Complete the table for the finite field

[diredragon]

Homework Statement

Let $({a, b, c}, *,+)$ be a finite field. Complete the field table for the operations $*$ and $+$
$\begin{array}{|c|c|c|c|} \hline * & a & b & c \\ \hline a & ? & ? & ? \\ \hline b & ? & ? & ? \\ \hline c & ? & ? & b \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|} \hline + & a & b & c \\ \hline a & a & ? & ? \\ \hline b & a & c & ? \\ \hline c & a & ? & ? \\ \hline \end{array}$

Homework Equations

3. The Attempt at a Solution [/B]
I figured that since it is a finite field the operation $+$ must be commutative and must have a symmetric table along the main diagonal.
$\begin{array}{|c|c|c|c|} \hline + & a & b & c \\ \hline a & a & a & a \\ \hline b & a & c & ? \\ \hline c & a & ? & ? \\ \hline \end{array}$
Now i have to use the information from the first table that $c*c = b$. I can use that by seeing that $b+b = c$ and than i can say:
$c*c = b$
$b+b = c$
$c*c + c*c = c$
$c*(c+c) = c$
and i can't figure what this tells me. Other than this i don't see a clue in solving this. Could you provide a helpfull clue?

 

[Dick]

I think there must be some errors in the question. The addition table is telling you that $a+b=a=a+a$. Adding the additive inverse of $a$ to both sides tells you $a=b$. ?

 

[Buffu]

I think there must be some errors in the question. The addition table is telling you that $a+b=a=a+a$. Adding the additive inverse of $a$ to both sides tells you $a=b$. ?

Doesn't that also tell that $a,b,c$ all are additive identity ? so does not that means you put whatever you want in any box ?

 

[SammyS]

Homework Statement

Let $({a, b, c}, *,+)$ be a finite field. Complete the field table for the operations $*$ and $+$
$\begin{array}{|c|c|c|c|} \hline * & a & b & c \\ \hline a & ? & ? & ? \\ \hline b & ? & ? & ? \\ \hline c & ? & ? & b \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|} \hline + & a & b & c \\ \hline a & a & ? & ? \\ \hline b & a & c & ? \\ \hline c & a & ? & ? \\ \hline \end{array}$

Homework Equations

3. The Attempt at a Solution [/B]
I figured that since it is a finite field the operation $+$ must be commutative and must have a symmetric table along the main diagonal.
$\begin{array}{|c|c|c|c|} \hline + & a & b & c \\ \hline a & a & a & a \\ \hline b & a & c & ? \\ \hline c & a & ? & ? \\ \hline \end{array}$
Now i have to use the information from the first table that $c*c = b$. I can use that by seeing that $b+b = c$ and than i can say:
$c*c = b$
$b+b = c$
$c*c + c*c = c$
$c*(c+c) = c$
and i can't figure what this tells me. Other than this i don't see a clue in solving this. Could you provide a helpful clue?

What you are given:
This is a finite FIELD.
Three elements: a, b, c .
Two binary operations: *, + .

In my opinion:

1. You can't assume that the * operation acts like multiplication.
2. You can't assume that the + operation acts like addition.​

In fact, it looks to be like the traditional roles are switched, with $\ a\$ being the zero element.

 

[rcgldr]

I'm wondering if + and * have been mistakenly swapped for this question.

 

[Dick]

What you are given:
This is a finite FIELD.
Three elements: a, b, c .
Two binary operations: *, + .

In my opinion:

1. You can't assume that the * operation acts like multiplication.
2. You can't assume that the + operation acts like addition.​

In fact, it looks to be like the traditional roles are switched, with a being the zero element.

Now that's a good point. They just say '+', not 'addition'. Sneaky.

 

[rcgldr]

1. You can't assume that the * operation acts like multiplication.
2. You can't assume that the + operation acts like addition.​

The symbols for mathematics, including fields, are defined to be + for "addition" and ⋅ (or * or x) for multiplication (at least for English usage). The question appears to have mistakenly swapped these symbols in the tables.

