Hi I don't know how to show that $\langle 2,x \rangle$ is not principal and the definition of a principal ideal is unclear to me. I need help on this, please.

The ring that I am talking about is $\mathbb{Z}[x]$ so $\langle 2,x \rangle$ refers to $2g(x) + xf(x)$ where $g(x)$, $f(x)$ belongs to $\mathbb{Z}[x]$.


I think it's relatively easy to see that $I=\langle 2,x \rangle = \{a_nx^n+\dots+a_1x+a_0; a_0\text{ is even}\}$.

Now, suppose that $I=\langle f(x) \rangle$ for some $f(x)\in I$.

If $f(x)$ is a constant polynomial, then $\langle f(x) \rangle$ contains only polynomials with even coefficients, and we do not get $x$.

If $f(x)$ is of degree at least 1, then $\langle f(x) \rangle$ contains only polynomials of degree at least 1, and we do not get $2$.

So $I$ is not of the form $\langle f(x) \rangle$.

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    $\begingroup$ Whwre did you get the idea that f(x) has to be of degree at least 1? $\endgroup$ – Person May 1 '11 at 9:56
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    $\begingroup$ @Person I considered two possibilities: either f(x) is constant, or it is not... $\endgroup$ – Martin Sleziak May 1 '11 at 9:57
  • $\begingroup$ It seems to me like you saw those possibilities from <2,x> as in considering the case when the generator is <2> and another when <x>. What if it was <2,x^2> would you create two cases, one for <2> and another for <x^2>? I'm trying to make sense of why you did that $\endgroup$ – Person May 1 '11 at 10:01
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    $\begingroup$ I guess the two possibilities can be discovered when you try to find the generator for the ideal. I don't think I'm able to explain it better or provide more insight. (For you example with $x^2$, the same two possibilities would work.) $\endgroup$ – Martin Sleziak May 1 '11 at 10:07
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    $\begingroup$ Maybe another way to look at it, which might help you: You're looking for polynomials that divide both 2 and x (see Jiangwei Xue's answer). $\endgroup$ – Martin Sleziak May 1 '11 at 10:20

I want to record a somewhat less elementary, but perhaps more conceptual answer.

Note first that $\langle 2 \rangle$, $\langle x \rangle$ and $\langle 2, x \rangle$ are all prime ideals of $\mathbb{Z}[x]$. Indeed, the quotients by these ideals are isomorphic, respectively, to $(\mathbb{Z}/2\mathbb{Z})[x]$, $\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$, which are all integral domains.

So in particular we have a proper inclusion of nonzero prime ideals

$0 \subsetneq \langle x \rangle \subsetneq \langle x, 2 \rangle$

in which the smaller ideal is principal. Now let $R$ be any integral domain and let $I \subset J$ be a proper inclusion of nonzero prime ideals, with $I$ a principal ideal. Then $J$ cannot be principal. Indeed, suppose $I = \langle x \rangle$ with $x$ a prime element. Suppose also $J = \langle y \rangle$. Then $x \in J$, so that there exists $a \in R$ with $x = ay$. Since $ay = x \in I$ and $I$ is prime, we have either $a \in I$ or $y \in I$. If $a \in I$, then $a = bx$, so $x = byx$ or $x(1-by) = 0$ in the domain $R$; since $x \neq 0$ we conclude $by = 1$, i.e., $y$ is a unit and therefore $J = R$, contradiction. Similarly if $y \in I$, then $y = bx$, so $x = abx$ and we conclude that $a$ is a unit and thus $I = J$, contradiction.

Added: A variant on the above argument is: if $0 \subsetneq I = \langle a \rangle \subsetneq \langle b \rangle = J$ with both $I$ and $J$ prime, then $a$ and $b$ are both irreducible elements and $b$ properly divides $a$, contradiction. This is technically a stronger fact because in an arbitrary domain a generator of a principal prime ideal is necessarily an irreducible element but the converse generally does not hold. However, the easiest way to show that an element $a \in R$ is irreducible is to show that $\langle a \rangle$ is prime, or equivalently that $R/\langle a \rangle$ is a domain. To show that a nonprime element $x$ is irreducible is more delicate.

