Let $R=\mathbb{Z}/2[x]$ and $I=R(x^{17}-1)$. Is there a nonzero element $y$ of $R/I$ with $y^2=0$?

Question

Let $R=\mathbb{Z}/2[x]$ and the ideal $I=R(x^{17}-1)$. Is there a non zero element $y$ of $R/I$ with $y^2=0$?

My approach: Suppose there is $y=\bar{f}(x)$ with $f(x)\in R$, then $\bar{f}(x)\not\in R(x^{17}-1)$ and $f(x)^2=(x^{17}-1)g(x)$. We have that $(x-1)|(x^{17}-1)g(x)$ and the 17th-cyclotomic polynomial also divides $(x^{17}-1)g(x)$ hence they divide $f(x)^2$ and since they are irreducible they both divide $f(x)$. So $(x^{17}-1)|f(x)$ which is a contradiction.

Is this correct or have I missed something?

Answer 1

This is correct, modulo the proof that the 17-th cyclotomic polynomial is irreducible. (note: it appears not to be!)

A slightly easier approach would be to show that $x^{17}-1$ has no repeated factors, by comparison with its derivative.

Answer 2

It doesn't look like 17th-cyclotomic polynomial is irreducible, so your approach falls down. Here's another go.

For $[y]^2=0$ ($[y]$ is $y \; mod \; I$), we must to have $y^2 \in R(x^{17} - 1)$.
If $(x^{17} - 1)$ has a repeated root, we can factor it as $x^{17} - 1 = g^2\prod f_i$. Then, $y = g^\prod f_i$ will have square $y^2 = g^2\prod f_i^2 = (x^{17} - 1)\prod f_i$, which leads to $y^2=0$.

We must hence show that $x^{17} - 1$ has no repeated roots. We can see this by taking the derivative : an equation has a repeated root if its derivative also has that root. The derivative of $x^{17} - 1$ is $17x^{16}$, whose only roots are $0$. Many, many zeroes. Zero, however, is clearly not a root of $x^{17} - 1$. Hence there are no shared roots and the problem is solved.