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Let $R=(\Bbb Z/5\Bbb Z$)[X]/ $\langle X^n \rangle $ with $n\in\Bbb N_+$. Show that $f\in R$ is a zero divisor if and only if $f_0=0$. And how many zero dividers exist?

I know that zero divispr means there exists non-zero $m$ such that $fm=0$ in $R$ and that $\Bbb Z/5\Bbb Z$ has no zero divisors, meaning $f_0$ or $m_0$ are equal to $0$ if $fm=0$.

How can i show that if $m_0\neq0$ then $f_0=0$?

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Every element of $R$ has a unique representative of degree less than $n$; this shows that $|R|=5^n$.

By construction you have $X^n=0$ in $R$. This shows that $X$ is a zero divisor, because $X\cdot X^{n-1}=0$, where $X^{n-1}\neq0$. But by the exact same argument every (non-zero) multiple of $X$ is a zero divisor; for every $g\in R$ you have $$(X\cdot g)\cdot X^{n-1}=X^n\cdot g=0\cdot g=0.$$ Of course an element $f\in R$ is of the form $X\cdot g$ if and only if $f_0=0$, so this shows that if $f_0=0$ then $f$ is a zero divisor.

Conversely, if $f\in R$ is a zero divisor then there exists some non-zero $m\in R$ such that $fm=0$, meaning that $fm$ is a multiple of $X^n$. Because $m$ is non-zero it is not a multiple of $X^n$. So for the product $fm$ to be a multiple of $X^n$ it must be the case that $f$ is a multiple of $X$, meaning $f_0=0$.

This shows that an element $f\in R$ is a zero divisor if and only if $f_0=0$. Because every $f\in R$ has a unique representative of degree less than $n$, this means the number of zero divisors of $R$ is $5^{n-1}$.

Do consider carefully what happens with this argument when $n=1$.

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  • $\begingroup$ Thank you ! How would i continue to find the units in R ? I think that the units in R are only 1,2,3,4 $\endgroup$ – ggok Dec 16 '16 at 2:03
  • $\begingroup$ Those are indeed units, but they are not all. A nice exercise in commutative algebra is to show that in a finite ring, every element is either a unit or a zero divisor. $\endgroup$ – Servaes Dec 16 '16 at 2:05
  • $\begingroup$ I think that would be way too advanced for my current level :D. Any easier way i could find them ? So according to you there are 4*$5^{n-1}$ units ? $\endgroup$ – ggok Dec 16 '16 at 2:08
  • $\begingroup$ Can i argument that every f which isnt a zero divisor is a unit ? thus $5^n-5^{n-1} = 4*5^{n-1}$ $\endgroup$ – ggok Dec 16 '16 at 2:10
  • $\begingroup$ Without context it is not an easy exercise. But let me give you a hint; for each element $r\in R$, consider the map given by multiplication by $r$, i.e. $$R\ \longrightarrow\ R:\ x\ \longmapsto\ xr.$$ What does it mean for this map to be surjective? And what does it mean for the map to be not injective? $\endgroup$ – Servaes Dec 16 '16 at 2:12
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Hint:

Any element in $R$ is the class mod. $X^n$ of exactly one polynomial of degree $<n$ in $\mathbf Z/5\mathbf Z$. Such a polynomial corresponds to a zero-divisor if and only if it is divisible by $X$, i.e. it is of the form $\;XP(X)$, with $\deg P<n-1$. These polynomials are determined by the sequence $(c_0,c_1,\dots , c_{n-2})$ of their coefficients. In other words, they're in bijection with $(\mathbf Z/5\mathbf Z)^{n-1}$.

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