Number theory - fields, multiplication table

[Rectifier]

The problem
Consider field ##(F, +, \cdot), \ F = \{ 0,1,2,3 \}##
With the addition table:

Find a multiplication table.

The attempt
Please read the most of my answer before writing a reply.

My solution was
$$
\begin{array}{|c|c|c|}
\hline \cdot & 0 & 1 & 2 & 3 \\\hline
0 & 0 & 0 & 0 & 0 \\\hline
1 & 0 & 1 & 2 & 3 \\\hline
2 & 0 & 2 & 0 & 2 \\\hline
3 & 0 & 3 & 2 & 1 \\\hline
\end{array}
$$

The solution in the book was

I don't really understand how they get ##2 \cdot 2 = 3## and ##3 \cdot 3 = 2## and ##3 \cdot 2 = 2 \cdot 3 = 1##. I know that multiplication in a field is commutative so a \cdot b = b \cdot a thus ##3 \cdot 2 = 2 \cdot 3 ## but I don't know why it is 1 and not 2.

Lets take one of these as example:
##2 \cdot 2## is 4 right? Then when we calculate the the remainder in respect to 4 ( as it is 4 elements in the field and a field is by definition a commutative ring, right? and we want the answer to be another element in the ring) so it becomes 0. That is ##R_4(4) = 0##. For ##3 \cdot 3 = 9## which produces the remainder ##R_4(9)=1##.
Could someone please help me with what I am doing wrong and how I should solve it instead? Please be specific. Tell me how I should tackle this problem.I am not asking for a solution but for the procedure for these kinds of problems.

 

[jedishrfu]

To solve these kinds of problem you must remember that multiplication is in fact multiple additions and so 3 x 3 means 3 + 3 + 3 and you must use the addition table to determine that 3 + 3 is 0
and hence 3 + 3 + 3 = (3 + 3) + 3 = 0 + 3 = 3

 

[Ray Vickson]

To solve these kinds of problem you must remember that multiplication is in fact multiple additions and so 3 x 3 means 3 + 3 + 3 and you must use the addition table to determine that 3 + 3 is 0
and hence 3 + 3 + 3 = (3 + 3) + 3 = 0 + 3 = 3

The fact that multiplication is repeated addition holds for ordinary numbers with the usual definitions. It may not---or at least, need not---hold for the field(s) in this question. I think that in this question there are several different possible multiplication tables, so there are really several different possible fields. (Note that the question asked for a multiplication table.)

 

[jedishrfu]

The fact that multiplication is repeated addition holds for ordinary numbers with the usual definitions. It may not---or at leas, need not---hold for the field(s) in this question. I think that in this question there are several different possible multiplication tables, so there are really several different possible fields. (Note that the question asked for a multiplication table.)

Good point!

The book answer looks like a shift operator.

 

[fresh_42]

There is only one solution. (As it is homework, I'll write only hints.)

• For all elements is ##-x=\,##?
• ##(3 \cdot 3 - 1 \cdot 1) = (3+1) \cdot (3-1)## gives one condition
• ##3 \cdot 2 = (2+1) \cdot 2## gives another condition

Now for ##3 \cdot 3 \in \{0,1,2,3\}## all but one can be ruled out by the fact that it has to be a field.

 

[SrayD]

This question does not look right. The use of 2 & 3 is the problem, mostly syntax, but does obscure semantics.The finite field of order 4 is a second degree extension of the finite field of order 2; so will work modulo 2. Means the addition table is incorrect as well, from mod 2 perspective. There must be a new element introduced that is the zero of x^2 + x + 1 = 0, over F2 = {0,1}. The addition table given is confusing in that it uses 2 & 3 when it should not, because you can see neither mod 2 nor mod 4 work with these symbols to give the addition table shown; they behave as the zeros of the polynomial x^2 + x + 1 = 0; usually given symbol as lower case alpha or something other than 2 or any other numeric symbol. I am thinking that maybe it was intended as a simplification, but just makes things worse. I could elaborate more, but do not want to answer question, being an assignment.