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

https://en.wikipedia.org/wiki/List_of_mathematical_symbols

 

[SammyS]

The symbols for mathematics, including fields, are defined to be + for "addition" and ⋅ (or * or x) for multiplication (at least for English usage). The question appears to have mistakenly swapped these symbols in the tables.

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

https://en.wikipedia.org/wiki/List_of_mathematical_symbols

As an Abstract Algebra exercise, it's not wise to make any assumptions regarding the roles of the two binary operations, without exploring what the given tables imply.

It's clear that the ' + ' operation is not consistent with being the "addition" operation. This is apparent from the given fragment of the ' + ' table.

Beyond this, OP (@diredragon) has demonstrated that the distributive law does not hold for ' * ' distributed over ' + ' .

The problem can be solved by thinking more abstractly about the two operations. Thinking this way will be most helpful for OP and we are here to help OP.

 

[rcgldr]

The symbols for mathematics, including fields, are defined to be + for "addition" and ⋅ (or * or x) for multiplication (at least for English usage). The question appears to have mistakenly swapped these symbols in the tables.

As an Abstract Algebra exercise, it's not wise to make any assumptions regarding the roles of the two binary operations, without exploring what the given tables imply.

It's clear that the ' + ' operation is not consistent with being the "addition" operation. This is apparent from the given fragment of the ' + ' table.

Beyond this, OP (@diredragon) has demonstrated that the distributive law does not hold for ' * ' distributed over ' + ' .

The problem can be solved by thinking more abstractly about the two operations. Thinking this way will be most helpful for OP and we are here to help OP.

My impression is that the symbols have been swapped in error. It's not normal for such problems to change the accepted meaning of mathematical symbols. The problem could have used different symbols that don't have an implied meaning, or noted that the symbols used do not conform to the symbols as defined in mathematics, including abstract algebra. For example, what if a problem asked for the solution to a = 5 - 3, but swapped the meaning of = and - ?

 

[diredragon]

What you are given:
This is a finite FIELD.
Three elements: a, b, c .
Two binary operations: *, + .

In my opinion:

1. You can't assume that the * operation acts like multiplication.
2. You can't assume that the + operation acts like addition.​

In fact, it looks to be like the traditional roles are switched, with $\ a\$ being the zero element.

The symbols for mathematics, including fields, are defined to be + for "addition" and ⋅ (or * or x) for multiplication (at least for English usage). The question appears to have mistakenly swapped these symbols in the tables.

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

https://en.wikipedia.org/wiki/List_of_mathematical_symbols

As an Abstract Algebra exercise, it's not wise to make any assumptions regarding the roles of the two binary operations, without exploring what the given tables imply.

It's clear that the ' + ' operation is not consistent with being the "addition" operation. This is apparent from the given fragment of the ' + ' table.

Beyond this, OP (@diredragon) has demonstrated that the distributive law does not hold for ' * ' distributed over ' + ' .

The problem can be solved by thinking more abstractly about the two operations. Thinking this way will be most helpful for OP and we are here to help OP.

My impression is that the symbols have been swapped in error. It's not normal for such problems to change the accepted meaning of mathematical symbols. The problem could have used different symbols that don't have an implied meaning, or noted that the symbols used do not conform to the symbols as defined in mathematics, including abstract algebra. For example, what if a problem asked for the solution to a = 5 - 3, but swapped the meaning of = and - ?

Neither operation is defined as usual addition or multiplication as their symbols might suggest. From the definition of a field and already filled table places the problem asks for the rest of the table to be filled based on that which was given.
$c*c=b$
$b+b=c$
$(c*c)+b=c$
$(c+b)*(c+b)=c$
Shouldn't this hold from the definition of a field. + is here a secondary operations and the general $a+(b*c) = a+b * a+c$

 

[rcgldr]

Neither operation is defined as usual addition or multiplication as their symbols might suggest.

But the symbols are defined in mathematics, including fields in general, that also apply to finite fields and abstract algebra with + meaning "addition" and ⋅ (or * or x) meaning "multiplication". The question calls it a finite field, and the term finite field defines meanings for the symbols + and ⋅ (or * or x), which the question violates. The question should have used symbols without defined meanings if part of the problem was to determine which table was addition and which was multiplication.

