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At the moment I am working on a problem that asks me to prove that the set $(2,x):= 2\mathbb Z[X]+X \mathbb Z[X]$ is an ideal, but not a principal ideal. Proving that it is an ideal is pretty easy, but I am not sure whether my way of showing it is not principal is completely correct.

What I did was assume $(2,x)$ is principal, then there is $A:=\sum_{i=0}^na_iX^i\in\mathbb Z[X]$ such that $A\mathbb Z[X]=(2,x)$ then I chose the polynomials $2,x\in(2,x)$, then $A$ must be a common factor of both. The only common factor is obviously $1$ so $(2,x)=1\mathbb Z[X]$ which is obviously false, therefore $(2,x)$ cannot be principal.

Would this approach be correct or did miss something? Help would be greatly appreciated!

Watson
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El Duderino
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    it seems correct – Baby elephant Sep 26 '16 at 13:44
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    That $1$ is the only common factor is not true (also $-1$ is). However the fact that the only common factors between $2$ and $x$ are invertible is the key and should be proved. It's easy, but not obvious, unless you have already proved that $\mathbb{Z}[X]$ is a unique factorization domain; however, the result is true with $\mathbb{Z}$ replaced by any domain $R$ and $2$ by any non invertible element in $R$, so resorting to a UFD argument is not the best idea. – egreg Sep 26 '16 at 13:45
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    When looking for weaknesses in your proofs, carefully consider those beginning with "it is obvious that..." What if you are glossing over the entire difficulty of the proof in that one phrase? – rschwieb Sep 26 '16 at 13:47
  • Related: https://math.stackexchange.com/questions/1335684 – Watson Sep 26 '16 at 16:12

1 Answers1

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You correctly state that $A(X)$ should be a common factor of $2$ and $X$. However the two facts you deem “obvious” aren't. Not difficult, but not obvious either.

Since $1\notin(2,X)$, $A(X)$ is not an invertible constant. Indeed, suppose $1=2P(X)+XQ(X)$; then evaluating at $0$ gives $1=2P(0)$, a contradiction. This is a special case of the statement that the polynomials in $(2,X)$ have even constant term, which is again proved by evaluating at $0$.

Thus $2=A(X)B(X)$, for some $B(X)\in\mathbb{Z}[X]$. Hence $A$ and $B$ have degree $0$. Since $\mathbb{Z}$ is a unique factorization domain, we conclude that $B(X)=1$ or $B(X)=-1$. Thus it's not restrictive to assume $A(X)=2$, because $-A(X)$ generates the same principal ideal as $A(X)$.

Then $X=2C(X)$, for some $C(X)\in\mathbb{Z}[X]$ and the degree of $C$ is $1$; thus $C(X)=pX+q$ and so $X=2pX+2q$ forcing $2p=1$, a contradiction.

If you replace $\mathbb{Z}$ with the ring $R=\mathbb{Z}[\frac{1}{2}]$, the ideal $(2,X)$ in $R[X]$ is instead principal (and “obviously” applies here).

egreg
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  • Ah ok I had not thought of arguing the point through the degree of the polynomials, this is a much more convincing argument than the one I made. – El Duderino Sep 29 '16 at 13:18