If $a \cdot a = 1$ then $a = 1$ or $a = -1$

Question

Let $F$ be a field.

And let $a \in F$ such that $a \cdot a = 1$.

Claim: $a = 1$ or $ a = -1$

Before, I begin with the proof I cannot understand what $ a = -1$ represents...

Is $a = -1$ representing the inverse of $a$?

I am capable of getting to $a = 1$ with no trouble, but how do you get the $-$ sign to show up in your proof?

Proof: This is my attempt at getting to $a=-1$.

If $a = 0$ then we have $0 \cdot 0 = 0 \neq 1$. Then this implies that $a \neq 0$ which implies that $ \exists b \in F,$ s.t. $a \cdot b = 1$, i.e. $b$ is the multiplicative inverse of $a$.

Knowing this fact prove as follows...:

$a \cdot a = 1 \implies$ $ (a \cdot a) \cdot b = 1 \cdot b \implies$ $a \cdot (a \cdot b) = 1 \cdot b \implies$ $a \cdot 1 = 1 \cdot b \implies$ $ 1 \cdot a =1 \cdot b \implies$ $a = b$.

But since $b$ is the multiplicative inverse of $a$ is this enough to say that $a = -1$?

Answer 1

Before, I begin with the proof I cannot understand what $a=−1$ represents...

You do not have to overthink thinks. The inverse of $a$ is $(-a)$. And $a=-1$ just means, that $a=-1$. Calculations work in every field the same. It does not matter if you have $\mathbb{R}$, $\mathbb{Q}$ or a finite field like $\mathbb{F}_2$, which is the field with two elements $0$ and $1$.

Finite fields can be confusing at first, there it would actually be $1=-1$ which looks odd at first.

But to make $(\mathbb{F}_2, +,\cdot)$ a field we have to calculate like this:

$0$ is the neutral element of $+$ and $1$ the neutral element of $\cdot$.

Hence $0+0=0$ and $1+0=1$ and $0\cdot 1=0$ and $1\cdot 1=1$. What is $1+1$? It has to be $1+1=0$! Because if we had $1+1=1$ we had $1=0$ after subtracting $1$ on both sides. But this contradicts that 0 and 1 are different, which they are by the axioms of a field. (Neutral elements of addition and multiplication are unique).

This means 1 is its own inverse and therfor $1=-1$ which is just notation at that point.

Just as a little "introduction".

To solve your question you can simply go like this:

$a^2-1=0\Leftrightarrow (a+1)(a-1)=0$

This is true in every field!

You might have shown, that $a\cdot b=0\Leftrightarrow a=0\vee b=0$ holds in a field.

Hence $a+1=0\vee a-1=0\Leftrightarrow a\in\{-1,1\}$.

Answer 2

A field is basically an additive group together with a multiplicative group that plays nicely with the additive one. The additive group is usually denoted with operator $+$, and the inverse function of that group is usually denoted with the operator $-$, so that $-a$ is defined to be the additive inverse of $a$ in the group. It may be the case that $-a = a$ (e.g. in any field of characteristic 2, this is true for all $a$). Indeed, $-$ may not have anything at all to do with the subtraction on $\mathbb{Z}$ that you're used to.

You need to show that $a^2 = 1$ implies $a = 1$ or $a = -1$; equivalently, if you're squeamish about minus signs, that $a = 1$ or $a + 1 = 0$. Those two formulations are exactly the same.

(And, to drive this home again, it's possible for $a = 1$ and $a = -1$ to be simultaneously true! In the field $\mathbb{F}_2$ of two elements, $1 = -1$; and it's true in any field of characteristic $2$.)