How many zero divisors there are in the ring $\mathbb{Z}_{7}[X] / (x^4+x^3-3)$?

What is the inverse element of $\overline{x+1}$?

I'm not sure where to begin, so I thought it might be a good idea to start with a basic definition:

Zero divisor is defined as an element $a\not = 0$ of a ring $R$ if $\exists b\not = 0$ such that $ab=0$ or $ba=0$.

Then I tried to get an idea of what the elements of $\mathbb{Z}_{7}[x] / (x^4+x^3-3)$ should look like. They should be the classes of the remainders of Euclidean division by $(x^4+x^3-3)$, so I think that my ring should look like this: $$\mathbb{Z}_{7}[x] / (x^4+x^3-3) = \{ [a_0 + a_1 x + a_2 x^2 + a_3 x^3] : a_i \in \mathbb{Z}_{7} \}.$$ Is this right?

At this point I'm not sure where to go. Can someone please guide me through the entire process or at least give me some advice?

  • $\begingroup$ Is $Z_7$ the $7$-adic or the field with $7$-elements? $\endgroup$ – Lior B-S Mar 18 at 13:31
  • $\begingroup$ The field with 7 elements. $\endgroup$ – Michele Mar 18 at 13:34

I answer assuming $\mathbb Z_7$ is the finite field of $7$ elements. The polynomial $f=x^4+x^3-3$ factors as $(x + 4) (x + 5) (x^2 + 6x + 3)$ to irreducibles. So by the Chinese Remainder Theorem we have $$\mathbb Z_7[x]/(f) \cong \mathbb Z_7[x]/(x+4)\times \mathbb Z_7[x]/(x+5) \times \mathbb Z_7[x]/(x^2+6x+3)\cong \mathbb Z_7\times \mathbb Z_7\times \mathbb F_{49},$$ where $\mathbb F_{49}$ is the finite field with $49$ elements. The zero divisors are the preimages of $(x,y,z)$ with at least one of $x,y,z$ being zero.

To have explicit description, you need to write down the CRT formula.

Edit: I thought the question was to find the zero divisors. To compute how many there are, we note that in a finite ring a nonzero element $x$ is either invertible or a zero divisor (since the multiplication-by-$x$ map is either injective and hence from finiteness also surjective, hence invertible, or it has a non-trivial kernel and so the element is zero divisor). Now an element is invertible if and only if it is invertible in each of the components in the factorization above. So we have $6*6*48= 1728$ invertibles and $7^4-1728-1= 672$ nonzero zero-divisors.

  • $\begingroup$ Thanks a lot! I'm not sure how to find the number of zero divisors of $\mathbb{Z}_{n}$ either, so I'm just going to ask it: is it correct to say that $m$ is a zero divisor of $\mathbb{Z}_{n}$ iff $gcd(m,n) > 1$? $\endgroup$ – Michele Mar 18 at 16:50
  • $\begingroup$ Hence the number of zero divisors should be $n-1-φ(n)$, where φ is the totient function. Is this right? $\endgroup$ – Michele Mar 18 at 16:58
  • $\begingroup$ I miss read the question. I thought you had to find all zero divisors. In this ring a nonzero element is either a zero divisor or invertible. So you may use the function field analogue of Eulers totient function. It is essentially what I sketched above. $\endgroup$ – Lior B-S Mar 19 at 5:57
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    $\begingroup$ Note that $F_{49}$ is not isomorphic to $Z/49 Z$. The former is a finite field with $49$ elements while the latter is not a field. Also note that any nonzero element in a field is invertible. $\endgroup$ – Lior B-S Mar 20 at 13:42
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    $\begingroup$ Oh ok, you're right. Now everything is clear. Thank you very much for your patience :) $\endgroup$ – Michele Mar 20 at 20:17

For the second question, note that $ x^4 + x^3 - 3 = x^3 (x + 1) -3 $.


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