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 $1\neq0$.

  • 1
    $\begingroup$ What is the group identity of $\{r \in R \mid r \neq 0\}$? $\endgroup$
    – Hayden
    Jun 6, 2020 at 6:45
  • $\begingroup$ Thanks! But how do we know $1\cdot0=0$, or that $1\neq0$? Maybe I'm missing something obvious... $\endgroup$
    – buffle
    Jun 6, 2020 at 6:47
  • 2
    $\begingroup$ @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 $1= 0$? $\endgroup$
    – Arthur
    Jun 6, 2020 at 6:49
  • $\begingroup$ Ahh it all makes sense now - thank you for your help. This is an embarrassing question now that think about it! $\endgroup$
    – buffle
    Jun 6, 2020 at 6:53

1 Answer 1


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.


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