0

I know proposition 7 in Dummit & Foote that says "Every nonzero prime ideal in PID is a maximal ideal" and I know that $F(x)$ is a PID.

But I was asked to show that: Every prime ideal of $F[x]$ is a maximal ideal of $F[x].$

My guess is that the question is asking me this(Every prime ideal and not every non zero prime ideal) because the zero ideal in $F[x]$ is just generated by zero so it comes from the zero polynomial which lies in $F$(any field is an integral domain and the zero ideal in an integral domain is a prime ideal) and in a field the zero ideal is always a maximal ideal, is my guess correct?

Emptymind
  • 1,901
  • @coffeemath I am sorry, I did not get how your comment answers my question, but I edited my question as I felt that it may be unclear. – Emptymind Oct 08 '22 at 05:40
  • 1
    What is the precise definition of "prime ideal" you are using? – coffeemath Oct 08 '22 at 05:49
  • 1
    Yes, obviously "every nonzero prime ideal is maximal" is intended. PIDs are precisely the UFDs satisfying this property (i.e. having Krull dimension at most $1).,$ See the linked dupe for more. – Bill Dubuque Oct 08 '22 at 07:12
  • @BillDubuque my question was why incase of polynomials in one indeterminate over a field the question said every prime ideal not every nonzero prime ideal and if my guess is correct or no – Emptymind Oct 08 '22 at 14:26
  • 1
    It's a typo/braino (inconsistent with the prior more general Prop. 7 you quote). It is incorrect without "nonzero". – Bill Dubuque Oct 08 '22 at 14:38
  • @BillDubuque so my justification for the zero prime ideal in this case was wrong also? – Emptymind Oct 08 '22 at 14:39
  • 1
    What you wrote is correct but it is not clear why you think this can justify said false claim. $(0)$ is prime in $F[x]$ but not maximal so that is a counterexample to what you were asked to show. – Bill Dubuque Oct 08 '22 at 14:43
  • Because the question was from a previous exam year so I was pretending that I am in the exam and that this will be my answer if I met this question @BillDubuque ....... also, if the statement is still correct then nothing is wrong with the exam question ...... right(the zero prime ideal is still a maximal ideal in this case and I think that is why the examiner did not put the word nonzero)? – Emptymind Oct 08 '22 at 14:47
  • The statement is not correct since my prior comment gives a counterexample. In one place you mention the field $F(x)$. Is that what you mean elsewhere (vs. $F[x]$?). Note $(0)$ is maximal in $R\iff R$ is a field, but $F[x]$ is not a field. – Bill Dubuque Oct 08 '22 at 14:58
  • My bad, in all of my writings, I only meant either $F$ or $F[x]$ ...... I did not mean to write $F(x)$ at all. @BillDubuque – Emptymind Oct 08 '22 at 15:31
  • @coffeemath I am using that a proper ideal $P$ of a commutative ring A with identity is prime iff when $ab \in P$ then either $a\in P$ or $b \in P$? – Emptymind Oct 08 '22 at 15:34
  • @Emptymind Then see Bill's comments above. – coffeemath Oct 08 '22 at 16:35

0 Answers0