Symmetric bilinear forms

1. Aug 4, 2013

DUET

When I start to read the the article called "symmetric bi-linear forms", I face the following sentence. But I don't understand what does the following sentence suggest. Could someone please help me here?

We will now assume that the characteristic of our field is not 2 (so 1 + 1 is not = to 0)

2. Aug 4, 2013

micromass

A field is a very general object. It is a set $F$ together with two functions $+:F\times F\rightarrow F$ and $\cdot:F\times F\rightarrow F$ satisfying:

• $a+(b+c) = (a+b)+c$
• $a+b=b+a$
• There exists a $0\in F$ such that $a+0 = 0+a = a$
• For each $a\in F$, there is an element $b\in F$ such that $a+b=b+a=0$
• $a\cdot(b\cdot c) = (a\cdot b)\cdot c$
• $a\cdot b = b\cdot a$
• There exists an $1\in F$ such that $1\neq 0$ and $a\cdot 1 = 1\cdot a = a$
• If $a\neq 0$, then there exists a $b\in F$ such that $a\cdot b =b\cdot a = 1$

The usual suspects, such as $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are fields. All of these fields satisfy that $1+1+1+1+1+....+1$ (n times) is nonzero. But this is not a general property of a field. Fields that have the property are said to be of characteristic 0.

For example, consider $F=\{0,1\}$ and define

$$1+0 = 0+1 = 1~\text{and}~1+1=0+0=0$$

and

$$1\cdot 0 = 0\cdot 0 = 0\cdot 1 = 0~\text{and}~1\cdot 1 = 1$$

This satisfies all the field axioms, but it has $1+1=0$. We say that this field has characteristic 2.

In general, if a field satisfies $1+1+1+...+1=0$ (n times). Then the field is said to have characteristic $n$. We can always show that $n$ is a prime number.

The main advantage of assuming that the field is not of characteristic $0$, is that we can divide by $1+1$. If we use the nice notation $2=1+1$, then $1/2$ exists.

3. Aug 4, 2013

DUET

Could you please explain it a bit more?

Specially the following.

What do you mean by "+" and "."?

4. Aug 4, 2013

micromass

The cartesian product $F\times F$ just means the set of all ordered pairs. Thus

$$F\times F = \{(a,b)~\vert~a,b\in F\}$$

For example $(0,0)\in \mathbb{R}\times\mathbb{R}$ and $(-1,2)\in \mathbb{R}\times \mathbb{R}$.

That $f$ is a function from $F\times F$ to $F$ just means that we associate with each element in $F\times F$, an element in $F$.

So given $(a,b)$ with $a,b\in F$, we associate an element $f(a,b)\in F$. Specifically, if $f$ is $+$, then to each ordered pair $(a,b)$ with $a,b\in F$, we associate $+(a,b)$ in $F$. We write this as $a+b$.

So to each two elements in $F$, we just associate a new element in $F$.