 

[jedishrfu]

There is only one solution. (As it is homework, I'll write only hints.)

• For all elements is ##-x=\,##?
• ##(3 \cdot 3 - 1 \cdot 1) = (3+1) \cdot (3-1)## gives one condition
• ##3 \cdot 2 = (2+1) \cdot 2## gives another condition

Now for ##3 \cdot 3 \in \{0,1,2,3\}## all but one can be ruled out by the fact that it has to be a field.

Where did you get the specifics of the problem? I didn't see these in the OP's post.

Don't answer if you feel it will give away the solution though.

 

[SrayD]

Sorry, is it a assignment question or a personal exploration problem?

 

[jedishrfu]

Sorry, is it a assignment question or a personal exploration problem?

An assignment from the looks of it, you should look at all PF posts as possible assignments unless others have posted comments and it then appears not to be or the cat is out of the bag.

 

[fresh_42]

Where did you get the specifics of the problem? I didn't see these in the OP's post.

Don't answer if you feel it will give away the solution though.

I don't understand your question. I thought the specifics are: Given a field and an addition table, what is the multiplication table?
So I just proposed to apply the distribution law without substituting the products by elements, just by writing them ##2\cdot 2 ,2\cdot 3## and ##3 \cdot 3## and using the addition table where possible (and of course ##1 \cdot 1 = 1##).

 

[jedishrfu]

I don't understand your question. I thought the specifics are: Given a field and an addition table, what is the multiplication table?
So I just proposed to apply the distribution law without substituting the products by elements, just by writing them ##2\cdot 2 ,2\cdot 3## and ##3 \cdot 3## and using the addition table where possible.

I didn't mean to imply anything it just seemed that you had listed specifics that I didn't see and figured I missed something. I guess I've forgotten that given a field implies a lot.

 

[fresh_42]

I didn't mean to imply anything it just seemed that you had listed specifics that I didn't see and figured I missed something. I guess I've forgotten that given a field implies a lot.

No, it was as simple as: Given three unknowns, ##2 \cdot 2,2 \cdot 3, 3 \cdot 3##, which equations are among them, That led me to the distribution law.

 

[SammyS]

The problem
Consider field ##(F, +, \cdot), \ F = \{ 0,1,2,3 \}##
With the addition table:
##
\begin{array}{|c|c|c|}
\hline \ + & 0 & 1 & 2 & 3 \\ \hline
0 & 0 & 1 & 2 & 3 \\\hline
1 & 1 & 0 & 3 & 2 \\\hline
2 & 2 & 3 & 0 & 1 \\\hline
3 & 3 & 2 & 1 & 0 \\\hline
\end{array}
##

Find a multiplication table.

The attempt
Please read the most of my answer before writing a reply.

My solution was
##
\begin{array}{|c|c|c|}
\hline \cdot & 0 & 1 & 2 & 3 \\\hline
0 & 0 & 0 & 0 & 0 \\\hline
1 & 0 & 1 & 2 & 3 \\\hline
2 & 0 & 2 & 0 & 2 \\\hline
3 & 0 & 3 & 2 & 1 \\\hline
\end{array}
##

The solution in the book was
##
\begin{array}{|c|c|c|}
\hline \cdot & 0 & 1 & 2 & 3 \\\hline
0 & 0 & 0 & 0 & 0 \\\hline
1 & 0 & 1 & 2 & 3 \\\hline
2 & 0 & 2 & 3 & 1 \\\hline
3 & 0 & 3 & 1 & 2 \\\hline
\end{array}
##
...

Could someone please help me with what I am doing wrong and how I should solve it instead? Please be specific. Tell me how I should tackle this problem.I am not asking for a solution but for the procedure for these kinds of problems.

First question to ask (and answer):
What is the definition of a field?

Notice that in your proposed definition of multiplication, 2 has no (multiplicative) inverse.

