# Definition of a Field of Characteristic $n$?

Let $V$ be a vector space over a field of characteristic not equal to $2$.
Prove that $\{u, v\}$ is linearly independent with $u, v$ being distinct if and only if $\{u+v, v-v\}$ is linearly independent.

Perhaps the text was meant for someone at a higher level. I never learned what a field of characteristic $n$ means. After googling for a while, I am still stuck. From various sources, it seems that the characteristic of $n$ is referring to the modulus $n$. If my understanding is correct, when the question say that $\{u, v\}$ is linearly independent, $0 = au + bv$, does $a,b$ equal $0$ or $0$ mod($n$)?

I tested this question many times. If $a, b = 0$, the condition of having a field of characteristic not equal to $2$ is trivial. On the other hand, if $a, b \equiv 0$ mod($n$), then the statement in the question is actually false. May someone kindly point me in the right direction? Thanks!

• One way to phrase this to avoid the term characteristic is to assume the field satisfies $x+x = 0 \Rightarrow x = 0$. Can you do the exercise then? – Tobias Kildetoft Aug 10 '13 at 20:13
• Do you know what a field is? – Thomas Andrews Aug 10 '13 at 20:14
• Some fields have the property that the cyclic additive group generated by $1$ is finite. If that happens, the least 'additive power' of $1$ that equals zero is called the characteristic of the field, and it's always prime. For practical purposes, all you really need for this exercise is that $2=1+1\neq 0$ in your field, meaning that $2^{-1}$ exists. – Jonathan Y. Aug 10 '13 at 20:15

A field of characteristic $n$ is a field such that

$$\underbrace{1+1+1+\cdots+1}_{n\text{ times}}=0$$

[where $1$ and $0$ play their "usual" roles as the mulitplication identity and the addition identity, respectively.]

The reason that fields of characteristic 2 must be eliminated is for (things like) the following example. Let $F_2$ be the field with only the two identity elements and $V=F_2\times F_2$. Then $\{(0,1),(1,0)\}$ is linearly independent.

However, observe that $0+1=1$ and $0-1=(1+1)-1=1$. Therefore, the sum of the vectors is $(1,1)$, and the difference of the vectors is also $(1,1)$; these are not linearly independent.

The theorem says this sort of trick is the only way that the property could fail.

• Nice explanation. Though the phrase "only the two identity elements" is a rather strange way to describe that field. – Tobias Kildetoft Aug 10 '13 at 20:26
• @ Mr. Stucky: However, I can do something similar with fields of other characteristics. For example, in a field of characteristic $3$, let $u = (2,1), v = (1,2)$; $u$ and $v$ are linearly independent, but $u+v = (0,0), u-v = (1,-1)$ are linearly dependent. So such failure can happen to many more characteristic $n$ .... Let me try Mr. Kildetoft's hint. – Andy Tam Aug 10 '13 at 20:46
• @AndyTam Those two vectors are not linearly independent. You even found a non-trivial linear combination that gives $0$ yourself. – Tobias Kildetoft Aug 10 '13 at 20:53
• Okay, my counter-example was wrong. Let me try again. TY – Andy Tam Aug 10 '13 at 21:01

Okay, this is my solution:

First, prove by contradiction that in a field of characteristic not equal to 2, $x + x = 0 \rightarrow x=0$. Suppose that $x \neq 0$. Then $x + x \neq 0$, contradicting the premise that $x + x = 0$.
($\rightarrow$) Since $\{u, v\}$ is linearly independent, $au + bv = 0$ if and only if $a = b = 0$. $c(u+v) + d(u-v) = (c+d)u + (c-d)v = 0$. From the system $c + d = 0$ and $c - d = 0$, we have $c + c = 0 \rightarrow c = 0$, and $d = 0$. If I use User142526's suggestion that $2 = 1 + 1 \neq 0$, then I can solve the system via its determinant.
($\leftarrow$) Suppose that $\{u+v, u-v\}$ is linearly independent, $c(u+v) + d(u-v) = 0$ if and only if $c = d = 0$. $c(u+v) + d(u-v) = (c+d)u + (c-d)v = au + bv = 0$. $a = 0, b = 0$.
For completeness, I will use Eric Stucky's example to show that the statement is not true in a field of characteristic $2$.