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Here is a soft question that I am dealing with. Please tell me if it's correct or not.

Suppose $A$ is a commutative ring with unity. Is $A$ a prime ideal of $A$?

I think the answer is true, because we know $I$ is an integral domain iff $R/I$ is an integral domain. But $A/A = 0.$ So it's an integral domain. So$A$ is a prime ideal in $A$ Tell me if I argument is right or wrong. Any help will be appreciated. Thanks

user631697
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    Usually when we consider prime ideals, we define them to be proper ideals, i.e. ideals that are not the whole of $A$. – rubikscube09 Aug 19 '19 at 17:31
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    Near-duplicate: https://math.stackexchange.com/questions/1326505/is-the-zero-ring-a-domain – Eric Wofsey Aug 19 '19 at 17:58
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    Also relevant: https://math.stackexchange.com/questions/427078/is-0-a-field – Eric Wofsey Aug 19 '19 at 18:01
  • So mostly it deoends on the convention. But I don't understand the argument behind whole ring not being a prime ideal. In the questions they discussed about 0 being any domain, while not about 0 being an integral domain. So I should consider 0 in general not an integral domain and the proceed further? – user631697 Aug 19 '19 at 18:13
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    That's not a soft question. – Paul Aug 19 '19 at 20:01
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    @user631697 you could call it a convention, but if so, this is really one of those conventions that has nearly 100% of the usage. I have never seen the alternative used even once in a text or paper. – rschwieb Aug 20 '19 at 02:35

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Commonly you will hear it said that it is a convention that prime ideals should be proper, or (equivalently) that integral domains should be nonzero. Actually with the correct definitions it is a theorem. We just have to state the definition of an integral domain in an "unbiased" way, referring to all $n$-ary multiplication operations $D^n \to D$, including $n = 0$:

Definition: An integral domain $D$ is a commutative ring such that $D \setminus \{ 0 \}$ is closed under $n$-ary multiplication for all $n \ge 0$.

Setting $n = 0$ gives that the set of non-empty elements $D \setminus \{ 0 \}$ must be closed under empty multiplication, which is $1$. This gives $1 \neq 0$ in an integral domain, not as a convention but as a natural extension of the definition. Equivalently:

Definition: A prime ideal $P$ of a commutative ring $R$ is an ideal such that $R \setminus P$ is closed under $n$-ary multiplication for all $n \ge 0$.

And again taking $n = 0$ gives that $1 \not \in P$.

Qiaochu Yuan
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It's usually taken as part of the definition that an integral domain must not be the zero ring.

For example, on Wikipedia (first sentence of the article, note the word “nonzero”), and in the standard references (Atiyah & Macdonald, Matsumura, Lang, etc.).

Hans Lundmark
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Proposition 13.5.1 $\textit{Let R be a ring. The following conditions on an ideal P of R are equivalent.}$ $\textit{ An ideal that satisfies these conditions is called a prime ideal.}$

$\textit{(a) The quotient ring R /P is an integral domain.}$

$\textit{(b) P$\neq$R, and if a and b are elements of R such that ab $\in$ P, then a $\in$ P or b $\in$ P.}$

$\textit{(c) P$\neq$ R, and if A and B are ideals of R such that AB C P, then A $\subseteq$ P or B $\subseteq$ P}$

(Algebra,by Michael Artin)

Here, (b) and (c) says that a unit ideal is never a prime ideal.

Vaishnavi C
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