This entry is about the notion of root in algebra. For the notion in representation theory see at root (in representation theory).
symmetric monoidal (∞,1)-category of spectra
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
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.
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.
Let $G$ be a finite subgroup of the multiplicative group $k^\ast$ of a field $k$. Then $G$ is cyclic.
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.
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 $m n$.
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 $k$.
Every finite field has a cyclic multiplicative group.
Last revised on May 20, 2023 at 11:08:02. See the history of this page for a list of all contributions to it.