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Attempt: Suppose $P$ a prime ideal, then $R/P$ is an integral domain, and so $R/P$ is a field, which implies that $R$ is maximal.

I am not sure if this is a sufficient proof. Furthermore, I do not understand how to use the condition "finite".

Decbelief
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  • Fields are integral domains, but the converse fails generally. There is one case where the converse holds.... – Randall Nov 12 '18 at 22:07
  • The statement is true in the more general case when $R$ is a commutative artinian ring. It's easy to show that an artinian domain is a field. – egreg Nov 12 '18 at 23:35

2 Answers2

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$\newcommand{ideal}[1]{\mathfrak{#1}}$ In the context of this question it may not be immediate that $R/\ideal{p}$ being a domain implies that $R/\ideal{p}$ is a field, since it is not true in general. For example, $(X) \subset \mathbb{Z}[X]$ verifies $\mathbb{Z}[X]/(X) \simeq \mathbb{Z}$ and the latter is a domain, but not a field. Hence $(X)$ is prime but not maximal.

You are however in the right track in the following sense: as your argument intends to show, prime ideals being maximal is equivalent to $R/\ideal{p}$ being a domain if and only if it is a field. One implication always holds, so we are left to prove that these domains are fields. Certainly the ring being finite should be helpful. I would encourage you to stop and give the problem another try. A (more general) solution follows:

Proposition. Let $R$ be a finite commutative integral domain. Then $R$ is a field.

Before proving this, note that it solves our problem: since $R/\ideal{p}$ is a finite and commutative domain, it immediately implies it is a field as per the proposition.

Proof. The ring $R$ is already commutative, so we are only left to see that it has multiplicative inverses. Let $a \in R$ be a nonzero element. Consider $$ b := \prod_{x \not \in \{0,a\}}x $$

the product of all elements except $0$ and $a$. Note that $R$ being a domain guarantees that $b \neq 0$ because all factors are nonzero. If $ab = a$, by cancellation (since $R$ is a domain) we get that $a = 1$ and so $a^{-1} = a$. Otherwise, $ab$ is nonzero (again, $R$ is a domain) and not equal to $a$. Hence,

$$ ab = a\prod_{x \not \in \{0,a\}}x = a \cdot (ab) \cdot \prod_{x \not \in \{0,a,ab\}}x = (ab) \cdot a \cdot \prod_{x \not \in \{0,a,ab\}}x $$

Once again by cancellation,

$$ 1 = a \cdot \prod_{x \not \in \{0,a,ab\}}x $$

which says that $a$ once again has a multiplicative inverse, thus completing the proof. $\square$

qualcuno
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Hint:

Show a finite integral domain $A$ is a field, considering, for any $x$ the multiplication by $x$ morphism: \begin{align} m_x:A&\longrightarrow A, \\ y&\longmapsto xy.  \end{align} What can you say of this morphism if $x\ne 0$?

Bernard
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