I have just begun studying finite fields today, and it is clear in GF(2) why 1+1=0. (I just show that 1+1 can't equal 1, or 1=0, which contradicts an axiom that states that 1 is not 0).

If we interpreted these symbols "1", "+", "1", "0" as we would in primary school, clearly this breaks arithmetic rules in Real numbers.

Given that, I have lost all confidence in how arithmetic can be applied in a finite field. How do I even know how to do basic arithmetic on GF(n) where n is prime? For example, for GF(7), how do I even know that 4+1=5? Can anyone show with just the 9 axioms of finite fields that 4+1=5?

Axioms: associativity of addition, additive identity, additive inverse, commutatitivity of addition, associativity of multiplication, multiplicative inverse, commutatitivity of mulitplication, distributive law

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    $\begingroup$ What you need to show is that if addition and multiplication are defined modulo $7$, then GF($7$) satisfies the axioms of a field, not the other way around. $\endgroup$
    – saulspatz
    Sep 12, 2019 at 0:20
  • $\begingroup$ Welcome to Mathematics Stack Exchange. Are you familiar with modular arithmetic (when $n$ is prime)? $\endgroup$ Sep 12, 2019 at 0:33
  • $\begingroup$ "clearly this breaks arithmetic rules in Real numbers." How so? Which rule? $1 + 1$ still equals $2$. It will always equal $2$. It's just that in $GF(2)$ we have a rule that $2=0$. In the real numbers we don't have a rule that $2=0$. So this isn't breaking the rules of real numbers. It's just that this is adding more rules that don't actual exist with the reals. $\endgroup$
    – fleablood
    Sep 12, 2019 at 1:06
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    $\begingroup$ Four is defined as $1+1+1+1$ and Five is defined by $1+1+1+1+1$. So $4+1 = (1+1+1+1) + 1 = 1+1+1+1+1=5$. That's all there is to it. $\endgroup$
    – fleablood
    Sep 12, 2019 at 1:08

3 Answers 3


This actually brings up a subtle point. What do we mean by $5$ in a finite field? Or if you choose to define $5$ in terms of $1 ~(5=1+1+1+1+1)$, then what do we mean by $1$?

One answer is to define $5$ in terms of equivalence classes. Say that two integers $m$ and $n$ are equivalent if $p \vert (m-n).$ First, you prove this really is an equivalence relation on the integers. Then you define $[m]+[n]=[m+n]$ and $[m][n]= [mn]$. So by $5$ we actually mean the equivalence class $[5]$.

You need to prove that your field operations are well-defined (you get the same answer no matter which representative of an equivalence class you choose) and that $[0]$ and $[1]$ really are the additive and multiplicative identities, as you'd expect. But once you've done that, you can see that $[4]+[1]=[5]$ (and usually we abuse notation by dropping the brackets) because we've defined it that way.

  • $\begingroup$ Thank you. This subtlety is exactly what I was looking for. Based on your answer, I understand that I could have a finite field GF(5) = {0,1, "element2", "element3", "element4"}, where the element "element2" itself doesn't even have to belong to N (element2 could be "a duck that quacks"), but rather, its "class number" or "subscript" has to be an integer..... (continued in next comment) $\endgroup$
    – Snowball
    Sep 12, 2019 at 2:07
  • $\begingroup$ And I know that since the "class values" or "subscripts" are obviously \in Z, thereby allowing me to do standard arithmetic operations 4+1=5 in my original question above, even outside of 0 to 4. So in GF(5), when we say 7=2, we’re really just saying that 7 and 2 are referring to the same element. Am I understanding you correctly? $\endgroup$
    – Snowball
    Sep 12, 2019 at 2:07
  • $\begingroup$ Yes. If you use equivalence classes as your definition, you're saying that $2$ and $7$ are equivalent to each other. $\endgroup$ Sep 12, 2019 at 5:53

There’s really no big deal. You are really working in integers (never in the reals), and whenever you get an answer that’s too big or too negative, subtract or add a multiple of your prime number $n$ (conventionally, we call the modulus $p$ in these cases). So, if you’re working modulo $7$, to add $4+1$, you do it in the integers $\Bbb Z$. Answer $5$. Is it at least $7$? No, so just leave it be. But to add $4+5$, the integer sum is $9$, so you subtract $7$ to get $2$, and in the system of integers modulo $7$, you have $4+5=2$. A standard way of writing this is $4+5\equiv2\pmod7$, which you read, “four plus five is congruent to two modulo seven”. This notation and terminology goes back to Gauss (1801), maybe even farther.


Beneath it all... we count.

And it doesn't matter what we count; we just count.

So we define $4$ as what we get if we add $1$ a "tick-tick-tick-tick" number of times. And we define $5$ as what we get if we add $1$ a "tick-tick-tick-tick-tick" numbers of times.

Well every time we add $4$ to $5$ we always count $9$ times. THat's because if you put "tick-tick-tick-tick-tick" and then put another "tick-tick-tick-tick" you get "tick-tick-tick-tick-tick-tick-tick-tick-tick" which we define as what $9$ is.

It's just that in the finite field $GF(2)$ you know that $1+1 = 0$.

So you count

1 tick $\to 1$

2 tick $\to 1+1 = 0$

3 tick $\to 0+1 = 1$

4 tick $\to 1+1 = 0$.

5 tick $\to 0+1 = 1$.

Okay that was "tick-tick-tick-tick-tick". Now to add "tick-tick-tick-tick

1 tick $\to 1+1 =0$

2 tick $\to 0+1 = 1$

3 tick $\to 1+1 = 0$

4 tick $\to 0+1 = 1$.

So $5 +4$ still equals $9$. We just have $5 = 1$ and $4 = 0$ and $9=1$.

The only rule we've changed is "adding one gives us a new number we never had before". That rule tells us $9\ne 1$ but without it $9=1$ is perfectly acceptable.


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