Find the characteristic of the field

[Hill]

Homework Statement

In a field K the equality $a^4=a$ is satisfied for all a. Find the characteristic of the field K.

Relevant Equations

$a^4=a$

My calculations:
$2a = (2a)^4 = 16a^4 = 16a$
$14a=0$
This allows for the characteristic to be 2 or 7. But, e.g., $3^4 = 81 = 4 ~(mod~ 7)$ rather than $3$. So, the only answer is $2$.

1. Is this derivation correct?
2. The hint in the textbook says to show that $2a=0$. How to do that?

 

[fresh_42]

My calculations:
$2a = (2a)^4 = 16a^4 = 16a$
$14a=0$
This allows for the characteristic to be 2 or 7. But, e.g., $3^4 = 81 = 4 ~(mod~ 7)$ rather than $3$. So, the only answer is $2$.

1. Is this derivation correct?

Yes.

2. The hint in the textbook says to show that $2a=0$. How to do that?

$2a=0$ is equivalent to $1+1=0$ so the hint is not really helpful.

I could only find something similar to your trick:
\begin{align*}
0&=a-a=a^4-a=a(a^3-1)\\
0&=3\cdot (3^3-1)= 2\cdot 3\cdot 13
\end{align*}
and
\begin{align*}
2&=2^4=16\\
0&=14=2\cdot 7
\end{align*}
Hence $2$ is the only prime that occurs in both equations.

 

[mathwonk]

If every non zero element is a root of X^3 -1 = 0, how many elements does the field contain?

 

[fresh_42]

If every non zero element is a root of X^3 -1 = 0, how many elements does the field contain?

Besides $1,$ the two roots of $x^2+x+1.$ That's where I got stuck with that approach.

 

[pines-demon]

Besides $1,$ the two roots of $x^2+x+1.$ That's where I got stuck with that approach.

I do not think there is a field where $a^2+a+1=0;\forall a\in K$, but I do not know how to prove it.

 

[fresh_42]

I do not think there is a field where $a^2+a+1=0;\forall a\in K$, but I do not know how to prove it.

There is a field with four elements and it has these roots as elements. Sure, it is of characteristic two, but how to prove it without going through the operation tables. That would be quite a detour and definitely longer than the two routes we already know.
https://de.wikiversity.org/wiki/Endlicher_Körper/4/Operationstafeln

 

[mathwonk]

The characteristic of a finite field K is a prime number dividing the cardinality of the field, since the field is a vector space over the prime field. So a field of cardinality 4 has characteristic 2. The characteristic of K is of course prime since it is the generator of the kernel of the map from Z to K taking the integer 1 to the multiplicative unit 1 of K. That generator is a prime integer since the kernel of map to a domain is a prime ideal.

 

[mathwonk]

actually the "operations table" approach is also quite easy. the field consists of {0,1,a,b} where a,b are roots of X^2+X+1 =0, hence a+b = -1. But since neither of a,b, is 0, then neither is -1, hence -1 = 1, and 2=0.

 

[Hill]

actually the "operations table" approach is also quite easy. the field consists of {0,1,a,b} where a,b are roots of X^2+X+1 =0, hence a+b = -1, and ab= 1. But since neither of a,b, or -1, is 0, then a+b is not any of a,b, or 0, hence it is 1. so a+b = 1 = -1, so 2=0 in our field.

Should not we also check the possibility of the X^3-1=0 having only two different solutions, a field {0,1,a}?

 

[mathwonk]

good question, but I don't believe my argument assumes that a,b are distinct.

 

[mathwonk]

ok here is an argument with no theory at all. since a^4 = a holds for all a in the field, it holds when a=-1, hence 1 = (-1)^4 = -1, so 2=0. QED.

 

[Gavran]

2. The hint in the textbook says to show that $2a=0$. How to do that?

The hint in the textbook says that you have to show that the characteristic is 2 by showing that 2a=0 and you have done it in

My calculations:
$2a = (2a)^4 = 16a^4 = 16a$
$14a=0$
This allows for the characteristic to be 2 or 7. But, e.g., $3^4 = 81 = 4 ~(mod~ 7)$ rather than $3$. So, the only answer is $2$.

 

[pines-demon]

ok here is an argument with no theory at all. since a^4 = a holds for all a in the field, it holds when a=-1, hence 1 = (-1)^4 = -1, so 2=0. QED.

Or $-a=(-a)^4=a^4=a\Rightarrow2a=0$.

 

[nuuskur]

The polynomial $x^4 - x$ has at most four roots as elements of $K$. Conclude that $K$ is finite and has at most four elements. The only candidates are thus $\mathbb F_2, \mathbb F_3, \mathbb F_4$. Which of these satisfies the given identity?