Can someone please help me in figuring out how all fields are integral domains but not all ID are fields? My course assumes IDs to be commutative with unity but fields require all elements to have a unit. I can't seem to grasp their concepts.
$\endgroup$ 04 Answers
$\begingroup$An integral domain is a field if an only if each nonzero element $a$ is invertible, that is there is some element $b$ such that $ab=1$, where $1$ denotes the multiplicative unity (to use your terminology), often also called neutral element with respect to multiplication or identity element with respect to multiplication.
An invertible element is also sometimes called a unit. (The terminolgy "have a unit" seems strange.)
So an integral domain is a field if each nonzero element is a unit, but this is something other than the unity element.
It seems your confusion is mainly due to the similarity of terms.
(To be precise, for a field you require that it has at least two element, while for an integral domain you might not in which the asertion at the start is not completly precise; but this is a detail.)
$\endgroup$ 8 $\begingroup$$\mathbb{Z}$ is an integral domain but not a field.
$\endgroup$ $\begingroup$Using the starting point of a Ring $\mathcal{R}$.
Both a Field and an Integral Domain (ID) require:
(i) $\mathcal{R}$ is commutative
(ii) $1\not=0$ (where $1$ is the multiplicative identity and $0$ is the additive identity)
So then, what is the difference between the two? Quite simply, in addition to the above conditions, an Integral Domain requires that the only zero-divisor in $R$ is $0.$
And a Field requires that every non-zero element has an inverse (or unit as you say). However the effect of this is that the only zero divisor in a Field is $0.$
And so it turns out that every Field is an Integral Domain but not every Integral Domain is a Field.
$\endgroup$ $\begingroup$Let's look at the definition of integral domains:
Which means for all $a, b$ of that ID: $(a \neq 0 \land b \neq 0) \rightarrow (ab \neq 0)$.
Which is different from the condition that separates commutative rings from fields, which is the existance of an inverse element under multiplication:
For a field $F$ and $a,b \in F$ $\exists a^{-1} (aa^{-1} = 1))$ with $1$ defined by this axiom $\forall a (1a = a)$.
It's probably easiest to show the difference with an example: $\mathbb {Z}$.
$\mathbb{Z}$ is an integral domain, because for any $a \in \mathbb{Z}$ there is no $b$ so that $((a \neq 0) \land (b \neq 0)) \land ab = 0$.
But is is no field, because for instance $5$ has no inverse $5^{-1} \in \mathbb{Z}$ so that $5 \times 5^{-1} = 1$.
