What does the notation of “1=/= 0” mean for “Let R be a ring with unity 1=/= 0”?

Question

Should I read it literally? Or does it mean that the multiplicative identity on the left hand side is not equal to the additive identity on the right hand side?

Answer 1

In general when discussing rings we like to use $1$ as notation for the multiplicative identity and $0$ as notation for the additive identity because this is what we are used to for $\mathbb{R}$ or $\mathbb{Z}$, even though in other rings they won't literally be $1$ and $0$. In this case, the notation $1\neq 0$ is a quick way to say the the multiplicative and additive identities are not the same.

Answer 2

It means $1\neq 0$, when people can't format well.

As usual, $0$ is what we call the additive identity of the ring and $1$ is what we call the multiplicative identity. There is a ring where $1=0$, it's just that it's the trivial ring $\{0\}$. This excludes that.

Answer 3

It mean that $R$ is a ring and that the neutral element of $+$ is different that the neutral element of the multiplication.

Answer 4

There does exist a ring where $1=0$. In fact, there's a single such ring, for if $1=0$ then for any $x \in R$ we have $x\cdot 1 = x \cdot 0 \Rightarrow x=0$. So, zero is the only element of this ring.

As you can imagine, such inconvenience can break a lot of proofs and computations. So, it is often assumed that $1\neq 0$ in a particular ring, literally.

Answer 5

I think the answers are clear. Here is one more just to be explicit. We are concerned with commutative rings with a multiplicative identity.

Suppose I have a set $S=\{a,b,c\}$ with commutative operations satisfying $a=a+c=b+b=ab$, $b=a+a=b+c=bb$ , $c=a+b=c+c=ac=bc=cc.$ Then one way or another you can check the rules for a ring and verify that the structure is a ring with additive identity $c$ and multiplicative identity $b$. A shorter way to say that is $0=c$ and $1=b$.

Without the rule you mention $T=\{x\}$ would count as a ring with $x+x=xx=x.$ Since $a+x=a$ for everything in the singleton set $T,$ $x=0$, similarly $ax=a$ gives $x=1.$

It turns out that $T$ is the only thing which follows all the ring rules except the requirement $0 \neq 1$. It is desireable to not have $T$ count as a ring. That uniqueness. Claim isnt that hard to prove but first you need some smaller results such as $0a=0$ . I do not recall for sure, but it might be cleanest to use $0\neq 1$ in an early proof.