Remark: If $R$ is a commutative Noetherian ring, then if $J$ is any nonzero principal prime ideal, there cannot be any nonzero prime ideal $I$ -- principal or otherwise -- with $0 \subsetneq I \subsetneq J$. This is a special case of Krull's Principal Ideal Theorem.

  • $\begingroup$ If you read the OPs questions in the comments you'll soon realize that all of what you wrote above is way over the OPs head. That's why I gave a very elementary answer (and even that required elaboration - see the comments). $\endgroup$ – Bill Dubuque May 13 '11 at 16:29
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    $\begingroup$ @Bill: Yes, I had already realized this. Other answers -- including optimally elementary ones -- have already been given, so this answer has a different purpose. $\endgroup$ – Pete L. Clark May 13 '11 at 16:39
  • $\begingroup$ @PeteL.Clark I recommend not using the letter $x$ to refer to both an arbitrary prime element of $R$ and the generator of $\mathbb{Z}[x]$. I know you are talking about different rings in the two different paragraphs, but this might still be confusing to the OP. $\endgroup$ – Steven Gubkin Oct 23 '14 at 13:49

Below is a $ $ complete $ $ proof from first principles - easily comprehensible to a high-school student.

We show $\rm\:(2,x)\ =\ (f)\ $ in $\rm\:\mathbb Z[x]\: $ yields a parity contradiction, by simply evaluating polynomials.

$\rm\ \ f\ \in\ (2,x)\, \Rightarrow\, f\ =\ 2\, G + x\, H.\: $ Eval at $\rm\: x = 0\ \Rightarrow\ \color{#0a0}{f(0)} = 2\,G(0) = \color{#c00}{2n}\:$ for some $\rm\: n\in \mathbb Z$

$\rm\ \ 2\ \in\ (f)\ \Rightarrow\ 2\ =\ f\, g\:\ \Rightarrow\ deg(f)\ =\ 0\:\ \Rightarrow\:\ \color{#c00}f\ =\ \color{#0a0}{f(0)}\ =\ \color{#c00}{2n}$

$\rm\ \ x\ \in\ (f)\ \Rightarrow\ x\ =\ \color{#c00}f\, h\ =\ \color{#c00}{2n}h.\,\ $ Eval at $\rm\ x = 1\ \Rightarrow\ 1\: =\ 2n\,h(1)\ \Rightarrow\ 1\:$ is even $\, \Rightarrow\!\Leftarrow$

Remark $\ $ The above proof works over any domain where $\,2\ne 0\,$ and $\rm\,2\,$ is not a unit. $ $ i.e. $\rm\:2\nmid 1.\:$ In particular, it works over any domain with a nontrivial sense of parity, i.e. having $\rm\:\mathbb Z/2\:$ as ring image, e.g. the Gaussian integers, or the rationals writable with odd denominator - see this post. Conversely, the result is false if $\rm\:2 = 0\,$ or a unit since then $\rm\,(2,x) = (x)\,$ or $\,(1)\,$ is principal.

Further, the proof still works if we replace $\,2\,$ by any element $\,c\,$ of the coefficient domain $\,D,\,$ yielding: $\ (c,x)\,$ is principal in $\,D[x]\iff c=0\,$ or $\,c\,$ is a unit.

