Let's consider the polynomial $x^d-1$. Theory tells us that it can have at most $d$ roots in (any extension of) a given field.
Here's my problem: let $A$ be the vector space spanned by $1,a,a^2,b,b^2,ab$ over $\mathbb{Z}_3$ (the integers mod $3$). $A$ is itself a field if we define $x^i*x^j=x^{i+j\pmod 3}$.
Choosing $d=2$ we should have that, at most, $2$ elements in $A$ should be equal to $1$...but I can count more than $2$ (at least $a^2,b^2,a^2b^2$). May somebody help me to point out the error in this reasoning?
This ring you have constructed is not a field, since we have (non-zero) zero-divisors. Note that $$ (a - 1)(a^2 + a + 1) = a^3 - 1 = 0 $$ The unique field (up to isomorphism) of size 27 is given by $$ F_{27} = \mathbb{Z}_3[t]/(t^3 + t^2 + t + 2) $$ The ring you have constructed can be thought of as $$ R = \mathbb{Z}_3[a,b]/((a^3-1)(b^3-1)) $$