 

[SrayD]

The problem is bad notation on the books part; should never use numeric symbols e.g. 2 & 3. Because the solution is adding a new element that solves the polynomial x^2 + x +1 = 0; which 2 & 3 do not, looking mod 2 or 4. The book is using '2' & '3' not as usual 2 & 3. With there notation, '2' solves x^2 + x + 1 = 0 mod2, which looks really weird; and why I'm saying the book is bad. Remember the finite field of order 4 is a 2nd degree extension of the field of order 2; you have the adjoin a zero of that polynomial; similar to extending real numbers to complex by adding i = (-1)^(.5). So, this book is using '2' & '3' as the new adjoined elements. This horrible notation from the book has led those down the wrong path; it's like trying to extend reals to complex by adjoining other real numbers, instead of some completely new "number". I hope this is getting through, I can reiterate if needed.

 

[fresh_42]

The problem is bad notation on the books part ...

This is not my opinion.

The way it is presented is, that there exists a field with four elements. It is automatically clear that the role of these elements cannot be the same as within the integers. Furthermore it is far better than naming four elements ##a,b,c,d## which is too abstract to keep in mind. Whereas the notation by numbers trains people to exactly define what is meant as needed everywhere in mathematics. We also often denote elements in ##\mathbb{Z}_5## and similar by numbers. I don't see any reason why it should be handled differently here.

There is everything alright with the book.

 

[Rectifier]

Thank you for all of your comments! I really appreciate the hints I have received,

First of all, it is a problem from a book (not an assignment that I should hand in, I am doing it just to get hang of the theory in the book as for all problems I have ever posted on PF but that does not really matter since there is almost no way to tell if a user posts an assignment disguised as a problem from a book so we should follow the guidelines :D ).

So, what I get from the comments is that I should somehow use the addition table to solve the problem (a question for later: what if we didn't have that addition table to start with?)

There is only one solution. (As it is homework, I'll write only hints.)

• For all elements is ##-x=\,##?

it is ##x## as it is an additive inverse and so are ##1,2,3##. I don't get your other hints though.

As@SammyS mentioned in #13, 2 didn't have an multiplicative inverse in my multiplication table which is wrong by definition (since there always exists an inverse in a field) so we know that my table is wrong without looking at the key table.

 

[fresh_42]

What am I missing here? I feel like I would be able to solve that problem by myself hadn't I misunderstood that crucial piece of theory.

Yes, by ##-x## I meant the number that solves ##x + (-x) = 0## And yes, ##x=-x## as you said (and deleted). I only mentioned it, because it makes your calculations easier. It means that you won't have to distinguish between addition and subtraction: ##x+y = x -y##. This way you won't have to calculate every "##-##" on its own but you may look up the addition in the table instead, e.g. ##2-1=2+1=3## or ##3-1=3+1=2##.

 

[Rectifier]

I know that by definition that additive inverse to 1 is its negative self right? I end up with ##1-1=0## which is by definition written as ##1 + (-1) = 0## and according to addition table (you take one term and add it to one another) I end up with two different versions of these additive inverses:
##1 + (-1) = 0## or ##1 -1 = 0## for definition and
##1 + 1 = 0## for the addition table :,(

 

[fresh_42]

I know that by definition that additive inverse to 1 is its negative self right? I end up with ##1-1=0## which is by definition written as ##1 + (-1) = 0## and according to addition table (you take one term and add it to one another) I end up with two different versions of these additive inverses:
##1 + (-1) = 0## or ##1 -1 = 0## for definition and
##1 + 1 = 0## for the addition table :,(

##1-1=0## is simply a short version of ##1 + (-1)=0##. It means the same. And the addition table shows you ##x+x=0## for any number.
This is the same as saying ##x=-x##.
Important here are two things:

1) Do not take any addition for granted. Look them up instead. Exception: ##x-x= 0## and ##x+0=x## by definition of ##0##.
2) Do not take the numbers ##2 \cdot 2 \,,\, 2 \cdot 3 \,,\, 3 \cdot 3## as if you knew the result. Here ##1## is the exception, because ##x\cdot 1 = x## by definition.

