# Zero divisor, nilpotent elements of quotient ring $\mathbb Z_2[x] / \langle x^8-1\rangle$

Consider the ring $R = \displaystyle\frac {\mathbb Z_2[x]}{\langle x^8-1\rangle}$.

i) Is $R$ a finite ring?

ii) Does $R$ have a zero divisor?

iii) Does $R$ have nilpotent elements?

ii) for zero divisor: $[(x^4+1)(x^2+1)(x+1)](x-1) = (x^8-1) = 0 \mod(x^8-1)$. Is this the right way of showing zero divisor?

And I have no idea how we can solve the other two.

Any help would be appreciated.

• i) You should prove that $R$ is a vector space over the field $\mathbb{Z}_2$ of dimension $8$. – Andrea May 5 '12 at 9:26
• ii) To show $a$ and $b$ are zero-divisors, it's not enough to show that $ab=0$. You also need to show that $a$ and $b$ are not zero themselves. – Chris Eagle May 5 '12 at 9:30
• Nilpotency of $f(x)+\langle x^8-1 \rangle\not=0$ is equivalent to existence of positive integer $n$ s.t. $(x^8-1)|f(x)^n$ but $(x^8-1)\nmid f(x)$. – KPK May 5 '12 at 9:38
• @Andrea how do I prove R is a vector space. – Faisal May 5 '12 at 10:23
• Note that modulo $2$ we have $x^2 -1 \equiv (x-1)^2 \mod 2$ so $x^8-1\equiv (x-1)^8 \mod 2$. That should help you. – Joel Cohen May 5 '12 at 12:32

1. In general, if $F$ is a field, and $f(x)$ is a polynomial of degree $n\gt 0$, then $$\frac{F[x]}{\langle f(x)\rangle}$$ is always a vector space of dimension $n$ over $F$; the reason is simply that $F[x]$ is a Euclidean domain, so given any $g(x)\in F[x]$ we can find $q(x),r(x)\in F[x]$ such that $$g(x) = q(x)f(x) + r(x),\quad r(x)=0\text{ or }\deg(r)\lt n.$$ That means that $g(x)+\langle f(x)\rangle = r(x)+\langle f(x)\rangle$, so every polynomial in $F[x]$ is congruent to a polynomial of degree strictly smaller than $n$. This in turn tells us that $1+\langle f(x)\rangle,\ldots,x^{n-1}+\langle f(x)\rangle$ span $F[x]/\langle f(x)\rangle$ as an $F$-vector space; and it is not hard to verify that they are linearly independent, so that the dimension over $F$ of the quotient is just $n$.
In particular, if $F$ is a finite field, then $F[x]/\langle f(x)\rangle$ would be a finite dimensional vector space over a finite field, and thus would be a finite set.
You just have the special case of $F=\mathbb{F}_2$ and $f(x) = x^8-1$.
2. Also, $F[x]/\langle f(x)\rangle$ is a domain (has no zero divisors) if and only if $\langle f(x)\rangle$ is a prime ideal; this occurs if and only if $f(x)=0$ or $f(x)$ is a prime element of the ring (because we are in a commutative PID), which occurs if and only if $f(x)$ is irreducible (because in a PID, prime elements are the same as irreducible elements).
3. An element $r(x)+\langle f(x)\rangle$ is nilpotent if and only if there exists $n\gt 0$ such that $f(x)|r^n(x)$, but $f(x)$ does not divide $r(x)$. Because we are in a UFD, this can only occur if $f(x)$ is not squarefree; indeed, if $f(x) = p_1(x)^{a_1}\cdots p_t(x)^{a_t}$ is a factorization of $f(x)$ into pariwise distinct irreducibles, then $f(x)|r^n(x)$ for some $n$ if and only if $p_i(x)|r(x)$ for each $i$ (by unique factorization); hence we need $p_1(x)\cdots p_t(x)$ to divide $r(x)$. But in order for $f(x)$ to not divide $r(x)$, we need $p_1(x)\cdots p_r(x)\neq f(x)$. That is, at least one of the $a_i$ must be greater than $1$; in other words, $f(x)$ must not be squarefree. Apply this to the case at hand.