  • $\begingroup$ The only issue I have with your proof is that you claim that f is a multiple of 2. After deg f = 0 shouldn't you say that f could be -1, 1, -2 or 2? Lastly why did you say that x must be 1? $\endgroup$ – Person May 1 '11 at 18:24
  • $\begingroup$ @Person: No, it cannot be $-1$ or $1$ because it lies in $(2,x)$. $\endgroup$ – Arturo Magidin May 1 '11 at 18:40
  • $\begingroup$ @per I don't say "x must be 1". Rather, I evaluate a polynomial equation at $\rm\:x=1\:,\:$ e.g. $\rm\:g(x)=h(x)\:$ evaluated at $\rm\:x=1\:$ yields $\rm\:g(1)=h(1)\:.\:$ My second step deduces that $\rm\:f\:$ is an even integer, and the final third step is one way of deducing a contradiction from the fact that this even integer must divide $\rm\:x\:$ in the ring $\rm\:\mathbb Z[x]\:.\:$ This may seem "obvious" but it requires rigorous proof. Perhaps your "shouldn't" remark stems from a desire to conclude the proof in a way different from my 3rd step, or by switching the order of the steps. $\endgroup$ – Bill Dubuque Jul 9 '11 at 16:04
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    $\begingroup$ @Bill: In my opinion writing "$x = 0$" for "evaluate the polynomial at $x = 0$" may cause confusion for some readers. You have a lot of equalities of ring elements and then an expression $x = 0$ which does not mean that the two elements $x$ and $0$ are equal in the ring (but rather in a quotient ring...). If you want to preserve the succinctness of the argument, maybe you could define a symbol for evaluation of polynomials and use that instead of $=$. $\endgroup$ – Pete L. Clark Jul 9 '11 at 16:32
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    $\begingroup$ As per my previous comment, it is possible that mentioning that evaluation of a polynomial at a ring element $a$ gives a homomorphism from $R[x]$ to $R$ and that this homomorphism may be identified with the quotient map $R[x] \rightarrow R[x]/(x-a)$ could be helpful. On the other hand beginning algebra students are famously not very comfortable with quotients... $\endgroup$ – Pete L. Clark Jul 9 '11 at 16:36

One way to see that $\langle 2,x \rangle$ is not principal is to note that $\mathbb{Z}[x]$ is a UFD(See example 3 in the wiki page), and both $2$ and $x$ are primes. So if the ideal is principal, then $2$ and $x$ will share a common divisor. Contradiction. It is not as down to earth as Martin's solution, but it is a way to look at the problem.

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    $\begingroup$ But $2$ and $x$ do share a common divisor, namely $1$. The point is that $1$ is not in the ideal generated by $2$ and $x$. This is in contrast to the situation in $\mathbb{Z}$ (also a UFD), where any two coprime elements generate the principal ideal $\langle 1 \rangle$. $\endgroup$ – Chris Eagle May 1 '11 at 9:38
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    $\begingroup$ You are right. I should have said non-unit common divisor. $\endgroup$ – Jiangwei Xue May 1 '11 at 10:08
  • $\begingroup$ Do you claim that using the UFD property somehow simplifies the proof? To show that $2$ and $\rm\:x\:$ have no nonunit common divisor in $\rm\:\mathbb Z[x]\:$ it suffices to show one of them is irreducible and does not divide the other. This can be done very simply, e.g. see my answer. Perhaps what you meant to say is that it follows from the fact that $2$ and $\rm\:x\:$ are nonassociate primes in $\rm\:\mathbb Z[x]\:,\:$ which is equivalent to what I said in terms of irreducibility. It doesn't invoke anything near the full power of the result that $\rm\:\mathbb Z[x]\:$ is a UFD. $\endgroup$ – Bill Dubuque Jul 9 '11 at 17:44

Suppose to the contrary that $(2, x) = \{2p(x) + xq(x) : p(x), q(x) \in \mathbb Z\}$ is a principal ideal where $(2, x) = (a(x))$ for some $a(x) \in \mathbb Z[x]$. Observe that $(2, x)$ is proper since the constant term must be even. Moreover it is immediate that $2 \in (a(x))$ and by definition there exists $p(x) \in \mathbb Z[x]$ such that $2 = p(x)a(x)$. But observe that $0 =\deg p(x)a(x) = \deg p(x) + \deg a(x)$ which implies that $\deg p(x) = \deg a(x) = 0$. Since 2 is prime, we it must follow that $a(x), p(x) \in \{\pm1, \pm2\}$. But if $a(x) = \pm1$ then $(a(x)) = R$ which is a contradiction to $(a(x))$ being a proper ideal. Hence it must follow that $a(x) = \pm2$. So $(a(x)) = (2) = (-2)$. But by construction it must also follow that $x \in (a(x)) = (2)$ so there must exist $q(x) \in \mathbb Z[x]$ such that $x = 2q(x)$. The only way this can happen is if $q(x) = \frac12 x$ which is impossible since $q(x)$ can only have integer coefficients. Hence we have arrived at a contradiction to our hypothesis that $(2, x)$ is a principal ideal.

