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I've been solving some problems from my abstract algebra course as training for the final exam, and I want to check if my solution to this one is correct:

Describe the units, the nilpotent elements and the zero divisors of the rings $\mathbb{Z}_4[X]$ and $\mathbb{Z}_6[X].$

This is the work I did:

For $\mathbb{Z}_4[X]$:

  • Units: this one's easy, since units in a polynomial ring $R[X]$ are the same as the units in $R$, so I conclude that the units of $\mathbb{Z}_4[X]$ are $\bar1,\bar3$ (the coprime elements to $4$).
  • Zero divisors: First, it's obvious that $\bar 2$ is zero divisor. Then, in order for a polynomial to get cancelled by multiply6ing it by a non-zero element, it must have all of it's coefficients $\bar0$ or $\bar2$. So I conclude that the zero divisors of $\mathbb{Z}_4[X]$ are all the polynomials with $\bar 0$ and $\bar 2$ as coefficients (not including $\bar 0$).
  • Nilpotent elements: in order for the power of a polynomial in $\mathbb{Z}_4[X]$ to be cancelled, the same reasoning from the zero divisors gives us that the nilpotent elements of $\mathbb{Z}_4[X]$ are the polynomials with coefficients $\bar0$ and $\bar 2$ (this time including $\bar 0$).

For $\mathbb{Z}_6[X]$:

  • Units: The units in $\mathbb{Z}_6[X]$ are the same of $\mathbb{Z}_6$, so they are $\bar 1$ and $\bar 5$ (the coprimes to $6$)
  • Zero divisors: using a similar reasoning to the one I used for $\mathbb{Z}_4[X]$, I end up concluding that the zero divisors in $\mathbb{Z}_6[X]$ are the polynomial with coefficients $\bar0$, $\bar2$, $\bar 4$; and the polynomials with coefficients $\bar0$, $\bar 3$ (not including $\bar 0$)(also not mixing coefficients, either it has $\bar2$, $\bar4$ or either it has $\bar3$).
  • Nilpotent elements: in order for a power of a polynomial to be cancelled, it's clear that it's leading coefficient must be multiple of $6$, as well as it's independent coefficient, but $6$ is $\bar 0$ in $\mathbb{Z}_6[X]$, so I end up concluding that the only nilpotent element in $\mathbb{Z}_6[X]$ is $\bar 0$.

Is my solution correct? If not, where did I go wrong? Any help will be appreciated, thanks in advance.

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2 Answers 2

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I think that parts with units and nilpotent elements are correct, but I'm not convinced about the part with zero divisors. It looks like you only included the case of multiplying our polynomial by something from $R$, not from $R[X]$. It is not clear why for polynomial like $2x^2+x+2$ it's not possible that multiplying it by some strange element of $R[X]$ won't result in some fancy cancelling and product equal to zero. (However I suppose the answer should be correct.)

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  • $\begingroup$ Thanks for your answer, I'll think about it a bit more to see if I find something that proves my point. $\endgroup$ Commented Jan 30, 2021 at 18:48
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Your conclusion regarding "units" is not correct.

Units: this one's easy, since units in a polynomial ring $R[X]$ are the same as the units in $R$, so I conclude that the units of $\mathbb{Z}_4[X]$ are $\bar{1}$, $\bar{3}$ (the coprime elements to $4$).

Note that, the units of any ring $R$ is always an unit of $R[X]$, because $R$ is embedded inside $R[X]$ as the constant polynomials. However, the converse is true only if $R$ is an integral domain (i.e. it has no zero divisors). So, if $R$ is not an integral domain, you cannot say that the only units of $R[X]$ are (precisely) the units of $R$. In other words, $R[X]$ might have units that are not in $R$.

For example, take $R = \mathbb{Z}_4$, so that $R[X] = \mathbb{Z}_4[X]$. Note that $R$ is not an integral domain ($2$ is a zero divisor, for example). See that the polynomial $2X + 1$ in $\mathbb{Z}_4[X]$ is a unit, which is not an unit of (not even present in) $\mathbb{Z}_4$.

Indeed, $(2X+1)(2X+1) = \bar{4}X^2+\bar{4}X+1 = 1$.

Similarly, $\mathbb{Z}_6[X]$ is not an integral domain, so you cannot say that the only units of $\mathbb{Z}_6[X]$ are $\bar{1}$ and $\bar{5}$.

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