# Notation for fields and the elements of the sets used to define them.

Context:

While trying to figure out how to prove that the characteristic of a field $$\mathcal{F}$$ is either $$0$$ or prime, I realized that I was conflating the elements of the field with integers, because of the notation.

Question:

Because articulating this question abstractly is beyond my ability, I'll illustrate my confusion with an example:

If $$\mathbb{Z}_5$$ is the set $$\{0,1,2,3,4\}$$, along with modular addition and multiplication (i.e. in this case, after performing the standard addition or multiplication, the least remainder after ordinary division by 5; e.g. $$4\cdot4 = 1$$ and $$3+4 = 2$$) defines a field. Here, $$4 + 1 = 0,$$ i.e. $$1$$ is the additive inverse of $$4$$.

Now, all the classical definitions I've seen usually define the additive inverse as follows (this example is taken from Wikipedia):

for every a in F, there exists an element in F, denoted −a, called the additive inverse of a, such that a + (−a) = 0.

Does the phrase "denoted -a" imply that, in the example provided above, we should denote "$$1 = -4$$"? Or are such definitions simply stating what they mean when they use the symbol (i.e. it is simply saying we should think of $$1$$ as the additive-inverse-of-4)? In other words, is "$$-4$$" an element of this field, even if it is not an element of the set used to define the field (does it even make sense to talk of "an element of a field" in this way)?

Just to be clear, if we were to instead denote the additive inverse of a by na, is it then appropriate to write something like $$1 = n4$$, for the above field? While I can understand what is meant by this, what bothers me is that $$n4$$ is not an element of any set described or defined, elsewhere.

Final Considerations

Up until here, I'd be tempted to more or less ignore the whole issue; however, while trying to work out a proof that a field's characteristic is either zero or prime, I found myself wanting to use integers to describe the operations, regardless of whether or not the elements of the set were, themselves, numbers. For example, given the field $$\mathbb{F}$$ defined by: $$\left\{ \begin{array}{lll} \mathbb{S} = \{a, b\}, & \begin{array}{c|c|c} \pmb{+} & \pmb{a} & \pmb{b} \\ \hline \pmb{a} & a & b \\ \hline \pmb{b} & b & a \end{array}, & \begin{array}{c|c|c} \pmb{\cdot} & \pmb{a} & \pmb{b} \\ \hline \pmb{a} & a & a \\ \hline \pmb{b} & a & b \end{array}, \end{array} \right.$$

it seems natural and useful to define an expression such as $$a + a + a$$ as $$3a$$; however, I'm not sure how to think about the relationship of the symbols introduced by this notation to the field itself (this becomes even more confusing if I use "1" to denote the multiplicative identity). This seems -- at least to my confused mind -- somehow related to the confusion expressed in the first example, above.

Is the use of 3 in "3a" and -4 in "1 = -4" nothing more than a convenient notation to describe the elements of the field? While I feel like that is all they are, I also feel that they carry with them some sort of implied structure (perhaps that of an abelian group?), that isn't accounted for, formally / explicitly.

tl;dr You are overthinking this.

When you write $$\mathbb{Z}_5$$ as $$\{0, 1, 2, 3, 4\}$$ those elements are not integers. They are just the names you have chosen for the five elements of that field. Of course those symbols also name integers, and the coincidence is not an accident. The coincidence together with the idea of "modulo $$5$$" tells you how to do the arithmetic in that field.

The wikipedia definition tells you that a standard way to name the inverse of the element $$a$$ is to write $$-a$$. Then in $$\mathbb{Z}_5$$ you write $$-1 = 4$$ since $$1 + 4 = 0$$. So "$$-1$$" is just another name for the element names "$$4$$". It's not some other element of the field you forgot to write down.

Every field has a multiplicative identity which is always called $$1$$ although it is not the integer 1. Then by repeated addition, for every positive integer every field contains an element you could reasonably call $$n$$. If those elements are all different from one another then the field contains a copy of the integers (even a copy of the rationals) and is said to have characteristic $$0$$. If there are repeats in the sequence of field elements named by the positive integers then ...

