# Prove that 1 = 0 at the trivial ring

I need to prove that zero is one at the trivial ring, but I don't have yet that one is a member of the trivial ring (the only constant at my zero ring is zero). So I thought to prove first that, if R is a ring with one (now I have the constant one inside the ring), then if the trivial ring has an identity, it must be the one (since the trivial ring is an R subring). After that I would prove that zero is an identify. I'm in trouble at the first step, that is, to show that one is trivial ring member. Could someone help me?

Edit  In my definition a ring doesn't necessarily have a multiplicative identity "one".

• Comments are not for extended discussion; this conversation has been moved to chat. – Aloizio Macedo Jun 16 at 17:52

You might be asking the wrong question. It seems like you're thinking of $$1$$ as the number or symbol. Don't try to show that $$1\in R$$. Instead, show that the element of $$R$$ has the required property:

We say that $$a\in R$$ is an identity if for all $$b\in R$$, $$ab=ba=b$$. Now, let's explore this quantified statement for $$R=\{0\}$$. Since $$R$$ has only one element, $$a=0$$. Now, let $$b\in R$$, since $$b$$ is in $$R$$, $$b=0$$. Since $$ab=0=b$$ and $$ba=0=b$$, it follows that $$a$$ has the properties of the identity, i.e., is the identity.

• And every element is invertible, nilpotent, idempotent etc because these properties are defined by (quantified) equalities of polynomial terms, but there is only one possible value the polynomials can take, so they are always equal. – Bill Dubuque Jun 14 at 17:59

After that I would prove that zero is an identify. I'm in trouble at the first step, that is, to show that one is trivial ring member.

Perhaps among all cases in which brute force is possible, this is one which it is truly acceptable to use brute force.

Obviously $$0$$ is the only candidate for the identity because it is the only element available.

Now just check: for all $$x\in R$$, that $$x\cdot 0=x$$ and $$0\cdot x=x$$. It will be the identity if and only if this holds true.

Here we go:

$$0\cdot 0=0$$

And we're done.

if R is a ring with one (now I have the constant one inside the ring), then if the trivial ring has an identity, it must be the one (since the trivial ring is an R subring).

If you insist that subrings of rings with identity$$\neq 0$$ must share their identity, then $$\{0\}$$ is never a subring of such a ring, according to that definition. The line of reasoning you were following would wind up: $$1\in \{0\}$$, therefore $$1=0$$, contradicting $$1\neq 0$$.

So if you wanted to prove it was still a ring with identity that happened to be contained in the ring, you could not rely on this relationship with the containing ring.

Okay. Your question is how to prove a trivial ring where $$0 + 0 = 0$$ and $$0\cdot 0 = 0$$ is a ring with $$1$$.

Well, is there an element $$e$$ so that $$a\cdot e = e\cdot a = a$$ for all elements $$a$$? Well, as there is only one element $$0$$ this is simply asking is there an element $$e$$ so that $$0\cdot e = e\cdot 0 = 0$$?

And the answer to that is... of course, $$0$$ is that element.

So if a trivial ring exists at all [1], then it is a ring with $$1$$ and $$1 = 0$$.

That's it. That's all there is.

[1] You never actually showed that $$\{0,+,\cdot\}$$ where $$0+0=0$$ and $$0\cdot 0 = 0$$ is a ring in the first place.

But it's trivial to do.: $$a+(b+c), (a+b)+c, a+b, b+a, a(bc), (ab)c, a(b+c), ab+ac, (a+b)c, ac + bc$$ can all only and must have one value of $$0$$ so $$+,\cdot$$ are associative, $$+$$ is a commutative, and multiplication distributes. And as $$a+0 = 0 + a = a=0$$ can only have one value, $$0$$, is an additive identity and the additive inverse of the only element. So $$\{0, +\}$$ is an abelian group.

I think that is better to clarify. My proof is a formal proof assisted by a computational verification system, so it's not possible to make new assertions about anything that has a previous definition. For example, I previously have that "ring with one" is a type and also a boolean function that check if I have the 5-tupla [T,+,º,zero,one] for a non-empty set T. The lemma I'm trying to prove is this one: I have that exists only one y that, for all x, we have xºyºx=x, where x,y are elements of the ring R [T,+,º,zero] (to have "one" is not a requirement to be a ring in this definition, that's already given). I need to prove that (a) R has no zero divisors (b) R is a ring with one (c) R, if not the trivial ring, is a division ring. The items (a) and (c) are ok but (b)... I divided it into two cases: (i) R has cardinality higher then 1 and that's ok (ii) R has cardinality 1, and that's the problem: it implies that R iś "[{zero},+,º,zero]". The first step, that is to prove that zero is the identity is ok to me. The problem is to prove that zero = one, and then the 5-tupla [T,+,º,zero,one] is true for the boolean function "ring with one" applied to any ring with the property "xºyºx=x", including the trivial ring. Just as some of you suggested, I'm about to conclude that the way things are defined, it's just not possible to prove this.

• If that is your question then ASK that as your question. Instead your question is "In the trivial ring, is $0$ the multiplicative identity?" And the answer to that is a trivial "Yes" and the proof is a trivial "$0*0=0$ so $0$ is the multiplicative identity". If you don't like that answer, than ask a different question. – fleablood Jun 16 at 15:29
• My question was not "In the trivial ring, is 0 the multiplicative identity?". The question title is as short as think it should be, but at the question by itself I made clear that "After that I would prove that zero is an identify. I'm in trouble at the first step, that is, to show that one is trivial ring member". About this, I think rschwieb already clarified the question. – André Camapum Jun 27 at 14:21
• $1$ has no meaning except as the multiplicative identity of a ring. If you assume $1$ is the $1$ that is in the real numbers you are already doomed because your ring isn't a subset of the reals or have anything to do with the reals. To ask if the real number $1$ is an element of the ring makes as much sense as asking of Babar, the elephant is a member of the ring. So first thing in asking if "$1=0$" figuring out what the heck "$1$" means. $1$ has to mean the multiplicative identity. So the question "Does $1=0$" has to mean "Is $0$ the multiplicative identity"? Nothing else makes sense. – fleablood Jun 27 at 15:01