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

There are also classic rules about a single element of the field being the additive identity 'a', where a + b = b, and another single element of the field being the multiplicative identity 'm', where m ⋅ b = b. Wiki words this a bit different:

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

In the OP's question, the second table for + violates the rule about a single additive identity for the field.

Consider this example of a field with 4 elements that follows the rules of the classic definition:

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

 

[diredragon]

But the symbols are defined in mathematics, including fields in general, that also apply to finite fields and abstract algebra with + meaning "addition" and ⋅ (or * or x) meaning "multiplication".

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

There are also classic rules about a single element of the field being the additive identity 'a', where a + b = b, and another single element of the field being the multiplicative identity 'm', where m ⋅ b = b. Wiki words this a bit different:

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

In the OP's question, the second table for + violates the rule about a single additive identity for the field.

The question intended only to test the knowledge of what fields are and how their properties relate to the tables. I do not know the exact name of the tables but in my language they are something like ,,Kaeily tables". I may have mispronaunced the name. If you are confused about + in the second table just use whatever symbol you wish and the question remains the same. It does not mean addition. It means some operation defined by symbol +. Properties are to be drawn from the fact that its a finite field and the given data.

 

[SammyS]

We call it a Cayley table .

Neither operation is defined as usual addition or multiplication as their symbols might suggest. From the definition of a field and already filled table places the problem asks for the rest of the table to be filled based on that which was given.
$c*c=b$
$b+b=c$
$(c*c)+b=c$
$(c+b)*(c+b)=c$
Shouldn't this hold from the definition of a field.? + is here a secondary operation and in general $a+(b*c) = a+b * a+c$

Yes.

You may want to include parentheses to avoid any ambiguity.

$a+(b*c) = (a+b) * (a+c)$​

You have identified the + operation as being the "secondary operation". The table for + shows that $({a, b, c}, +)$ does not form a Group, because of the behavior of $\ a \,.\$ It's not the identity for + and it does not have an inverse for the + operation.

You should be able to determine the (only possible) identity element for + .

Can you ?

Added in Edit:
Actually, for the general case of the distributive property, you should use other symbols, rather than $\ a,\, b,\, c,.$

$x+(y*z) = (x+y) * (x+z)$​

 

[diredragon]

We call it a Cayley table .Yes.

You may want to include parentheses to avoid any ambiguity.

$a+(b*c) = (a+b) * (a+c)$​

You have identified the + operation as being the "secondary operation". The table for + shows that $({a, b, c}, +)$ does not form a Group, because of the behavior of $\ a \,.\$ It's not the identity for + and it does not have an inverse for the + operation.

You should be able to determine the (only possible) identity element for + .

Can you ?

Added in Edit:
Actually, for the general case of the distributive property, you should use other symbols, rather than $\ a,\, b,\, c,.$

$x+(y*z) = (x+y) * (x+z)$​

The only possible identity element for + is $c$ since $b+b=c$ so the table for + looks like:
$\begin{array}{|c|c|c|c|} \hline + & a & b & c \\ \hline a & a & a & a \\ \hline b & a & c & b \\ \hline c & a & b & c \\ \hline \end{array}$
How would i complete the * table? I know that $c$ can't be the identity so that leaves $a$ or $b$.

 

[Dick]

You know $a$ is the 'zero' for the multiplication $+$. What does that tell you?

 

[diredragon]

You know $a$ is the 'zero' for the multiplication $+$. What does that tell you?

That it is the identity element for the $*$?
So in general, for finite field $(A, *, +)$ if some element $x$ obeys $a+x = x$ and $x + a = x$ than that element $x$ is the identity for the operation $*$?

 

[Dick]

That it is the identity element for the $*$?
So in general, for finite field $(A, *, +)$ if some element $x$ obeys $a+x = x$ and $x + a = x$ than that element $x$ is the identity for the operation $*$?

Yes, prove that using the distributive property.

 

[diredragon]

Yes, prove that using the distributive property.

