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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.

Jessy White
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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.)

quid
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    I think one usually requires that an integral domain is nonzero so that for a commutative ring $A$, the ideal $A$ is not prime. –  Nov 25 '14 at 17:38
  • Yes, the stress in my parenthetical came out differently than I meant it. It is certain that a field has at least two elements, while an ID might not though it is also common. – quid Nov 25 '14 at 17:41
  • @quid: IDs are non-trivial, this requirement is universally accepted. – Martin Brandenburg Nov 25 '14 at 20:16
  • @MartinBrandenburg please see 'Elements of Abstract Algebra' by Allan Clark. – quid Nov 25 '14 at 20:45
  • So for Allan Clark integral domains don't have a field of fractions in general? The rank of a free module is not well-defined? The statement "$R/P$ is an integral domain $\Leftrightarrow$ $P$ is a prime ideal" only holds if $P \neq R$? Or is $R$ a prime ideal for him? Is $1$ a prime number? etc. – Martin Brandenburg Nov 25 '14 at 20:46
  • @MartinBrandenburg indeed, only the nontrivial ones have. To be precise I am not certain now how things are set up, and I do not get the page where it is defined. It might be that all have something that is called a field of fractions, and this "field of fractions" is a field if the domain is non-trivial. Anyway, I repeat my invitation to check it. It then says frequently nontrivial domain. (And to be sure I do not claim it is a good idea to do this. I only contest your "universally"; that it is common was said by me before you even commented.) – quid Nov 25 '14 at 20:54
  • @MartinBrandenburg what now? Do you maintain or retract your claim? – quid Nov 26 '14 at 12:46
  • Both ID and Fields have no non-zero divisors(ie, $ab=0 \implies$ a=0 or b=0). But in a ID it is not necessary that all elements have to be a unit(ie, the element has its inverse) whereas it is necessary in a Field. (eg. $\mathbb{Z}$, there does not exist a and b such that $ab=0$, and also there does not exist $c$ such that $2*c=1$ in $\mathbb{Z}$) – Jayanth Kumar Jan 31 '17 at 07:10
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$\mathbb{Z}$ is an integral domain but not a field.

Nicolas Bourbaki
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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.

Premez
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Let's look at the definition of integral domains:

An integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero.

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 $5 \times 5^{-1} = 1$.

Nearoo
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