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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1 Answer
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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
answered Apr 10, 2019 at 11:42
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$\begingroup$ $p(x)|2\Rightarrow p(x)=\pm1,\pm2$ can you continue? $\endgroup$ Commented Apr 10, 2019 at 11:46
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1$\begingroup$ So $\langle1\rangle$ is the whole $\mathbb{Z}[x]$ and $\langle2,x\rangle=\{2p(x)+xq(x):p(x),q(x)\in\mathbb{Z}[x]\}$. Can $3\in\langle2,x \rangle$? $\endgroup$ Commented Apr 10, 2019 at 12:01
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1$\begingroup$ If $3=2a(x)+xb(x)$ 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 $\endgroup$ Commented Apr 10, 2019 at 12:15
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1$\begingroup$ Pretty nice...now is clear. I really appreciate it . $\endgroup$ Commented Apr 10, 2019 at 12:18