Let $(A, +, *)$ be a finite field. If: $(\exists t)$ $(\forall x \in A)$ $x*t = t*x = t$ than $y$ is an identity element for operation $+$.
Proof:
Using the distributive law:
$t*(x+y) = t$
$t = t*x+t*y$
$t = t + t$
This is true only is $t$ is neutral element for the $+$.
Let's suppose that it is in fact the neutral element. Then:
$(x+t)*y = x*y + t*y = x*y + t = x*y$
Is this valid? And id the second part necessary or is the first one enough? I kinda thought that the first was lacking as the same result could come up if $t$ is also the zero element of $+$ as well it being of $*$.

 

[Dick]

Let $(A, +, *)$ be a finite field. If: $(\exists t)$ $(\forall x \in A)$ $x*t = t*x = t$ than $y$ is an identity element for operation $+$.
Proof:
Using the distributive law:
$t*(x+y) = t$
$t = t*x+t*y$
$t = t + t$
This is true only is $t$ is neutral element for the $+$.
Let's suppose that it is in fact the neutral element. Then:
$(x+t)*y = x*y + t*y = x*y + t = x*y$
Is this valid? And id the second part necessary or is the first one enough? I kinda thought that the first was lacking as the same result could come up if $t$ is also the zero element of $+$ as well it being of $*$.

That's looks correct, except I think you have reversed the $*$ and $+$ back to their conventional meanings.

 

[diredragon]

That's looks correct, except I think you have reversed the $*$ and $+$ back to their conventional meanings.

I meant for it to be general with + the primary and * the secondary operation. It's more relatable this way as well :)

 

[Dick]

I meant for it to be general with + the primary and * the secondary operation. It's more relatable this way as well :)

I know. It's really hard to think with them backwards. And I do think the first part is enough.

 

[SammyS]

I meant for it to be general with + the primary and * the secondary operation. It's more relatable this way as well :)

(Added in Edit: I have deleted some of this post & struck through some. Better ideas have prevailed. No need to take up extra space.

Maybe use two other symbols for the given operations such as # for * and @ for +. I have converted the tables for these.

So @ is the "secondary operation" often referred to as multiplication and $\ a\$ is the identity element for #, the "primary operation", a.k.a. addition..

Also, @ distributes over # as in:

x@(y # z) = (x@y) # (x@z)​

.

 

[Dick]

Maybe use two other symbols for the given operations such as # for * and @ for +.

Maybe. I went with diredragon's approach in post 20. Work it out with + for addition and * for multiplication and then reverse them when you are finished.

 

[SammyS]

Maybe. I went with diredragon's approach in post 20. Work it out with + for addition and * for multiplication and then reverse them when you are finished.

Yes, that is perfectly reasonable. In fact, as I put my post (22) together I made several mistakes and had to edit and re-edit.
(I may edit it again using again with diredragon's approach - to basically summarize what his progress so far.)

 

[diredragon]

Maybe use two other symbols for the given operations such as # for * and @ for +. I have converted the tables for these.

The * table becomes $\ \displaystyle \begin{array}{|c|c|c|c|} \hline \# & a & b & c \\ \hline a & ? & ? & ? \\ \hline b & ? & ? & ? \\ \hline c & ? & ? & b \\ \hline \end{array}$

The + table which you completed becomes $\ \displaystyle \begin{array}{|c|c|c|c|} \hline @ & a & b & c \\ \hline a & a & a & a \\ \hline b & a & c & b \\ \hline c & a & b & c \\ \hline \end{array}$

So @ is the "secondary operation" often referred to as multiplication and $\ a\$ is the identity element for #, the "primary operation", a.k.a. addition..

Also, @ distributes over # as in:

x@(y # z) = (x@y) # (x@z)​

.

Maybe. I went with diredragon's approach in post 20. Work it out with + for addition and * for multiplication and then reverse them when you are finished.

