112
$\begingroup$

The characteristic of a ring (with unity, say) is the smallest positive number $n$ such that $$\underbrace{1 + 1 + \cdots + 1}_{n \text{ times}} = 0,$$ provided such an $n$ exists. Otherwise, we define it to be $0$.

But why characteristic zero? Why do we not define it to be $\infty$ instead? Under this alternative definition, the characteristic of a ring is simply the “order” of the additive cyclic group generated by the unit element $1$.

My feeling is that there is a precise and convincing explanation for the common convention, but none comes to mind. I couldn't find the answer in the Wikipedia article either.

$\endgroup$
  • 7
    $\begingroup$ What happens if your field is so large it has a number larger than $1+1+1+\ldots$ infinitely many times? $\endgroup$ – Asaf Karagila Jan 12 '12 at 22:41
  • 3
    $\begingroup$ I've always thought Erdos numbers should follow the same convention. If you have never published with anyone with a finite Erdos number, then you have Erdos number $0$... $\endgroup$ – user1729 Jan 12 '12 at 22:47
  • 7
    $\begingroup$ @user1729: but then what Erdös number should Erdös have? $\endgroup$ – Cheerful Parsnip Jan 12 '12 at 23:05
  • 4
    $\begingroup$ @Jim: $\infty$? $\endgroup$ – Asaf Karagila Jan 12 '12 at 23:09
  • 67
    $\begingroup$ @JimConant: Paul Erdős would have an Erdős number of 2, since he's not collaborated with Erdős, but he has collaborated with someone who collaborated with Erdős. $\endgroup$ – Clive Newstead Jan 12 '12 at 23:41
76
$\begingroup$

There are two orderings of the set $\mathbb N = \{0,1,\dots\}$:

  • magnitude $a \leq b$
  • divisibility $a\mid b$ (i.e. $\exists c. b = a c$)

They are mostly compatible - usually when $a \mid b$, it holds $a \leq b$.

Some definitions are phrased using "greater than" ordering, while in fact the "divisibility" ordering is the real essence.

For example, the greatest common divisor of $a$ and $b$ might be defined as the greatest number which is a common divisor of both $a$ and $b$. Characteristic of a ring $R$ might be defined as smallest number $n>0$ which satisfies $n \cdot 1 = 0$.

Under such commonly taught definitions, it seems natural that $\operatorname{gcd}(0,0)=\infty$ and $\operatorname{char} \mathbb Z = \infty$.

However, those definitions implicitly rely on ideals, and are better phrased using divisibility order. The incompatibility is then more visible: $0$ is the largest element in divisibility order, while it is smallest in magnitude order. Magnitude has no largest element, and often $\infty$ is added to cover this case.

So let's formulate the definitions again, but this time using divisibility ordering.

  • The greatest common divisor of two numbers $a,b$ is greatest number (in sense of $\mid$) that is a divisor of $a$ and $b$ (i.e. is smaller than $a$ and $b$ in divisibility ordering). This is prettier - $\operatorname{gcd}$ is now the $\wedge$ operator in lattice $(\mathbb N, \mid)$; it also forms a monoid, with $0$ as identity element. Additionally, the definition can be adapted to any ring.
  • The characteristic of a ring $R$ is the smallest number $n$ (in sense of $\mid$) that satisfies $n \cdot 1 =0$. As a bonus, compared to previous definition, we can remove the $n>0$ restriction: zero is always a valid "annihilator" but it is often not the smallest one. Now we get $\operatorname{char} \mathbb Z = 0$.

Characteristic is a "multiplicative" notion, like gcd. If you have a homomorphism of rings $f: A \to B$, it must hold $\operatorname{char} B \mid \operatorname{char} A$. For example, you cannot map ${\mathbb Z}_2$ to ${\mathbb Z}_4$ - in a sense, ${\mathbb Z}_2$ is "smaller" than ${\mathbb Z}_4$. "Bigger" rings have "more divisible" characteristic, their characteristics are greater in the sense of divisibility. And the "most divisible" number is 0. Another example is $\operatorname{char} A \times B = \operatorname{lcm}(\operatorname{char} A, \operatorname{char} B)$.

In a bit more abstract language: given any ideal $I \subseteq \mathbb Z$, we associate to it the smallest nonnegative element, under the divisibility order. By properties of $\mathbb Z$, every other element of $I$ is a multiple of it. Let's call this number $\operatorname{min}(I)$.

We can now define $\operatorname{gcd}(a,b)=\operatorname{min} ((a) + (b))$, and $\operatorname{char} R = \min (\ker f)$, where $f \colon \mathbb Z \to R$ is the canonical map.

The definition of $\operatorname{min}(I)$ works for any PID, it does not require magnitude order. In any PID, $I = (\operatorname{min}(I))$.

