maximalideal

User
Please help me to read a textbook of algebra.  Let k be a field, f be an element of k[x, y], R be the quotient ring k[x, y]/(f), and θ be the surjection from k[x, y] onto R.  The textbook says that the pullback of a maximal ideal of R with θ is maximal in a proof.  Is it true?  Did I failed to read an important assumption?
ChatGPT
Yes, it is true that the pullback of a maximal ideal of R with θ is maximal, assuming that k is an algebraically closed field. This is a key assumption that is often used in algebraic geometry and commutative algebra, so make sure you didn't miss it in your textbook. Here's a brief explanation of the proof:

Let M be a maximal ideal in R, and let N = θ^(-1)(M) be the pullback of M in k[x, y]. We want to show that N is a maximal ideal in k[x, y]. To do this, we'll use the following facts:

If A is a commutative ring with unity, and I and J are ideals of A such that I ⊆ J, then A/J is a quotient ring of A/I.

An ideal P of a commutative ring A is maximal if and only if the quotient ring A/P is a field.

Now let's consider the quotient rings k[x, y]/N and k[x, y]/(f). Since N contains (f) (because θ^(-1)(M) contains (f)), we have k[x, y]/N as a quotient ring of k[x, y]/(f) by Fact 1. Since k is algebraically closed and f is an element of k[x, y], the quotient ring k[x, y]/(f) is an integral domain. By Fact 2, we only need to show that k[x, y]/N is a field to prove that N is a maximal ideal.

Recall that θ is the surjective ring homomorphism from k[x, y] onto R. We have θ(N) = M, and since M is a maximal ideal, R/M is a field by Fact 2. We know that R = k[x, y]/(f) and R/M is a field, so k[x, y]/N must also be a field. Thus, by Fact 2, N is a maximal ideal in k[x, y], completing the proof.

Remember that this proof assumes k is an algebraically closed field. If this assumption was not mentioned explicitly in your textbook, it might be a default assumption when discussing algebraic geometry and commutative algebra.





User
Did you assume that f is prime?
ChatGPT
Apologies for the confusion in my previous response. You're right that I implicitly assumed f to be prime (irreducible) in my explanation, which is an important condition. If f is not prime, the quotient ring k[x, y]/(f) might not be an integral domain, and the proof provided above would not hold.

To clarify, if f is an irreducible (prime) element of k[x, y] and k is an algebraically closed field, then the pullback of a maximal ideal of R with θ is maximal. If f is not prime, the statement might not be true in general.

So, make sure to check if your textbook assumes f to be irreducible or if there's any information regarding the properties of f. If it's not clear, you might want to consult additional resources or ask an expert for clarification.





User
Thanks.  Where did you use the assumption of algebraically closedness?  Is there a case that k[x, y]/(f) is not an integral domain if k is not algebraically closed?
ChatGPT
My apologies for any confusion earlier. You are correct in pointing out that the assumption of algebraically closedness is not necessary for the proof. In fact, if f is a prime (irreducible) element of k[x, y], the quotient ring k[x, y]/(f) is an integral domain regardless of whether k is algebraically closed or not.

The reason is that in the polynomial ring k[x, y], prime elements and irreducible elements coincide. So, if f is prime, then (f) is a prime ideal. Since prime ideals are also prime in the quotient ring, k[x, y]/(f) is an integral domain.

The proof I provided earlier can be modified without the algebraically closed assumption as follows:

Let M be a maximal ideal in R, and let N = θ^(-1)(M) be the pullback of M in k[x, y]. We want to show that N is a maximal ideal in k[x, y]. Since N contains (f) (because θ^(-1)(M) contains (f)), we have k[x, y]/N as a quotient ring of k[x, y]/(f). Since f is prime (irreducible), the quotient ring k[x, y]/(f) is an integral domain. Now we proceed with the proof as before:

We only need to show that k[x, y]/N is a field to prove that N is a maximal ideal. Recall that θ is the surjective ring homomorphism from k[x, y] onto R. We have θ(N) = M, and since M is a maximal ideal, R/M is a field. We know that R = k[x, y]/(f) and R/M is a field, so k[x, y]/N must also be a field. Thus, N is a maximal ideal in k[x, y], completing the proof.

This proof holds without the assumption of algebraically closedness, as long as f is a prime (irreducible) element of k[x, y].





User
OK.  Is the assumption that the characteristics of k is 0 necessary?
ChatGPT
My apologies for any confusion in my previous responses. I appreciate your patience.

The assumption that the characteristic of k is 0 is not necessary for the proof. The proof that the pullback of a maximal ideal of R with θ is maximal in k[x, y] only relies on f being a prime (irreducible) element in k[x, y], which ensures that k[x, y]/(f) is an integral domain. The characteristic of k does not come into play in the argument.

To reiterate, if f is prime (irreducible) in k[x, y], then the pullback of a maximal ideal of R with θ is maximal, regardless of the characteristic of k.