Prove $I$ is non-principal ideal of $\mathbb{Z}[x]$?

Question

I'm new to algebra and got stuck with concept of ideals.

The question is to prove that

$$I = \left\{ {{a_0} + {a_1}x + \cdots + {a_n}{x^n} \mid {a_i} \in \mathbb{Z},{a_0} \in 2\mathbb{Z}} \right\} $$ is a non-principal ideal of $\mathbb{Z}[x]$.

I tried to show for any $f \in \mathbb{Z}\left[ x \right]$ the principal ideal $fR \ne I$.

So I considered three different cases:

1) $f$ is an odd constant polynomial.

2) $f$ is an even constant polynomial.

3) $f$ is a non-constant polynomial.

Then I had, $$\begin{array}{l} aR = a\left( {{\alpha _0} + {\alpha _1}x + \cdots {\alpha _n}{x^n}} \right)\\ bR = b\left( {{\alpha _0} + {\alpha _1}x + \cdots {\alpha _n}{x^n}} \right)\\ \left( {x + 1} \right)R = {\alpha _0} + \left( {{\alpha _0} + {\alpha _1}} \right)x + \left( {{\alpha _1} + {\alpha _2}} \right){x^2} + \cdots + \left( {{\alpha _{n - 1}} + {\alpha _n}} \right){x^n} \end{array} $$

where ${\alpha _i} \in \mathbb{Z}$ and $a$ is an even constant and $b$ is an odd constant.

But I cannot distinguish any difference between $fR$ and $I$ with above expressions.

Or is it just enough to say that

$$\begin{array}{l} 2 \cdot R = 2{\alpha _0} + 2{\alpha _1}x + \cdots + 2{\alpha _n}{x^n}\\ \therefore 2{\alpha _1},2{\alpha _2}, \cdots ,2{\alpha _n} \ne \mathbb{Z}\\ \\ 1 \cdot R = {\alpha _0} + {\alpha _1}x + \cdots + 2{\alpha _n}{x^n}\\ \therefore {\alpha _0} \ne 2\mathbb{Z}\\ \\ \left( {1 + x} \right) \cdot R = {\alpha _0} + \left( {{\alpha _0} + {\alpha _1}} \right)x + \cdots + \left( {{\alpha _{n - 1}} + {\alpha _n}} \right){x^n} + {\alpha _n}{x^{n+1}}\\ \therefore {x^{n+1}} \notin I \end{array} $$

Your goal is to show two sets are unequal. One way to do this is to find an element in one set that is not an element of the other set. — , Jun 06 '15 at 16:48
@user26857 . I thought that (x+1)R would give us sufficient reason why it cannot be equal to I, but i would be nicer to any polynomial of degree > n is not included in I — hmmmmm, Jun 07 '15 at 06:35

Sasha Patotski · Accepted Answer · 2015-06-06T20:55:56.943

2

You can note that $2\in I$. So, if $I$ is principal, say $I=(f)$, then $2$ should be divisible by $f$. But there are not too many polynomials that divide $2$ (namely, you have $1$ and $2$), and none of the ideals $(1),(2)$ equals $I$.

edited Jun 06 '15 at 20:55

answered Jun 06 '15 at 17:02

Sasha Patotski

7,351

score 1 · Answer 2 · answered Jun 06 '15 at 17:09

Notice $\rm\,I = (2,x).\,$ We show $\rm\:I\ =\ (f)\ $ in $\rm\:\mathbb Z[x]\: $ yields a parity contradiction.

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

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

$\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 $\, \Rightarrow\Leftarrow$

score 0 · Answer 3 · answered Jun 06 '15 at 17:02

I propose you the following hinted way:

Suppose $\;I=\langle\; f(x)\;\rangle\;,\;\;f(x)\in\Bbb Z[x]\;$ . Since $\;\deg(kf)\ge\deg f\;\;\forall \;k(x)\in\Bbb Z[x]$ , we get that all the elements in $\;I\;$ are either the zero polynomial or else polynomials of degree at least $\;\deg f\;$ .

Since $\;2\in I\;$ the above means $\;\deg f=0\;$ and thus $\;f(x)=k\;,\;\;k\in\Bbb Z\;$ is a constant polynomial . But we also have that $\;x+2\in I\;$ , so it must be that $\;x+2=kg(x)\;$ , for some $\;g(x)\in\Bbb Z[x]\;$ .

End the argument now.

Prove $I$ is non-principal ideal of $\mathbb{Z}[x]$?

3 Answers3