(I dislike saying the ideal $\{0\}$ is "generated" by $0$; although this is true, it also generated by empty set. We do not say that $(2)$ is generated by $0$ and $2$.)

| cite | improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ I am reminded of an answer I wrote many weeks back, trying to motivate the definition of gcd/lcm. But this answer makes the idea even clearer. :-) $\endgroup$ – Srivatsan Jan 14 '12 at 1:59
  • $\begingroup$ Great answer :) ! I would be so grateful if you can expand on "Additionally, the definition can be adapted to any ring." I can't find GCD of rings other than special case PID, such as this question. Also according to this question there are some times no GCD. Thanks, I want to learn! $\endgroup$ – Santropedro Mar 26 '17 at 14:39
  • 1
    $\begingroup$ @Santropedro en.wikipedia.org/wiki/… says that the notion of GCD can be defined in any commutative ring, but in general, it is not guaranteed to exist or to be unique. (I have no idea how useful the notion is in general, I've always use it on PIDs only.) $\endgroup$ – sdcvvc Apr 7 '17 at 0:05
  • $\begingroup$ @sdcvvc Wow, thank you very much for that! i'll look into it. $\endgroup$ – Santropedro Apr 7 '17 at 0:59
74
$\begingroup$

Given a ring $R$ there is a unique ring homomorphism $\varphi:\mathbb Z\to R$. The characteristic of $R$ is the (canonical, non-negative) generator of $\ker \varphi$.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ also (though this is probably obvious) note that we know $ker(\varphi)$ is always a principle ideal since $\mathbb{Z}$ is a PID so we can always find said generator. $\endgroup$ – Deven Ware Jan 12 '12 at 22:56
  • 4
    $\begingroup$ The important thing is that the characteristic is actually an ideal of $\mathbb Z$. That we pick a representative generator to "name it" isn't really relevant. $\endgroup$ – Thomas Andrews Jan 13 '12 at 14:53
  • 1
    $\begingroup$ I have a slight preference for "non-negative" rather than "nonnegative", but certainly both are used. $\endgroup$ – Michael Hardy Jan 30 '12 at 4:38
34
$\begingroup$
  1. Consider the following statement:

    Let $n\geq 0$. The characteristic of $R$ is $n$ if and only if ($ka=0$ for all $a\in R$ $\iff$ $n|k$).

    The statement holds for positive characteristic, but it also holds for characteristic $0$, since $0$ is the only multiple of $0$. This would not hold for any ring if we defined the characteristic to be $\infty$. This definition also makes sense for rings without $1$.

  2. For rings with unity, the definitions follows as indicated by lhf: the characteristic of $R$ is the nonnegative generator of the kernel of the canonical map from $\mathbb{Z}$ to $R$.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ I think you need an "iff" in the latter half of your definition; as it is, this says that $\Bbb Z_4$ has characteristic $2$ (and also char $4$). $\endgroup$ – Mario Carneiro Mar 17 '15 at 3:55
  • $\begingroup$ \mid is correct for set-builder notation, but it is incorrect for divisibility statements. Since I'm not doing "does not divide", the claim that \mid looks better when negated is irrelevant. $\endgroup$ – Arturo Magidin Aug 14 '18 at 22:04
16
$\begingroup$

Recall that an R-algebra is a ring A containing a central image of the ring R. This image is $\,\cong$ R/I so it is characterized by the kernel I. For example, if R = $\mathbb Z$ then an R-algebra is simply a ring A, and the kernel $\rm\ I = (n)\ $ characterizes the canonical image of $\mathbb Z$ in A, via $\rm 1\mapsto 1_A.\,$ Therefore we say that A has characteristic n because n characterizes the canonical image of $\:\mathbb Z\:$ in A.

Remark $\ $ For more general notions of "characteristic rings" see below - excerpted here.

W.D. Burgess; P.N. Stewart. The characteristic ring and the "best" way to adjoin a one.
J. Austral. Math. Soc. 47 (1989) 483-496. $\ \ $

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

It has something to do with the classification of cyclic groups. If $(R,+)$ is an infinite cyclic group generated by $1,$ then it is isomorphic to $(\mathbb Z,+)$ (additive group of integers). If it is a finite cyclic group, then it is isomorphic to $(\mathbb Z_n,+)$ (additive group of integers modulo $n$). Looking deeper into it, in both cases, it is isomorphic to $\mathbb Z/\ker{(f)},$ where $f$ is the group homomorphism defined by $f:(\mathbb Z, + )\to (R, + )$ with $f(n)=\underbrace{a+\cdots+a}_{n \text{ times}}$ for all $a\in R$ and all $n\in \mathbb Z.$ In the finite case, $\ker{(f)}$ consists of all multiples of $n$, i.e. $\ker{(f)}=\{kn\mid k\in\mathbb Z\}.$ In the infinite case, $\ker{(f)}$ consists of all multiples of $0$, i.e. $\ker{(f)}=\{0\}.$ Therefore, it makes sense to define the characteristic correspondigly. See also characteristic zero instead of characteristic infinite.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Welcome to MSE. Your answer adds nothing new to the already existing answers. $\endgroup$ – José Carlos Santos Aug 11 '19 at 7:47
  • $\begingroup$ Thanks! In my opinion, the new thing in my answer is that it gets along without "divisibility", "ideals", "ring homomorphism", or "orders in rings", which are required in the existing answers. The characteristic of a ring does not require the multiplication operation of the ring, only the addition. Therefore, my answer reduces the question to its core, dealing only with the addition operation. Therefore, my answer gets along with group-theoretical concepts. $\endgroup$ – apio Aug 11 '19 at 10:21
-1
$\begingroup$

One way to see why this definition is natural is to consider that if $$\underbrace{1 + 1 + \cdots + 1}_{n \text{ times}}$$ never vanishes for any positive integer $n,$ then it makes sense to say that the ring has characteristic $0$ since it takes no number of units to vanish under its additive operation.

In the end, though, it's just a convenient stipulation.

| cite | improve this answer | |
$\endgroup$
-2
$\begingroup$

Not only does adding 1 infinitely times in a field not make sense, but also if adding 1 a positive number of times never yields zero, then certainly adding it zero times yields zero, thus it only seems natural to refer to such a field as "characteristic zero".

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ @Santropedro He said "IF adding 1..." $\endgroup$ – user50229 May 9 at 3:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.