I don't get it how calculate $\langle2,x\rangle$. I need that because I want to use that these ideals cannot be generated by one single element. And concludes that $\mathbb{Z}[x]$ is not a PID.
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HINT
If $\langle 2,x \rangle=\langle p(x)\rangle$ then $p(x)|2 $ & $p(x)|x$ which is a contradiction
1123581321
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1why is a contradiction? – José Marín Apr 10 '19 at 11:45
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$p(x)|2\Rightarrow p(x)=\pm1,\pm2$ can you continue? – 1123581321 Apr 10 '19 at 11:46
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but $p(x)|x$ means that there exist a $g(x)$ such that $p(x)g(x) = x$ and we can set $g(x) = p(x)^{-1}x$ becuase $p(x)$ is a constant also the reciprocal an then $g(x)$ is a polynomial? – José Marín Apr 10 '19 at 11:51
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maybe you have written something wrong, $p(x)|x$ means $x=p(x)g(x)$ so $\deg p(x)=1 \text{ or }0\Rightarrow p(x)=\pm1,\pm x$ – 1123581321 Apr 10 '19 at 11:54
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Can you conclude that $p(x)=\pm1$? – 1123581321 Apr 10 '19 at 11:56
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it semms to me that $p(x) = ±1$ is a common solution. – José Marín Apr 10 '19 at 11:56
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right! Now we have $\langle 2,x\rangle=\langle 1\rangle=\mathbb{Z}[x]$, is this a contradiction? – 1123581321 Apr 10 '19 at 11:57
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I don't know.. the thing is that I don't understand $<1>$ neither $<2,x>$ linear combination under the ring in discussion or what? – José Marín Apr 10 '19 at 11:59
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1So $\langle1\rangle$ is the whole $\mathbb{Z}[x]$ and $\langle2,x\rangle={2p(x)+xq(x):p(x),q(x)\in\mathbb{Z}[x]}$. Can $3\in\langle2,x \rangle$? – 1123581321 Apr 10 '19 at 12:01
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I see.. so it is something like the generators of vector spaces... thank you. – José Marín Apr 10 '19 at 12:10
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Yes, can you show that $3\not\in \langle2,x\rangle$ in order to reach a contradiction? – 1123581321 Apr 10 '19 at 12:11
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intuitively yes but not in a formal writing... – José Marín Apr 10 '19 at 12:13
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1If $3=2a(x)+xb(x)$ for some $a(x),b(x)\in\mathbb{Z}[x]$ then for $x=0$ we have $3=2a(0)$ so $2|3$ which is the contradiction – 1123581321 Apr 10 '19 at 12:15
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1Pretty nice...now is clear. I really appreciate it . – José Marín Apr 10 '19 at 12:18