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Context: Let $K$ be a field. Then $K[x,y]$ is not a PID because $x$ is irreducible but the quotient $K[x,y]/(x)$ is isomorphic to $K[y]$, which is not a field.

So there must be an ideal in there which is not principal.

Question: What is an example of such an ideal?

MyNameIs
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    The obvious guess is $(x)+(y)=(x,y)$. – egreg Jul 27 '16 at 10:52
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    True, that one seems to work. If $(x,y)=(g)$ for some $g\in K[x,y]$, then $g$ would divide both $x$ and $y$. Since $g$ divides $x$, $g$ can have no $y^k$ for $k\geq 1$ because that would contribute a power of $y$ to $x$. By the same token, $g$ can have no $x^k$ for $k\geq 1$. Consequently, $g\in K$. But then $(g)=K[x,y]$, a contradiction. – MyNameIs Jul 27 '16 at 10:57

3 Answers3

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Hint: if $R$ is a domain and not a field, take $r\in R$ such that $r\ne0$ and $r$ is not invertible. Then the ideal $(r,x)$ in $R[x]$ is not principal.

egreg
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  • P Vanchinathan's example being the special case $R=K[y], r=y$ ... – Georges Elencwajg Jul 27 '16 at 12:50
  • @GeorgesElencwajg Yes, of course, but pointing to a more general solution might reveal helpful. – egreg Jul 27 '16 at 13:06
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    Dear @egreg Of course! My comment was not a criticism but, on the contrary, wanted to draw the reader's attention to its generality, in that it contains Vanchinathan's answer. In a word, yours is a great answer! – Georges Elencwajg Aug 10 '16 at 18:20
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Consider the function $\phi\colon K[x,y]\to K$ defined by $\phi(f) = f(0,0)$ (that is evaluation at $(0,0)$). Check that it is a homomorphism.

So its kernel consists of polynomials of two variables vanishing at origin; that is they have no constant terms. This is not a principal ideal (in fact all maximal ideals of this ring are non principal).

P Vanchinathan
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Apply the Lemma below with $\rm\,C = K[y],\,$ noting $\rm \,0\neq y\,$ is not a unit.

Lemma $\, $ If $\ \rm \color{#90f}{0\neq c\in domain}\ C\ $ then $\rm \ (c,x) = (f)\ $ in $\rm \,C[x]\,\Rightarrow \,c\,$ is a unit in $\,\rm C.\ $

${\bf Proof}_{\,1}\ $ If $\,(c,x)\!=\!(f)\,$ then $f\mid c\Rightarrow \deg f \!=\! 0$ $\Rightarrow\!\color{#0a0}{f(1)=f(0)},\,$ so eval $\,(c,x)=(f)\,$ at $\,x\!=\!0\,\Rightarrow\, \color{#c00}{(c)}=\color{#0a0}{(f(0))};\ $ eval at $\,x\!=\!1\Rightarrow (1)= (\color{#0a0}{f(1)})\! = \!(\color{#0a0}{f(0)}) = \color{#c00}{(c)}\Rightarrow \color{#c00}c\mid 1\Rightarrow c\,$ unit.

$\bf Proof_{\,2}\ $ Below is an element-ary form of above proof (it evaluates elements vs. ideals)

$\rm\ \ f\ \in\ (c,x)\ \Rightarrow\ f = c\, g_1 + x\,h_1.\, $ Eval at $\rm\, x = 0\ \Rightarrow\ \color{#0a0}{f(0)} = cg_1(0) = \color{#0a0}{cd},\,\ d\in C$

$\rm\ \ c\ \in\ (f)\ \Rightarrow\ c\ =\ f\, g, \ \,\color{#90f}{hence}\,\ \ \deg f\:\! =\:\! 0\:\Rightarrow\: \color{#c00}f = \color{#0a0}{f(0) = cd}$

$\rm\ \ x\ \in\ (f)\ \Rightarrow\, x\ =\ \color{#c00}f\, h\ =\ \color{#0a0}{cd}\:\!h.\, $ Eval at $\rm\ x\! =\! 1\ \Rightarrow\ 1 = cd\,h(1)\ \Rightarrow\ c\,$ is a unit in $\rm\,C$

