nLab root

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

Context

Algebra

higher algebra

universal algebra

Contents

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 $\mathbb{H}$ 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 $3^{rd}$) 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

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

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 $f \colon \mathbb{C} \to \mathbb{C}$, 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 $k^\ast$ 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 $k^\ast$ of a field $k$. Then $G$ is cyclic.

Proof

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

This is enough to force $G$ to be cyclic. Indeed, write $e = p_1^{r_1} p_2^{r_2} \ldots p_k^{r_k}$. Since $e$ is the least common multiple of the orders of elements, the exponent $r_i$ is the maximum multiplicity of $p_i$ occurring in orders of elements; any element realizing this maximum will have order 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 $x y$ has order $m n$.

Proof

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

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

Corollary

Every finite field has a cyclic multiplicative group.

Last revised on July 1, 2015 at 08:42:48. See the history of this page for a list of all contributions to it.