I was given the following Group S= {0,1,2,3}

with the following '+' and '×':

given '+' and 'x'

and now I need to prove that it's a field, so I need to prove the following:

1.Associativity of addition and multiplication: seems immediate from the tables.

2.Commutativity of addition and multiplication: seems immediate from the tables.

3.Additive and multiplicative identity: 0,1 from S

4.Additive inverses: every element is the inverse of itself

5.Multiplicative inverses: 1x1=1, 2x3=1, still immediate from the table

6.Distributivity of multiplication over addition: a · (b + c) = (a · b) + (a · c) - in this one I'm not really sure if pointing out to the table is a "proof", is there a formal way to prove it?


  • $\begingroup$ This is the table of finite group of order $4=2^2$. Learn more here and here. $\endgroup$
    – DaifM
    Apr 11 at 11:14
  • 2
    $\begingroup$ I'm impressed if you can just look at a composition table and see associativity in it immediately. $\endgroup$ Apr 11 at 11:22
  • $\begingroup$ You have to prove associativity, too. You can't just point at the table. There are $64$ cases for addition and $64$ for multiplication. $\endgroup$
    – saulspatz
    Apr 11 at 11:24
  • $\begingroup$ @saulspatz You're right, is there a formal way to prove it instead of writing down 64 cases? $\endgroup$ Apr 11 at 11:26
  • 1
    $\begingroup$ Prove associativity along the same lines a Zuy has suggested for distributivity. Try to eliminate as many cases as you can by logic, and then check the rest explicitly. $\endgroup$
    – saulspatz
    Apr 11 at 11:29

Another way to go about it is to show that the tables correspond to something you already know is a field.

Since your mystery structure has $4$ elements, there is only one field it can be, namely the field with four elements, which can be constructed as $$ \mathbb F_4 = \mathbb F_2[x]/\langle x^2+x+1\rangle $$

Its four elements are $\{0,1,x,x+1\}$. You'll need to decide how they correspond to your $\{0,1,2,3\}$ -- but it is actually obvious from the tables that your $0$ is the zero of the field, and your $1$ is the one of the field. So the only choice you really have is which of your $2$ and $3$ correspond to $x$ and $x+1$.

In fact it turns out you can do it either way around and still get the same result. (Because swapping $x$ for $x+1$ is an automorphism of $\mathbb F_4$). So draw up the tables for $\mathbb F_4$ and write 3,2 instead of $x, x+1$, and see that you get exactly the tables in your question.

The attraction of this approach is that it gives you the tedious associativity and distributivity checks for free.

If you don't know enough of the theory of finite fields but still know the quotient above gives you a ring, that will still be enough to get associativity and distributivity; you then just need to check explicitly for multiplicative inverses in the multiplication table.

However, if you don't even know the quotient notation above, then it is likely that the purpose of the exercise is to make you actually walk through the tedious case-by-case verification once in your life, such that you'll properly appreciate the tricks you'll learn later ...


I don't think so, but one easily sees that it suffices to check the cases for $a=2,3$ and $(b,c)=(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)$. The cases for $a=0,1$ are trivial, and the other cases for $(b,c)$ follow by commutativity of $+$. Observation by @saulspatz: If $b=c$, you get $b+c=0$, so that the cases $(b,c)=(1,1),(2,2),(3,3)$ may also be skipped. The remaining $6$ cases are easily dealt with.

  • 1
    $\begingroup$ One doesn't need to check the cases where $b=c$, since from the addition table, $x+x=0$. So $a\cdot b+a\cdot b=0=a\cdot0=a(b+b)$ $\endgroup$
    – saulspatz
    Apr 11 at 11:23

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