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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 $\mathbb{Z}[x]$.

7 Answers 7

42

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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    I spent the all afternoon trying to show that! You saved my evening! Thanks a lot for your very good and intuitive explanation!2015-04-29
29

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.

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    @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.2014-10-23
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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.

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    See also [this answer.](http://math.stackexchange.com/a/110758/23500)2012-02-19
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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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    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.2011-07-09
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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.

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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.

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    Thank you, Now I understand it2017-10-13
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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?

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    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.2011-05-01