Every prime ideal is maximal [duplicate]

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Problem: Show that if R is a finite ring, then every prime ideal of R is maximal.

My attempt: Let I be a prime ideal of R. Then, by definition of a prime ideal, ab ∈ I implies a ∈ I or b ∈ I for every a, b ∈ R. Since R is a finite ring, there exists an ideal J such that I ⊆ J ⊆ R.

Question: Please help me on this. The definition of a maximal ideal is as follows: "The ideal I is said to be a maximal ideal of R if for all ideals J of R such that I ⊆ J ⊆ R, either J=I or J=R."

I have to apply the fact that R is a finite ring. But I don't know how I can satisfy I ⊆ J ⊆ R...

marked as duplicate by Dietrich Burde, Jyrki LahtonenAug 21 '15 at 19:07

• What are the characterisations of prime and maximal ideals in terms of $R/I$? – Daniel Fischer Aug 21 '15 at 19:02
I can sketch a proof, then you can make all the details: Suppose $I$ is a prime ideal $\Rightarrow$ $\frac {R}{I}$ is a domain. Notice that every finite domain is a field then $\frac{R}{I}$ is a field so ${I}$ is maximal.