representation, ∞-representation?
symmetric monoidal (∞,1)-category of spectra
It is well known that you cannot divide by zero, lest you be doomed to triviality. Conversely, in a field, you can divide by anything except zero. But this rule can be misleading, since it's possible that (even) an ordinary number can be zero when you don't expect it! The characteristic of a field states when (if ever) this happens.
It is straightforward to generalise from fields to other rings, and even rigs. See also characteristic zero.
Let $K$ be a rig (possibly a ring, possibly a commutative ring, possibly even a field). Then there exists a unique homomorphism $\phi_K\colon \mathbb{N} \to K$ to $K$ from the initial rig, which is the rig $\mathbb{N}$ of natural numbers. The kernel of $\phi_K$ is an ideal of $\mathbb{N}$, which (by a well-known property of $\mathbb{N}$) is a principal ideal with a unique generator. This generator is the characteristic of $K$, denoted $\char K$.
If $K$ is a ring, then we may use $\phi_K\colon \mathbb{Z} \to K$ instead, where $\mathbb{Z}$ is the ring of integers. However, in this case, the kernel will usually have two generators, in which case we pick the positive one to get the same result as above.
If $n$ is a natural number, then we suppress mention of $\phi_K$ to think of $n$ as an element of $K$. If $K$ is a ring, then we do the same for a negative integer $n$. We then have that $n = 0$ in $K$ if and only if $n$ is a multiple of $\char K$.
The characteristic of a field must be either zero or a prime number. Basically, this is because the kernel of $\phi_K$, for $K$ a field, must be a prime ideal.
Every rig with positive characteristic is in fact a ring, since we have $\char K - 1 = -1$. In other words, any rig other than a ring must have characteristic zero (although many rings also have that characteristic).
If there is any homomorphism at all between two fields, then they have the same characterstic. In other words, any extension of a field keeps the same characteristic.
If $n$ is a positive natural number, then the characteristic of $\mathbb{N}/n = \mathbb{Z}/n$ is $n$. This ring is always a commutative ring, and it is a field if and only if $n$ is prime, in which case it is the prime field $\mathbb{F}_n$. More generally, every finite field has positive prime characteristic.
For $n = 0$, $\mathbb{N}/0 = \mathbb{N}$, $\mathbb{Z}/0 = \mathbb{Z}$, and the prime field $\mathbb{F}_0 = \mathbb{Q}$ (the field of rational numbers) are no longer all the same, but they still all have characteristic $0$. Every ordered field has characteristic $0$. The real numbers and complex numbers each form fields of characteristic $0$.
Recently the concept of the characteristic has been extended to E-∞ rings
Last revised on February 2, 2016 at 17:09:36. See the history of this page for a list of all contributions to it.