Beware $ $ Our use of $\rm C$ is a $\rm\color{#90f}{domain}$ to infer $\,\deg c=0,\ f\mid c\Rightarrow \deg f = 0\,$ is crucial to the proof. This generally fails for non-domains, and so does the Lemma, e.g. in $\,\Bbb Z/6[x]\!:\ (2,x) = (2+3x)\,$ by $\,(2,x)=(2+3x,x)=(2+3x),\,$ by $\,2+3x\mid x,\,$ by $\,(2+3x)(3+2x)=x.\,$ Notice here we have positive degree $ f\mid c,\,$ i.e. $\,2+3x\mid 2\,$ [by $\,-2(2+3x) = 2$].

Remark $ $ It is often true that evaluation serves to reduce many arithmetical properties of polynomial rings to corresponding properties in their coefficient ring, e.g. see here and its links for application of such methods to factoring polynomials.

See here for general results characterizating when a polynomial ring or semigroup ring R[S] is a PIR (principal ideal ring), e.g. we have

Theorem $\ $ TFAE for a semigroup ring R[S], with unitary ring R, and nonzero torsion-free cancellative monoid S.

  1. $\ $ R[S] is a PIR (Principal Ideal Ring)
  2. $\ $ R[S] is a general ZPI-ring (i.e. a Dedekind ring, see below)
  3. $\ $ R[S] is a multiplication ring (i.e. $\rm\ I \supset\ J \Rightarrow\ I\ |\ J\ $ for ideals $\rm\:I,J\:$)
  4. $\ $ R is a finite direct sum of fields, and S is isomorphic to $\mathbb Z$ or $\mathbb N$

A general ZPI-ring is a ring theoretic analog of a Dedekind domain i.e. a ring where every ideal is a finite product of prime ideals. A unitary ring R is a general ZPI-ring $\iff$ R is a finite direct sum of Dedekind domains and special primary rings (aka SPIR = special PIR) i.e. local PIRs with nilpotent max ideals. ZPI comes from the German phrase "Zerlegung in Primideale" = factorization in prime ideals. The classical results on Dedekind domains were extended to rings with zero divisors by S. Mori circa 1940, then later by K. Asano and, more recently, by R. Gilmer. See Gilmer's book "Commutative Semigroup Rings" sections 18 (and section 13 for the domain case).

Bill Dubuque
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    Which is exactly a (hard to read) proof of my hint, isn't it? – egreg Jul 27 '16 at 14:44
  • @egreg Please take your incessant rude comments elsewhere. – Bill Dubuque Jul 27 '16 at 14:48
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    Rude? Where? You come by two hours after my answer and give essentially the same. Next thing you do is downvoting my answer. Who's being rude? – egreg Jul 27 '16 at 15:10
  • @egreg I had not read your hint. The above proof was composed long before this site existed. The reason it is written out in such great detail is that I have seen many students struggle with the details, so I think it is worthwhile to teach how to do these things in a very simple way using only evaluation (the above is comprehensible to a high-school student). TIP polite critiques are usually much more constructive. – Bill Dubuque Jul 27 '16 at 15:15
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    You should read other answers before giving yours. This one is a clear duplicate and the downvote to mine is completely unjustified (that you gave it is clear from you avoiding to deny you did). My opinion that your proofs are generally hard to read is every day stronger, so what? Is it impolite to criticize your style? I see no impoliteness in my short comment. – egreg Jul 27 '16 at 15:24
  • @egreg I have no interest whatsoever in your opinions or speculations on these subjective nonmathematical matters. Please have the courtesy to stop pinging me on these off-topic matters. – Bill Dubuque Jul 27 '16 at 15:53
  • @BillDubuque Quick question: $g,g_1,h,h_1\in D[x]$, wheres the other variables have degree $0$, thus are in $D$ (or, it looks like you use $C$), correct? – User7238 Jan 04 '22 at 00:43
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    @User7238 That lone $D$ should be $C$. The hint is meant as intuition for the elementary proof in the Lemma, where $,c\in C.,$ I will update it shortly to make the proof even more conceptual (meant to do so long ago). – Bill Dubuque Jan 04 '22 at 02:43
  • @BillDubuque I see! Thank you so much for taking the time to respond, I really appreciate it! – User7238 Jan 04 '22 at 02:54