• Love the tl;dr -- was pretty certain that was the case, but knowing it doesn't help, when one is confused. For example, when you say "Then by repeated addition, for every positive integer every field contains an element you could reasonably call $n$" what is actually meant by this? $n$ clearly doesn't belong to the set used to describe the field, so does this mean there is another set always implied by the field? Nov 15, 2020 at 20:33
• It means that when I refer to a positive integer $n$ as an element of a field $F$, that is shorthand for the expression 1 + 1 + ... + 1 ($n$ times), where the 1 refers to the multiplicative identity of the field, and the + is the addition operation of the field. Formally, there is a canonical map from the integers to any field $F$, and when I refer to an integer $n$ as being an element of $F$, I really mean the image of $n$ under that map. It is too pedantic to write everything out completely formally all the time, so we use shorthand like this.
– Ted
Nov 15, 2020 at 21:08
• @Ted thank you! I think I'm getting closer. Where could I find more about this "canonical map from the integer to any field"? Does it have a name? Because it isn't clear to me how all the integers are mapped onto the $\{a, b\}$ example provided, above; rather, it seems that we're suddenly pulling in the integers to simplify the notation. I know it seems pedantic, but at my stage of understanding, it is helpful to distinguish what is useful notation and what is part of the formal definition. eg Your answer clarified that "-4" is just a synonymous notation for $1$, not an element of the field. Nov 16, 2020 at 0:56
• The canonical map sends 0 to the additive identity ($a$ in your example), 1 to the multiplicative identity ($b$ in your example), then 2 to the sum of the multiplicative identity with itself (so $b+b = a$), then 3 to $b+b+b=b$, then 4 to $b+b+b+b=a$, and so on. Then -1 is mapped to $-b = b$, -2 to $-b-b = a$, etc. So in your $\{a,b\}$ example, the canonical map sends all even integers to $a$ and all odd integers to $b$.
– Ted
Nov 16, 2020 at 1:28

For every a in F, there exists an element in F, denoted −a, called the additive inverse of a, such that a + (−a) = 0.

This actually means: For every $$a \in F$$ there exists a unique element $$b \in F$$ such that $$a+b=0$$. Since the element $$b$$ is unique for a given $$a,$$ the set $$\{ (a,b) \in F \times F \mid a+b=0 \}$$ is a function (I hope that you know the set-theoretic definition of function) which will be denoted with a prefix minus sign, i.e. $$-a$$ denotes the unique element in $$F$$ such that $$a+(-a)=0.$$

For example, in $$\mathbb{Z}_5 = \{ 0,1,2,3,4 \}$$ (with operations modulo 5) the notation $$-3$$ denotes the element $$2$$ since $$3+2=0.$$ It does not really denote the negative integer $$-3$$ which is not in $$\{0,1,2,3,4\}.$$

Is the use of 3 in "3a" and -4 in "1 = -4" nothing more than a convenient notation to describe the elements of the field?

That is correct.

We define a mapping $$\mathbb{N}\times F \to F,$$ given by $$(n,x) \mapsto \underbrace{x+\cdots+x}_{n\text{ copies of }x}.$$ For simplicity, and since there's no risk of confusion, as KCd points out, we denote this with $$nx$$.

For example, in $$\mathbb{Z}_5,$$ we have $$7x = x + x + x + x + x + x + x$$. It can be shown that this equals $$2x = x+x,$$ which in turn equals $$2x$$ as the product of the elements $$2$$ and $$x$$ in $$\mathbb{Z}_5.$$ (Thus, there is no risk of confusion.)

This notation can be extended to negative $$n \in \mathbb{Z}$$ by $$nx=-(-n)x,$$ where the inner minus sign is ordinary negative on the integers, and the outer/front minus sign is negative in $$F$$.

For example, in $$\mathbb{Z}_5,$$ we have $$(-7)x=-(7x)$$ by definition, and therefore can skip the parentheses: $$-7x$$ is unambiguous.

• Wow! Thank you for taking the time to put together such a clear exposition. Obviously, I don't fully agree about the "no risk of confusion" :D, but you're right: once what is being done is clearly stated, there is no risk. To confirm then, strictly speaking, it doesn't make sense to say "3 is an element of the Field" when using notation like "a + a + a = 3a". There really isn't any ambiguity about the meaning of "3a" but the use of 3 in this way did lead to it having an ambiguous status within the definition of a field; it's basically additional convenient notation, right? Again, thank you! Nov 16, 2020 at 1:06
• @RaxAdaam. There of course is an ambiguity about whether $3$ in $3a$ is the element in $F$ so that the product is $F\times F \to F$ or the natural number so that the "product" is $\mathbb{N}\times F \to F$ and means $a+a+a$. But the good thing is that it doesn't matter at the end; the results are the same. Therefore we can forget the notational ambiguity. It's similar to ambiguity of associativity: does $xyz$ mean $(xy)z$ or $x(yz)$? In a ring like $\mathbb{R}$ it doesn't matter so we can skip the parentheses. Nov 16, 2020 at 11:06
• Yeah, I see the ambiguity is ultimately unimportant; it was understanding how the integers were inserting themselves into any field that was bothering me. I think it's fair to say that they don't properly belong to the field (unless actually part of the set used to build it), according to the definition, but they are a useful additional notation. Thanks again for your patience and taking the time to write such a thorough and clear response. Nov 16, 2020 at 15:55
• @RaxAdaam. We can make the integers be symbols in our field by the mapping $\mathbb{N} \to F,\ n \mapsto n1 = 1+\cdots+1\ (n \text{ ones}).$ The will then just be alternative notations for the already existing elements. For example, in $\mathbb{Z}_5,$ the symbol $7$ will be another name for $2,$ and in your example $\{ a,b \},$ the symbol $7$ will be another name for $b$. Nov 16, 2020 at 16:16

