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On $\mathbb{Z}[x]$ polynomial rings is module $M=(x,3)=x\mathbb{Z}[x]+3\mathbb{Z}[x]$ cyclic?

Note: $(x,3)$ means generated by x and 3

I have no idea how i should start. I only know cyclic means generated by only one element. I would be appreciated for any help.

BERAT
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    Assume that $(x,3)=(p(x))$ for some $p\in\mathbb{Z}[x]$. Then, there should be $q\in\mathbb{Z}[x]$ such that $3=q(x)p(x)$. This forces $p$ to have degree $0$ and divide $3$. Therefore, $p$ must be either $\pm1$ or $\pm3$. You could ignore the sign, since $-1$ is a unit. The case $p=1$ cannot be, since the homomorphism that sends $x$ to $3$ sends all elements of $(x,3)$ to multiples of $3$, while it sends $1$ to $1$, which is not a multiple of $3$. The case $p=3$, since the homomorphism sending $x$ to $1$ sends all elements of $(p)=(3)$ to multiples of $3$, but sends $x\in (x,3)$ to $1$. – plop Jun 23 '21 at 23:29
  • I am not sure i understand you. Which homomorphism are we talking about? – BERAT Jun 23 '21 at 23:40
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    $f:\mathbb{Z}[x]\to\mathbb{Z}$ defined by $f(p)=p(3)$, for all $p\in\mathbb{Z}[x]$, in one case, or $f(p)=p(1)$ in the other case. – plop Jun 23 '21 at 23:42
  • So there wont be such $p(x)$ because non of the options cant be. So M isnt cyclic. Am i correctly understand you? – BERAT Jun 23 '21 at 23:46
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    Well, all proofs that something is not true consist in showing that no option can be. Yes, we are proving that something doesn't happen. However, there are other details in the argument that are specific to this problem. – plop Jun 23 '21 at 23:49

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