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I can prove that the ideal $(4, 2x, x^2)$ in $\mathbb Z[x]$ is not principal. But I failed to prove that this cannot be generated by two elements.

It's really difficult for me. Would you give me a hint.

user26857
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ryoto
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2 Answers2

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Note that $(4, 2x, x^2)=(2,x)^2$. Set $J=(2,x)$. We have $\mathbb Z[x]/J\simeq\mathbb F_2$ (the field with two elements). Show that $4, 2x, x^2$ are linearly independent in $J^2/J^3$ over $\mathbb F_2$, so $\dim_{\mathbb F_2}J^2/J^3\ge3$. If $J^2$ is two-generated, then the quotient $J^2/J^3$ is also two-generated, a contradiction.

user26857
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Here is one approach: If $(4,2x,x^2)$ were generated by two elements in $\mathbb{Z}[x]$, it would be generated by two elements in $\mathbb{Z}[x]/(x^3,8)$. The latter is a finite ring and we only need to check finitely many cases.

Andrew Dudzik
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  • You have two rings R and R/I, and you're saying that elements of R/I generate an ideal in R. What does this mean? Perhaps you mean lifts of those elements? Is it independent of the lift chosen? – RghtHndSd Apr 14 '15 at 16:21
  • @RghtHndSd The opposite; elements in $R$ generate an ideal in $R/I$. – Andrew Dudzik Apr 14 '15 at 16:22
  • Ok, so the "it" after the comma does not refer to noun in the previous clause, but rather the image of the ideal under the quotient homomorphism which is never mentioned. Hopefully you see the reason for my confusion and will clarify this. – RghtHndSd Apr 14 '15 at 16:24