$\langle 2,x \rangle$ is a non-principal ideal in $\mathbb Z [x];\, $ $D[x]$ PID $\iff D$ field, for a domain $D$

Question

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 $\mathbb{Z}[x]$.

Answer 1

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 non-zero polynomials in $\langle f(x) \rangle$ have degree at least $1$, and we do not get $2$.

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

Answer 2

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.

Answer 3

Apply the Lemma below with $\rm\,D = \Bbb Z\,$ and nonunit $\rm\,c = 2\in \Bbb Z$

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

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

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

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

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

$\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 $\rm\,D$

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Corollary $ $ If $\rm\,D\,$ is a domain then $\rm\,D[x]\,$ is a PID $\rm \iff D\,$ is a field.

Proof $\ (\Rightarrow)\ $ By the lemma if $\rm\,c\neq 0\,$ then $\rm\,(c,x)\,$ principal $\rm\Rightarrow c\,$ is a unit, so $\rm\,D\,$ is a field.
$(\Leftarrow)$ well-known - by $\rm D$ field $\rm\Rightarrow D[x]$ has a Euclidean division (with remainder) algorithm.

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

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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?

Answer 8

We prove by contradiction.

Suppose that $(2,X)$ is a principal ideal. Then, we have $(2,X)=(f(X))$ for some polynomial $f(X) \in \mathbb{Z}[X]$.

Therefore, we have \begin{equation} f(X)\cdot g(X) = 2 \end{equation} for some polynomial $g(X) \in \mathbb{Z}[X]$.

By inspecting the degree, we have \begin{equation} \text{deg}(f(X)\cdot g(X)) = \text{deg}(f(X)) + \text{deg}(g(X)) = \text{deg}(2) = 0. \end{equation}

Hence, we must have \begin{equation} \text{deg}(f(X)) = 0 = \text{deg}(g(X)), \end{equation} since the degree of polynomials cannot be negative.

This gives us that both $f(X)$ and $g(X)$ are constant polynomials (integers). Therefore, we have either $f(X)=\pm 1$ or $f(X)=\pm 2$.

Suppose that $f(X)=\pm 1$, then we have \begin{equation} \pm 1 = 2\alpha(X) + X\cdot\beta(X) \end{equation} for some polynomials $\alpha(X),\beta(X)\in\mathbb{Z}[X]$.

Let the coefficient of the constant term in $\alpha(X)$ be an integer $x\in\mathbb{Z}$, then observe that the coefficient of the constant term of $2\alpha(X)$ must be an even number. Hence, we obtain $1=2x$. But, there is no integer $x$ that satisfies this equation. Therefore, $f(X)\neq\pm 1$.

We are left with $f(X)=\pm 2$. Similarly, observe that the coefficient of the term $X$ of $2\alpha(X)$ must be an even number while the coefficient of the term $X\cdot \beta(X)$ can be either odd or even. Hence, $f(X)\neq \pm 2$ also.

We have ruled out all possibilities for $f(X)$. Therefore, we conclude that $(2,X)$ can never be a principal ideal of $\mathbb{Z}[X]$.