# A field is a commutative division ring

Herstein's Topics in Algebra defines a ring $$(R,+,\cdot)$$ as having the following properties:

• $$(R,+)$$ is an abelian group, and its identity is denoted by $$0$$.
• $$(R,\cdot)$$ is a semigroup, which means multiplication is associative and $$R$$ is closed under it.
• Distributivity: $$(a+b)c=ac+bc$$ and $$a(b+c)=ab+ac$$.

Also, the book adopts the convention that rings do not need a unit element $$1$$ for which $$1\cdot r=r\cdot1=r$$ for all $$r\in R$$.

Then it proceeds to define "commutativity" and "division ring." These ideas are fine for me. But here is the definition that confuses me:

A field is a commutative division ring.

My understanding is that a commutative division ring is a ring in which

• the multiplication is commutative
• the set $$\{r\in R:r\neq0\}$$ forms a group under multiplication.

But the field axioms from my beginning analysis course state that a field needs to have a multiplicative identity $$1$$, and $$1$$ cannot be the same as the additive identity $$0$$.

Is this a meaningful difference? If so, how should I reconcile this two definitions? From Herstein's definition, it doesn't look like a field even needs to have $$1$$, much less have $$1\neq0$$.

• What is the group identity of $\{r \in R \mid r \neq 0\}$? Jun 6, 2020 at 6:45
• Thanks! But how do we know $1\cdot0=0$, or that $1\neq0$? Maybe I'm missing something obvious... Jun 6, 2020 at 6:47
• @buffle $1\cdot 0 = 0$ is true because $a\cdot 0 = 0$ is true for any $a$ in the ring, and $1$ is an element of the ring. $1$ is defined as the identity element of (the multiplicative group) $\{r\in R\mid r\neq 0\}$, and $0$ isn't contained in there, how could we possibly have $1= 0$? Jun 6, 2020 at 6:49
• Ahh it all makes sense now - thank you for your help. This is an embarrassing question now that think about it! Jun 6, 2020 at 6:53

Let $$1$$ denote the group identity (with respect to multiplication) of $$\{ r \in R \mid r\neq 0\}$$, which exists by hypothesis. Being an element of $$R \setminus \{0\}$$, we immediately have $$1 \neq 0$$. By definition, $$1$$ is a/the multiplicative identity.