Note: This immediately implies that $\mathbb Z[x]$ is not a PID.


We have $\langle 2, x \rangle = \{ a_n x^n + \cdots a_1 x + a_0 \mid a_0 \in 2 \mathbb{Z}, a_1, \ldots, a_n \in \mathbb{Z} \}$.

Suppose that $\langle 2, x \rangle$ is principal. Then there is some polynomial $f(x)$ in $\langle 2, x \rangle$ such that $\langle f(x) \rangle = \langle 2, x \rangle$. Therefore $x \in \langle 2, x \rangle = \langle f(x) \rangle$ and $2 \in \langle 2, x \rangle = \langle f(x) \rangle$.

It follows that $2 = f(x) g(x)$, $g(x) \in \mathbb{Z}[x]$. Therefore $f(x)=c \in \mathbb{Z}$.

Since $x \in \langle f(x) \rangle$, $c$ must be $1$ or $-1$ (for example, if $c=2$, then $\langle f(x) \rangle = \langle c \rangle = \{a_n x^n + \cdots a_1 x + a_0 \mid a_0, \ldots, a_n \text{ are even}\} \neq \langle 2, x \rangle$ which is a contradiction to our assumption).

But the ideal of $\mathbb{Z}[x]$ generated by $1$ or $-1$ is $\mathbb{Z}[x]$. Since $\langle f(x) \rangle \neq \mathbb{Z}[x]$, we obtain a contradiction again.

  • $\begingroup$ The proof given by you is very good. But I am not able to understand, last sentence. In the question it is not given that <f(x)> # z[x], so How can it be a contradiction $\endgroup$ – Tortoise Oct 13 '17 at 18:15
  • $\begingroup$ @SurendraJ, $\langle f(x) \rangle = \langle 2, x \rangle = \{a_n x^n + \cdots + a_1 x + a_0 \mid a_0 \in 2 \mathbb{Z}, a_1, \ldots, a_n \in \mathbb{Z}\}$. Therefore $\langle f(x) \rangle \neq \mathbb{Z}[x]$. $\endgroup$ – LJR Oct 13 '17 at 20:11
  • $\begingroup$ Thank you, Now I understand it $\endgroup$ – Tortoise Oct 13 '17 at 20:14

An ideal $\langle a_1, \dots, a_k \rangle$ is the smallest ideal containing these elements, explicitly the set of all linear combinations $r_1a_1+\dots + r_ka_k$ where the $r_i$ are arbitrary elements from the ring.

A principal ideal is an ideal that can be generated by a single element.

So first of all, you have to say which ring you are looking at to have a definite question.

Now, you could comment on whether you understand this definition of principal ideal.

If the ideal $\langle 2,x \rangle$ were principal, the generator would have to divide 2. What are the integer polynomial divisors of 2?

  • $\begingroup$ I think that definition that you gave sounds like something that he would've said but it's not really enough of a criteria to show that something is a principal ideal. The way he did it was try to prove by contradiction that there's no generator <z> where <z> = <2,x> but he got stuck on that step. $\endgroup$ – Person May 1 '11 at 9:11
  • $\begingroup$ But of course, it is enough to show something if it is well-defined. $\endgroup$ – Phira May 1 '11 at 9:13
  • $\begingroup$ The integer polynomial divisors of 2? 2 factors as a(x)b(x) = 2*1. I guess you're trying to say that the generator has to be <2> or <1>? Are all principal ideals supposed to be prime? $\endgroup$ – Person May 1 '11 at 9:47
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    $\begingroup$ Yes, the generator has to be 2 or 1. Now, if you can exclude the two cases, you see that there is no single generator. Second question: No, principal ideals are not prime in general. Think of the ideal $\langle 6 \rangle$ in $\mathbb Z$ and the reason why they are called prime ideals. $\endgroup$ – Phira May 1 '11 at 11:59

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