With that and the equations I gave you, ##3 \cdot 3 - 1 \cdot 1 = 3 \cdot 3 - 1 = (3+1) \cdot (3-1) ## and ##3 \cdot 2 = (2+1)\cdot 2 = 2 \cdot 2 + 2 \cdot 1##
you can consecutively deal with the cases ##3 \cdot 3 = 0## , ##3 \cdot 3 = 1## , etc. and see which one will lead to impossible solutions and which doesn't. Btw. I did not take 3=2+1 as naturally given, I looked it up!

 

[Rectifier]

Thank you so much for your help so far!

##1-1=0## is simply a short version of ##1 + (-1)=0##. It means the same. And the addition table shows you ##x+x=0## for any number.
This is the same as saying ##x=-x##.

So in an addition table there is no addition but only subtraction?
1 - 3 = 1 - 3 + 4 (as it is mod 4) = 1 + 1 = 2
1 - 1 = 0 and so on

EDIT:
It does not seem to work for 2 - 3 = 2 - 3 + 4 = 3 but the table gives 1.

##3 \cdot 3 - 1 \cdot 1

How do you get that equation? Are you deriving it from the fact that ##3 \cdot 3 = 1## \cdot 1## and thus ##3 \cdot 3 - 1 \cdot 1 =0## ?

EDIT 2:
I think that my head has had enough of math for today. I will sleep on this information and hopefully I will understand it tomorrow.

 

[fresh_42]

So in an addition table there is no addition but only subtraction.

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.

1 - 3 = 1 - 3 + 4 (as it is mod 4) = 1 + 1 = 2
1 - 1 = 0 and so on

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.

 

[rcgldr]

Why is this called a field? Since the only defined operations are addition and multiplication (no subtraction, no division), shouldn't this be called a ring?

 

[SammyS]

Why is this called a field? ...

It satisfies the axioms for a field.

 

[fresh_42]

The addition table is based on binary exclusive or. For example 3 + 2 = 11 + 10 = 1.

The multiplication is based binary multiplication using xor instead of add, modulo 7 (binary 111). For example 3*3 mod 7 = 11*11 mod 111 = 101 mod 111 = 2.

If you insist on the ring structure, what are its maximal ideals?

 

[rcgldr]

If you insist on the ring structure, what are its maximal ideals?

With only addition defined, it wasn't clear that the ring would also be a field until I understood the math behind the addition and multiplication tables.

The addition table is based on binary exclusive or. For example 3 + 2 = 11 + 10 = 01 = 1.

The multiplication is binary multiplication using xor instead of add, modulo 7 (binary coefficient polynomial x^2 + x + 1).
2*2 mod 7 = 10*10 mod 111 = (000 + 100) mod 111 = 100 mod 111 = 011 = 3
3*3 mod 7 = 11*11 mod 111 = (011 + 110) mod 111 = 101 mod 111 = 010 = 2

This would be a finite field. Subtraction is also xor, so effectively the same as addition. The division table can be generated from the multiplication table. Every non-zero number is a power of 2, 2^0 = 1, 2^1 = 2, 2^2 = 3, so 2 (x + 0) would be a primitive (alpha) for the field. This also works with 3, 3^0 = 1, 3^1 = 3, 3^2 = 2, so 3 (x + 1) would also be a primitive (alpha).

I'm wondering if the students in this class were supposed to be aware of the exclusive or operator.

 

[SrayD]

This is not my opinion.

The way it is presented is, that there exists a field with four elements. It is automatically clear that the role of these elements cannot be the same as within the integers. Furthermore it is far better than naming four elements ##a,b,c,d## which is too abstract to keep in mind. Whereas the notation by numbers trains people to exactly define what is meant as needed everywhere in mathematics. We also often denote elements in ##\mathbb{Z}_5## and similar by numbers. I don't see any reason why it should be handled differently here.

