e$. But of course $g^m=1$ for all $g\in G$, so $e\le m$, and thus $e=m$.

This is enough to force $G$ to be cyclic. Indeed, write $e=p_1^{r_1}{p}_2^{r_2}\dots p_k^{r_k}$. Since $e$ is the least common multiple of the orders of elements, there is (for each $i$) an element whose order is divisible by $p_i^{r_i}$, and some power $y_i$ of that element will have order exactly $p_i^{r_i}$. Then $y={\prod }_i{y}_i$ will have order $e=m$ by the following lemma and induction, so that powers of $y$ exhaust all $m$ elements of $G$, i.e., $y$ generates $G$ as desired.

Lemma

If $m,n$ are relatively prime and $x$ has order $m$ and $y$ has order $n$ in an abelian group, then $xy$ has order $mn$.

Proof

Suppose $(xy{)}^k=x^k{y}^k=1$. For some $a,b$ we have $am-bn=1$, and so $1=x^{kam}{y}^{kam}=y^{kam}=y^k{y}^{bn}=y^k$. It follows that $n$ divides $k$. Similarly $m$ divides $k$, so $mn=\mathrm{lcm}(m,n)$ divides $k$, as desired.

Clearly there is at most one subgroup $G$ of a given order $n$ in $k^{*}$, which will be the set of $n^{\mathrm{th}}$ roots of unity. If $G$ is a finite subgroup of order $n$ in $k^{*}$, then a generator of $G$ is called a primitive $n^{\mathrm{th}}$ root of unity in $k$.

Corollary

Every finite field has a cyclic multiplicative group.

Related concepts

• root

• square root

• Newton's method