Spivak's Calculus Chapter 28-1 (Fields)

Question

Question in Textbook

The question defines a set F and then defines the operations on F through a table. In this chapter, Spivak outlines the properties that operators on a field must have. In the example in textbook he says that that the verifications of the conditions are simple, case by case checks. However, would that not be extremely long and tedious going through every case for every element in set F? Is there a simpler way to do this?

Answer 1

So, if you know that this field happens to be the field $\mathbb{Z}/3\mathbb{Z}$ (you don't need to know what that means), then maybe you can use the general argument that $\mathbb{Z}/p\mathbb{Z}$ (again, don't need to know what that means) is a field.

However, I don't think the axioms are as hard to verify as you might think they are. For instance, the existence of additive identity means that one of the columns in the addition table looks the same as the column listing the elements being added. The existence of additive inverses is the same as saying that every column in the addition table has a $0$ in it. Commutativity means that flipping the multiplication table along the diagonal gives the same table. So, this is quicker to manually verify than you might think. Does that help?

Answer 2

The Field is only !$3$! elements. It is very easy and short to check everything.

Axiom 1: Addition is commutative. If $a = b$ then $a+b= a+a=b+a$ so we only have to check if $a\ne b$ and there are only three such pairs. $0 + 1 = 1=1+0; 0+2=2=2+0; 1+2=0=2+1$. That's it. We checked them all.

Axiom 2: Addition is associative. The check $(a+b) + c = a+(b+c)$ is $27$ cases of $a,b,c$ but we can cut that down with commons sense. If $a,b$ or $c=0$ then $(a+b) + 0 = a+b = a +(b+0)$ and $(a+0)+c= a+c = a+(0+c)$ and $(0+b) +c = b+c = 0 + (b+c)$. (See axiom 3). If none of $a,b,c=0$ then that is only $8$ cases but still we must have two of them equal or three of the equal and as addition is commutative:

$(a+a) + a = a+(a+a)$ and $(a+a)+b=a+(a+b)$ can be checked with $4$ cases. And $(a+b) +a= a+(a+b)=a+(b+a)$. And $(b+a)+a=(a+b)+a=a+(a+b)=(a+a) + b=b+(a+a)$. so that associativity.

Axiom 3: $0+a = a+0= a$. Just three things to check. $0+0=0;1+0=1;2+0=0$ that's it we checked them ALL.

Axiom 4: even element has an additive inverse. Only three to check. $0+0=0$ so $0$ has an inverse. $1+2 = 0$ so $1$ has an inverse. $2+1=0$ so $2$ has inverse.

Multiplication axioms are similar.

that was not a lot to do.

Showing no order can exist is figure: $0 = 1 + 1 + 1$ If $1 > 0$ then $1+1 > 0+1$ and $1+1+1 > 0+1+1$ and if $1< 0$ .... well same thing.