There is everything alright with the book.

I will have to disagree with you about notation. Yeah of course the symbols '1', '2', etc are not integers in this case, well they are through a homomorphism of course. Also, you have to go with some abstract notation on this one. Your adjoining a new element to extend the base field, that is not a homomorphic image of an integer. Again, it's like extending reals to complex, normally use 'i' as symbol for root of x^2 + 1 = 0; same in this case need symbol for root of x^2 + x + 1 = 0 mod2; using '2' & '3' for these you can do, but bad notation. With all the time and confusion on this question it should be kind of obvious that there is an issue with notation obscuring understanding. Of course we are all entitled to our opinions, but I hope that you at least see what I am concerned about. Also, I see none of my advice about solving the problem is being discussed, no talk of finite field extension etc; so no point for me to waste time on this. I do hope Rectifier42 is not dissuaded.

 

[SrayD]

I will have to disagree with you about notation. Yeah of course the symbols '1', '2', etc are not integers in this case, well they are through a homomorphism of course. Also, you have to go with some abstract notation on this one. Your adjoining a new element to extend the base field, that is not a homomorphic image of an integer. Again, it's like extending reals to complex, normally use 'i' as symbol for root of x^2 + 1 = 0; same in this case need symbol for root of x^2 + x + 1 = 0 mod2; using '2' & '3' for these you can do, but bad notation. With all the time and confusion on this question it should be kind of obvious that there is an issue with notation obscuring understanding. Of course we are all entitled to our opinions, but I hope that you at least see what I am concerned about. Also, I see none of my advice about solving the problem is being discussed, no talk of finite field extension etc; so no point for me to waste time on this. I do hope Rectifier42 is not dissuaded.

Sorry, meant just Rectifier, not Rectifier42.

 

[fresh_42]

[deleted, because too offending]

@SrayD Just a few questions?

- Do you think your view is helpful in the given context?
- What else do you know about the book, that I don't? Intentions? Further content? Who it is addressed to?
- What do you know about the OP's private circumstances, that you can afford such a judgement?
- Are you prepared to fully explain what you mean from the scratch?
- Did you offer any constructive alternatives?

To say the OP's book is "terrible" is in my opinion simply irresponsible. Nothing less.

 

[SrayD]

[deleted, because too offending]

@SrayD Just a few questions?

- Do you think your view is helpful in the given context?
- What else do you know about the book, that I don't? Intentions? Further content? Who it is addressed to?
- What do you know about the OP's private circumstances, that you can afford such a judgement?
- Are you prepared to fully explain what you mean from the scratch?
- Did you offer any constructive alternatives?

To say the OP's book is "terrible" is in my opinion simply irresponsible. Nothing less.

I think this has gotten out of hand and not benefiting Rectifier. I apologize if I upset anyone. Also, never said book is "terrible", said the notation for this problem was horrible; especially for those just learning the material. Any slights were imagined.

 

[rcgldr]

symbol for root of x^2 + x + 1 = 0 mod2; using '2' & '3' for these you can do, but bad notation.

In my work with Galois fields (mostly for error correction code), it's common to describe n term binary (0 or 1) coefficient polynomials as hex values, for example, x^4 + x^3 + x + 1 = binary 11011 = hex 1b (some textbooks use octal). The unknown here is what has been presented to the students prior to this assignment, such as the concept of exclusive or for binary field addition, or representing such polynomials as hex or octal values. In this case you have a 2 bit field modulo the binary coefficient polynomial x^2 + x + 1 (hex 7), (where alpha can be x + 0 (hex 2)or x+1 (hex 3)). I gave an answer back in post #25.

Wiki article about finite field with 4 elements. Note that a (alpha) can be either x+0 or x+1. In the original post here, alpha is x+0 or hex 2. Within the tables, the "+" in the expression "1+a" is effectively exclusive or.

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

 

[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.