Nilpotent element $a^2$ in a polynomial quotient ring

Asked Apr 19 '18 at 16:25

Active Apr 19 '18 at 16:25

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$\textbf{Problem Statement : }$ Let $R : = \mathbb{Z}/2\mathbb{Z}[x]$ and let $I$ be the ideal $I = R \cdot(x^{17} -1)$ Is there a non-zero element a of the quotient ring $A = R/I$ such that $a^2 = 0$?

I do not believe there is, as if there was, then $a^2$ would have to be contained within the ideal. However $x^{17} -1$ is reducible and is equal to $(x-1)(\sum_{i=0}^{16} x^i)$, which are both irreducible so there can be no element $a \notin I$ but $a^2 \in I$. Am I on the right track?

asked Apr 19 '18 at 16:25

blanchey

By $\Bbb Z/2\Bbb Z[x]$ you mean $\Bbb Z_2[x]$, right? That is, $\Bbb Z/2\Bbb Z[x] = (\Bbb Z/2\Bbb Z)[x] = \Bbb Z_2[x]$? – Robert Lewis Apr 19 '18 at 16:32
1

Yes, that is correct @RobertLewis – blanchey Apr 19 '18 at 16:34
OK, thanks for clarifying, Cheers! – Robert Lewis Apr 19 '18 at 16:34

Nilpotent element $a^2$ in a polynomial quotient ring

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