# Number theory - fields, multiplication table

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1. Sep 14, 2016

### Rectifier

The problem
Consider field $(F, +, \cdot), \ F = \{ 0,1,2,3 \}$

Find a multiplication table.

The attempt

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 dont 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.

2. Sep 14, 2016

### Staff: Mentor

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

3. Sep 14, 2016

### Ray Vickson

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.)

4. Sep 14, 2016

### Staff: Mentor

Good point!

The book answer looks like a shift operator.

5. Sep 14, 2016

### Staff: Mentor

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.

Last edited: Sep 14, 2016
6. Sep 14, 2016

### 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.

7. Sep 14, 2016

### Staff: Mentor

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.

8. Sep 14, 2016

### SrayD

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

9. Sep 14, 2016

### Staff: Mentor

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.

10. Sep 14, 2016

### Staff: Mentor

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 $1 \cdot 1 = 1$).

11. Sep 14, 2016

### Staff: Mentor

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.

12. Sep 14, 2016

### Staff: Mentor

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.

13. Sep 14, 2016

### SammyS

Staff Emeritus
What is the definition of a field?

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

14. Sep 14, 2016

### 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.

15. Sep 14, 2016

### Staff: Mentor

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.

16. Sep 14, 2016

### Rectifier

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?)

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.

Last edited: Sep 14, 2016
17. Sep 14, 2016

### Staff: Mentor

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 $3-1=3+1=2$.

18. Sep 14, 2016

### 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 :,(

19. Sep 14, 2016

### Staff: Mentor

$1-1=0$ 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!

20. Sep 14, 2016

### Rectifier

Thank you so much for your help so far!

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.

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.

Last edited: Sep 14, 2016