Let's solve it and make it clear ditching + and - and using ⊕ and ⊗.
Let $({a, b, c}, ⊕, ⊗)$ be a finite field. Complete the Cayley's tables for operations $⊕$ and $⊗$
$\begin{array}{|c|c|c|c|} \hline \ ⊕ & a & b & c \\ \hline a & ? & ? & ? \\ \hline b & ? & ? & ? \\ \hline c & ? & ? & b \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|} \hline \ ⊗ & a & b & c \\ \hline a & a & ? & ? \\ \hline b & a & c & ? \\ \hline c & a & ? & ? \\ \hline \end{array}$

From the $⊗$ table we see that since it's a finite field $⊗$ is commutative and hence symmetrical on both sides of the main diagonal.
Furthermore from the table it is seen that $b⊗b=c$ and we conclude that $c$ must be the identity element. From there we complete.
$\begin{array}{|c|c|c|c|} \hline \ ⊗ & a & b & c \\ \hline a & a & a & a \\ \hline b & a & c & b \\ \hline c & a & b & c \\ \hline \end{array}$

Element $a$ from the table $⊗$ is a destructive element which means it's the identity of the table $⊕$. From the table it is seen that $c⊕c=b$ which means $c$ is not the inverse element of $c$ because it does not yield $a$ which is the identity. Therefore it is $b$ and $c⊕b=a$. For element $b$ by the same logic the inverse is $c$.
The complete table is given as:
$\begin{array}{|c|c|c|c|} \hline \ ⊕ & a & b & c \\ \hline a & a & b & c \\ \hline b & b & c & a \\ \hline c & c & a & b \\ \hline \end{array}$

 

[Dick]

Let's solve it and make it clear ditching + and - and using ⊕ and ⊗.

And that's it.

 

[SammyS]

Let's solve it and make it clear ditching + and - and using ⊕ and ⊗.
Let $({a, b, c}, ⊕, ⊗)$ be a finite field. Complete the Cayley's tables for operations $⊕$ and $⊗$
...

Excellent choice and excellent result. (No need for me to quote the entire post.)

There are several (equivalent) ways to describe a field. One which I like goes like this:

Given a set (finite set for a finite field) and two binary operations on that set: $\ (A, \oplus, \otimes) \$. This combination is a field, provided that:

1. The set, $\ A\,$, together with one of the operations (in this case, $\ \oplus\$) forms a commutative group. Even if the elements are not numbers, we generally call this "addition" and refer to its identity element as the "zero" for this field. Let's call it $\ \hat 0 \$ here.
2. The set, $\ A\backslash \hat 0 \,$, together the other operation, $\ \otimes\,$, forms a commutative group. Even if the elements are not numbers, we generally call this "multiplication" and refer to its identity element as "unity" for this field.
   
3. The distributive law holds for multiplication over addition. $\ x\otimes (y \oplus z) = (x\otimes y) \oplus (x\otimes z) \,$, for all $x, y, z$ in $A$ .

I see Dick has also given his seal of approval . :-)

 

[rcgldr]

There are several (equivalent) ways to describe a field.

From wiki: "a field is a set on which are defined addition, subtraction, multiplication, and division, which behave as they do when applied to rational and real numbers. " ... "a field is a set on which are defined an addition a + b, and a multiplication a ⋅ b that behave similarly as they behave for rational numbers and real numbers, including the existence of an additive inverse −a for all elements a, and of a multiplicative inverse b⁻¹ for every nonzero element b. This allows to consider also the so-called inverse operations of subtraction a − b, and division a / b ... "

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

"a finite field is a set on which the operations of multiplication, addition, subtraction and division are defined and satisfy certain basic rules. "

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

From the error correction textbooks I have:

"A ring is a set of elements for which two operations are defined. One is called addition and denoted a + b and the other is called multiplication and denoted ab (or a · b as shown later), even though these operations may not be ordinary addition or multiplication ... " "A field is a commutative ring with a unit element (multiplicative identity) in which every non-zero element has a multiplicative inverse." - Error Correcting Codes - Peterson and Weldon.

"...a field is a set of elements in which we can perform addition, subtraction, multiplication, and division without leaving the set." "... two binary operaions, called addition "+" and multiplication "·" are defined." - Error Control Coding - Shu Lin - Daniel J. Costello, Jr.