How many $n$th roots does $0$ have?

Question

Do we say that $0$ has $n$ $n$th roots, all nondistinct, or only one?

I don't think it makes any difference, but I'm curious what the convention is.

Answer 1

The terminology varies a bit between people and fields, but what I would say is that $x^n=0$ has one root (namely $0$) of multiplicity $n$.

If we explicitly say that we count that root "with multiplicity", of course, there are $n$ of it.

Answer 2

As many as roots of the polynomial equation $X^n=0$, that is, $n$.

Answer 3

There is, as is visible, no universally agreed upon convention. However, I would argue in favor of considering by an large the number of $n$-th roots of an element $a$ (in some structure) the cardinality of the set of solutions of $X^n = a$, so that in the complex and real numbers, and in any other field, $0$ is the unique root of $0$.

The alternative convention is arguably convenient over the complex numbers, however, already in the real numbers the situation is less clear (though one might still present reasonable arguments in its favor).

Yet roots are also considered in other structures.

• Roots of unity, so roots of $1$, in finite fields play an important role, as do for example quadratic residues; for the Legendre symbol $0$ is treated separately, corresponding to the convention that it has a unique square-root modulo $p$, and typically when one wonders about the number of roots of unity in a finite field one will consider the cardinality of the set (see an MO question on that subject).

• Further, looking beyond fields, the situation become even more cumbersome for other conventions. For example in $Z/4Z$ we have $0$ and $2$ as square-roots of $0$. Should $0$ still be counted twice? Should $2$ also be counted twice, after all $(X+2)^2 = X^2$?

• Still further, it is common to consider roots of (certain) matrices. There is an infinitude of real $2 \times 2$ matrices whose square is the $0$-matrix. They are all square-roots of the $0$-matrix, among them one finds the $0$-matrix itself.

In each case, it seems the cardinality of the set of solutions makes some sense, while the assigning of multiplicities becomes rather more unclear (I might miss something though).