Show that $\mathbb{Z}[x]=\lbrace \sum_{i=0}^{n}{a_ix^i}:a_i \in \mathbb{Z}, n \geq 0 \rbrace$ is not a principal ideal ring. I know the definition of principal ideal ring is that every ideal is generated by a single element. So my aim here is to find an ideal which is not generated by a single element. But I fail to locate such ideal. Can anyone help me in finding such ideal?


Hint: What about the ideal $\langle2,x\rangle$?

To answer your comment, a principal ideal is just a "set of multiples" of the generating element. Thus every element of a principal ideal has the generator as a divisor, and usually has similar properties. For example, in $\mathbb{Z}[x]$, the ideal $\langle2\rangle$ is the set of polynomials where every coefficient is even, and $\langle x\rangle$ is the set of polynomials of degree greater than $1$. However, $\langle 2,x\rangle$ is made up of all polynomials with even constant term. It doesn't feel as if polynomials with this property have a shared divisor (If they did, what would it be?) so this ideal feels like it's not principal.

In a way, there are somehow too many restrictions on the elements of the ideal to be generated by a single element. (Degree is greater than one AND constant term is even.) To illustrate the idea of "too many restrictions" further, consider the ideals $I_2=\langle 2,x\rangle$, $I_3=\langle 4,2x,x^2 \rangle$, $I_4=\langle 8,4x,2x^2,x^3\rangle, \dots$ It is a fact (that I recall being reasonably hard to prove) that $I_n$ is generated by $n$ elements and no fewer. This is because of the inrease in the number of restrictions of what elements of $\mathbb{Z}[x]$ can be in each ideal.

  • $\begingroup$ well, what motivates you to think of this ideal ? I mean intuitively, how you know that this ideal is not principal? $\endgroup$ – Idonknow Mar 25 '13 at 20:22
  • $\begingroup$ @Idonknow I just updated my answer to address your comment, hope it helps! $\endgroup$ – Tom Oldfield Mar 25 '13 at 20:33
  • $\begingroup$ @Idonknow The idea is that there is nothing "in-between" being a constant and being a formal variable. (Constants can only be multiples of other constants, and monic polynomials of degree $\ge1$ cannot be multiples of anything but other monic polynomials of degree $\ge1$.) $\endgroup$ – blue Mar 25 '13 at 20:36
  • $\begingroup$ @Idonknow See my answer for one way to motivate the search for a counterexample. $\endgroup$ – Math Gems Mar 25 '13 at 20:46

Alternatively, if $R$ is a PID, then any (non-zero) prime ideal is maximal (i.e. it is dimension $1$). Note that $(x)$ is a prime ideal of $\mathbb{Z}[x]$ which is not maximal--this follows because $\mathbb{Z}[x]/(x)\cong\mathbb{Z}$ is a domain but not a field.

This further shows that if $R[x]$ is a PID, then $R$ is a field. The converse is also true, because $R[x]$ will posses a euclidean function ($\deg$).

  • $\begingroup$ Non-zero prime ideals :) $\endgroup$ – Ragib Zaman Mar 25 '13 at 22:43
  • 1
    $\begingroup$ @RagibZaman Yeah, yeah, yeah :) $\endgroup$ – Alex Youcis Mar 25 '13 at 22:43

Hint $\ $ A nonzero ideal $\rm\,I\,$ in a PID is generated by any element of $\rm\,I\,$ having the least number of prime factors among all elements of $\rm\,I.\:$ Hence if $\rm\,I\,$ contains nonassociate primes then $\rm\: I = (1)$. Counterexamples abound in $\rm\,R = \Bbb Z[x],\:$ i.e. it is easy to find primes $\rm\:p,q\in R\:$ with $\rm\:(p,q)\ne (1).$


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