Number theory - fields, multiplication table

[Rectifier]

There is of course addition. That's how it's made and meant. And of course you can read subtractions out of it, too.
Row 2 (number $1$) and column 3 (number $2$) gives you $3$ in the table, which means $1+2=3$
If you read it inside out, you get a subtraction: Search $3$ in the row that is label with $2$, i.e. the third row, and look up the label of the column, here the second column (where we have found the $3$), that is labeled by $1$. Then you have performed $3-2=1$.
So the table gives you both, depending how you read it.
My remark that for all $x$ hold $x=-x$ means essentially that all minus signs can be treated as plus signs in this special case here.

Please immediately forget $\mod 4$ here. It has nothing to do with $\mod 4$ or $\mathbb{Z}_4$. Nothing.
I even would have added the additional exercise to show, that this field is different from $\mathbb{Z}_4$. The latter cannot be a field!
Do you know why? If you answer this question, then this will help you a lot to rule out the false possibilities for $3 \cdot 3$.
(Hint: $3 \cdot 3 = 0$ cannot be. Why?)

So all you have is the elements, the addition between them and the laws that a field has to obey:
associativity, distributivity, etc.

It has been a half years time and I still haven't figured out a way to solve this lovely problem. I have even tried replacing 0,1,2,3 with a,b,c,d and $+$ and $\cdot$ with arbitrary operations but I feel that this is getting out of hand, I need you help once again.

I tried to read all comments in this thread but I have a hard time following the discussion about bad notations and other topics most of the time. Let's focus on solving this problem now and not discussing the technicalities (thank you c: ). I have pulled enough hairs out of my head over this problem already.

$\mathbb{Z}_4$ is not a field since 4 is not a prime (there is a definition in my old lecture notes). I have also learned that if an element e has an inverse (multiplicative inverse) the gcd(m,e)=1 should be 1 but I guess that that is only the case with integers ##\mathbb{Z}_m#. We know that all elements in this field have an inverse but the problem is to find the elements that are their multiplicative inverse.

Here is the definition of a field from my old lecture:

I didn't get your tip about $3 \cdot 3$, unfortunately. I don't know if 3 + 3 = 0 has anything to do with that. Namely;
$(3 + 3) +(3 + 3) = 0 + 0 \Rightarrow (3+3+3) + 3 = 0 \Rightarrow 3 \cdot 3 + 3 = 0 \Rightarrow 3 \cdot 3 = -3$ ?

 

[rcgldr]

The confusion seems to be trying to relate addition and multiplication. For this field, 2·a is not a+a and 3·a is not a+a+a.

The axioms for a field do not include 0·a = 0 or (-1)·a = -a. The wiki article claims states these are consequence of the axioms, but doesn't show a derivation.

https://en.wikipedia.org/wiki/Field_(mathematics)#Elementary_consequences_of_the_definition

Also if a != b, then for c != 0, c·a != c·b. Show by contradiction: assume a != b, c != 0, then if c·a = c·b, there is an inverse d = 1/c => d·c·a = d·c·b => a = b, which is a contradiction.

Addition is XOR, rewriting the table using 2 bit binary numbers:

\begin{array}{l | c|c|c|c |} + & 00 & 01 & 10 & 11\\
\hline 00 & 00 & 01 & 10 & 11 \\
\hline 01 & 01 & 00 & 11 & 10 \\
\hline 10 & 10 & 11 & 00 & 01 \\
\hline 11 & 11 & 10 & 01 & 00 \\
\end{array}

Since this is a field, then for the multiplication table, 0·a = 0, and 1·a = a, so that takes care of all but the four bottom right products, 2·2, 2·3, 3·2, 3·3. Consider the table row for 2·a: 2·0 = 0, 2·1 = 2, 2·2 = ?, 2·3 = ?, and for 3·a: 3·0 = 0, 3·1 = 3, 3·3 = ?, 3·2 = ? . Since every non-zero number a has a inverse 1/a, then two of the four possible products 2·2 = ?, 2·3 = ?, 3·2 = ?, 3·3 = ? have to equal 1. Since 2·3 = 3·2, then it would seem that 2·3 = 3·2 = 1, leaving two remaining possible products for 2·2 and 3·3. As answered by Dick, you can use trial and error to determine the products by checking to see is the axioms fail for a trial set of products, or using if a != b, then for c != 0, c·a != c·b, then there's only one possible value left in the row for 2·a and the row for 3·a.

It turns out that this is the same as multiplication done via GF(4) (Galois Field) (binary math modulo x^2 + x + 1), wiki link:

https://en.wikipedia.org/wiki/Finite_field#Field_with_four_elements

In the wiki table, a can be either 2 or 3, and + means XOR. Using a = 2, and 2 bit numbers:

\begin{array}{c | c|c|c|c |} · & 00 & 01 & 10 & 11\\
\hline 00 & 00 & 00 & 00 & 00 \\
\hline 01 & 00 & 01 & 10 & 11 \\
\hline 10 & 00 & 10 & 11 & 01 \\
\hline 11 & 00 & 11 & 01 & 10 \\
\end{array}

Being able to produce a multiplication table from an addition table for a finite field only works in specific cases, such as this one or integer type field modulo some prime number. For a binary field GF(8), there's one addition table, but 2 possible multiplication tables, depending if the field is modulo x^3 + x + 1 or x^3 + x^2 + 1. For GF(16), there are 3 possible multiplication tables, and for GF(256), 30 possible multiplication tables.

In this case, for a field with 4 elements, there's only one addition table and one multiplication table:

https://en.wikipedia.org/wiki/Field_(mathematics)#A_field_with_four_elements

 

[Dick]

It has been a half years time and I still haven't figured out a way to solve this lovely problem.

Just concentrate on the part of the table mixing 2 and 3. You already know what the rest must be. Now use the properties of the identity to conclude that every row and column must contain one of each number (like sudoko). Then you can cut it down to only two possibilities. Either i) 2*3=1 (2 and 3 are each others inverse) or ii) 2*2=1 and 3*3=1 (they are their own inverses). Now work with examples the distributive law to show one of the possibilities violates the distributive law. When you find that you know the other possibility must be the correct one.