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For any $a \in F$, $a0 = 0$.

I know this proof works: $a0 = a(0+0) = a0 + a0$; taking the additive inverse of $a0$, we get $a0=0$.

But how about this? $a0 = a(1+(-1)) = a + (-a) = 0$

The reason I'm skeptical is because not all fields have -1 as an element. I tried to work around this issue by saying that "-1" exists in all fields in the sense that all fields have 0,1 and "-1" stands for the additive inverse of "1." For example, {0,1} is a field with 1 + 1 = 0: here, 1 is "-1" (it's the additive inverse). To make sense of this idea, I redid the proof without thinking of numbers; instead I used variables to represent an element of a field and its additive inverse. For example, let j be an element of a field and w its additive inverse:

$a0 = a(j+w) = aj + aw$

Where do I go from here? How do I show that $aj$ and $aw$ are additive inverses of each other?

beginner
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  • Why don't you want one of the usual proofs, e.g., this one? And don't mix up other exercises under this title. – Dietrich Burde Jan 27 '21 at 17:56
  • You can think in "2" and "3" as [2] and [3], because $[2]\cdot [x_2]=[2x_2]$ by definition of the equivalence relation. You can also see this as $2[x_2]=[x_2]+[x_2]=[2x_2]$ – PrV Jan 27 '21 at 18:01
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    "Not all fields have $-1$ as an element. I tried to work around this issue by saying that ... '-1' stands for the additive inverse of '1'" Of course. When we talk about arbitrary fields and we write "0" and "1" we do not mean the usual $0$ and $1$ from the real numbers... the exact form that these take can be wildly different than what we are used to. There should be no ambiguity here though since it is understood from context that $0$ is in reference to the additive identity and $1$ the multiplicative identity of the field, $-1$ also is unambiguous as the additive inverse of $1$. – JMoravitz Jan 27 '21 at 18:01
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    A field is guaranteed to have an additive identity, a multiplicative identity, and additive inverses for each element. We customarily give the name $0$ to the additive identity, $1$ to the multiplicative identity, and given an element $a$ we give the name $-a$ to its additive inverse. There is nothing dangerous about referring to $-1$. It doesn't bother us that $-1$ might equal $1$, as in the case of $\Bbb F_2$, there just happen to be multiple ways in which we can write the value... just like how we could write $9$ as $3^2$ or as $3\times 3$ or as $1+1+1+1+1+1+1+1+1$ – JMoravitz Jan 27 '21 at 18:04
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    All fields have a $-1$ by definition. – Randall Jan 27 '21 at 18:05
  • @JMoravitz Thank you! – beginner Jan 27 '21 at 18:12

2 Answers2

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First, all fields DO have $-1$ as an element: every field has $1$ as an element; every field has an additive inverse for every element in the field; and therefore every field has $-1$ as an element.

But as you seem to have intuited, your new proof does require knowledge that $(-1) \cdot a = -a$. And you ask how to prove this, so here is a chain of equivalences each of which only uses one of the field axioms: \begin{align*} (-1) \cdot a = -a &\iff (-1) \cdot a + a = 0 \\ &\iff (-1) \cdot a + 1 \cdot a = 0 \\ &\iff ((-1) + 1) \cdot a = 0 \\ &\iff 0 \cdot a = 0 \end{align*} As you can see, you can deduce either of these two statements from the other, and so you'll need to prove one of them first before you can deduce the other.

Your question has a pretty simple proof that $0 \cdot a = 0$, so that seems like the one to prove first.

Lee Mosher
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In your notation, let $j$ be the multiplicative identity and $w$ be the additive inverse of the mulitplicative identity. Then it works out as $$a\cdot 0 = a (j+w)= aj + aw = aj-aj=0.$$ Here we have used the fact that $aw=-aj$ which needs the fact that $$0=a(w+j)=a\cdot 0.$$ Do you see the cirularity? Better to stick to the standard proof that you stated.

Favst
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  • Thank you. I understand the first proof I stated. So this proof $a0 = a(1+(-1)) = a1-a1 = 0 is not valid, right? I saw it on math stack exchange and was trying to wrap my head around it. – beginner Jan 27 '21 at 18:02
  • @beginner As far as I can tell, it leads to circularity because you end up needing to use what you are proving. – Favst Jan 27 '21 at 18:02