I thought the zero divisors were the polynomials of the form ax^2+bx+2c where a,b,c=0 or 1. But I've since figured out that this quotient ring is not local, so there exists other zero divisors like x+3 since (x+3)(x^2+2). I have found a total of 12 zero divisors now (including x+1, x^2+1,x^2+3), but can someone help determine if there are more?
The quotient ring is obtained by taking the ring of polynomials Z4[x]Z4[x] and quotienting out the ideal generated by x3−x2−2x3−x2−2, 2x22x2, and 2x2x. This means that in the quotient ring, any element equivalent to x3−x2−2x3−x2−2, 2x22x2, or 2x2x is considered zero.
Let's denote the equivalence class of f(x)f(x) in the quotient ring by [f(x)][f(x)].
Now, we want to find elements a(x)a(x) and b(x)b(x) in the quotient ring such that a(x)≠0a(x)=0 and b(x)≠0b(x)=0, but [a(x)][b(x)]=[0][a(x)][b(x)]=[0]. In other words, we are looking for non-zero elements whose product is equivalent to the zero element in the quotient ring.
Here are some candidates:
These are examples of zero divisors in the quotient ring Z4[x]/(x3−x2−2,2x2,2x)Z4[x]/(x3−x2−2,2x2,2x).
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