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The following is on page 4 of Atiyah and MacDonald's commutative algebra book. Example 1: Let $A=k[x_1,..,x_n]$, k a field. Let $f \in A$ be an irreducible polynomial. By unique factorization, the ideal $(f)$ is prime. Then they go on to write that in $Z$ every ideal is of the form $(m)$ where $m>=0$ and that $(m)$ is prime if and only if m is $0$ or prime, that $(p)$ is maximal if p is prime, and that $Z(p)$ is the field of p elements.

Right after the above, the last sentence on page 4 of Atiyah and MacDonald's commutative algebra says "The same holds for example 1 for n=1 but not for n>1". Did they mean that for $k[x_1]$ every prime ideal is maximal? If yes, is this because of the following? The irreducible polynomials in $k[x_1]$ are the primes and $x_1$, and the ideals generated by them are prime, and these are all the ideals that are generated. They are maximal too because nothing divides $x_1$ except itself, and nothing divides any prime except itself.

In general, would it be right to say that in a principal ideal domain all prime ideals are maximal? I reckon $k[x_1,x_2]$ is not a principal ideal domain since $(x_1,x_2)$ has two generators. Is this the reason why in $k[x_1,x_2]$ not all prime ideals are maximal? For example, $(x1) \subset (x_1,x_2)$.

user2371765
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Two principal ideals satisfy $(a) \subseteq (b)$ iff $b|a$, hence we find that principal ideals generated by irreducible elements are indeed maximal.

In a PID irreducible and prime elements agree. Thus nonzero prime ideals are precisely those ideals generated by irreducible elements and therefore maximal.

$K[X]$ is a PID with irreducible elements given by irreducible polynomials.

Jonas Linssen
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  • Thank you. Is the following right? I reckon k[x1,x2] is not a principal ideal domain since (x1,x2) has two generators. Is this the reason why in k[x1,x2] not all prime ideals are maximal? For example, (x1)⊂(x1,x2). – user2371765 Apr 12 '20 at 16:16
  • $K[X_1,X_2]$ is indeed not a principal ideal domain, since $(X_1,X_2)$ is not a principal ideal, ie. is not generated by one element. Tbh I don’t know whether having each nonzero prime ideal being maximal already implies that the ring is a PID. I think in this instance the argument is more like „$(X_1,X_2)$ is a nonmaximal prime ideal“ as an explicit counterexample rather than invoking results about PIDs – Jonas Linssen Apr 12 '20 at 16:23
  • Thank you. Do you know whether it is true that $(x_1,x_2) = (x_1) \cup (x_2)$ in $K[x_1,x_2]$? I am doubtful about $2x_1+3x_2$. It lies in LHS, but probably not in RHS. – user2371765 Apr 12 '20 at 18:13
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    You are right. This is one instance of why unions of ideals need not be ideals... In other words: this identity does not hold – Jonas Linssen Apr 12 '20 at 19:25