The notation $$-a$$ means "the element $$b$$ that fits the equation $$a + b = 0$$". So in the integers modulo $$5$$, $$-1 = 4$$, $$-2 = 3$$, $$-3 = 2$$, $$-4 = 1$$, and $$-0 = 0$$.

It is perfectly fine to write $$na$$, for a positive integer $$n$$, to mean the sum of $$a$$ with itself $$n$$ times. There is no inconsistency between the "$$3$$" in $$3a$$ as $$a + a + a$$ and the "$$3$$" in $$3a$$ as the product of the element $$3$$ in the field and $$a$$ because $$a + a + a = 1\cdot a + 1 \cdot a + 1 \cdot a = (1 + 1 + 1)\cdot a = 3a.$$

By the way, it is a terrible idea to limit your viewpoint of the integers modulo 5 as being only the symbols $$0,1,2,3,4$$ with a wraparound addition and multiplication rule modulo $$5$$. You should accept $$7$$, $$11$$, and $$-46$$ as perfect reasonable integers modulo $$5$$ that happen to be the same thing in the integers modulo $$5$$ as $$2$$, $$1$$, and $$4$$, respectively.

After all, you should be perfectly comfortable with writing the same fraction in many ways: $$1/2 = 2/4 = 3/6 = (-6)/(-12)$$, and so on. It would be extremely inconvenient if you were only allowed to work with fractions in reduced form (try to define addition and multiplication when the inputs and the output can only be written as reduced form fractions). Similarly, it is terrible to restrict yourself to only working with the integers modulo $$5$$ as $$0,1,2,3,4$$. The integers modulo $$5$$ can be represented by all integers, but there is a specific rule about when they get identified: $$a$$ and $$b$$ are the "same integer mod $$5$$" when $$a - b$$ is a multiple of $$5$$, just like the fractions $$a/b$$ and $$c/d$$ are the "same fraction" when $$ad = bc$$ as integers.

• Thank you very much for taking the time to reply. Your comment "There is no inconsistency between the "3" ... as the product of the element 3 in the field and ... " points directly to what that example was meant to illustrate: 3 is not an element of that field, is it? Isn't the field the set $\{a, b\}$ along with the operations, as defined? A similar question applies to your comments about accepting 7, 11, and -46 in the field $\mathbb{Z}_5$ described (I do understand that the chosen symbols are irrelevant, but in any case only 5 symbols are members of the set, no?). Nov 15, 2020 at 20:23
• I think that it can be a good idea to have different views of $\mathbb{Z}_5$. One is to see it as equivalence classes of integers modulo 5. Another is to view it as the numbers $0,1,2,3,4$ with a "wraparound" addition. There might be even more ways to see it, each with its pros and cons. Nov 15, 2020 at 20:30
• The point here is: it is convenient to have multiple symbols to represent the same object, as long as you know the rules that tell you when two symbols are equal.
– Ted
Nov 15, 2020 at 21:02
• Consider angles: do you demand that the only possible radian angles are numbers in the interval $[0,2\pi)$, or do you accept that every real number can represent an angle, but we have the relation that numbers $\theta$ and $\varphi$ describe the same radian angle measure when $\theta - \varphi$ is an integral multiple of $2\pi$? Trigonometric identities like $\sin(2x) = 2\sin x \cos x$ lose some significance if you only allow numbers in $[0,2\pi)$ to be put inside a trig function. Similarly, the integers mod $5$ has size $5$ but each element can be written in many ways, just like fractions!
– KCd
Nov 15, 2020 at 21:56
• It is a very bad idea to conceive of the integers mod $5$ only as a set $\{0,1,2,3,4\}$ with a wacky wraparound addition and multiplication rule for the same reason it is a very bad idea to only allow fractions that are in reduced form or to only allow radian angles that are in the interval $[0,2\pi)$.
– KCd
Nov 15, 2020 at 21:58