Roots for polynomial of coefficients in non integral domain

Question

By plugging $0,1,2,3,4,5$ into the first polynomial, the zeros are $1,2,4,5$. Since $\mathbb{Z}_6$ is not a field, this polynomial has more the roots than its degree. For $x^2+1$, there are no zeros in $\mathbb{Z}_6$. But, we cannot think of extension of $\mathbb{Z}_6$ because it is not a field. So, should we say just that there is no root in $\mathbb{Z}_6$? Can't we discuss anymore about the roots of $x^2+1$ by using some similar construction of field extension for $\mathbb{Z}_6$?

Answer 1

In $\mathbb{Z}_6$, $-1 \equiv 5 \pmod 6$ is not a quadratic residue (relatively simple to check this).

In $\mathbb{Z}_6$ the polynomial $x^2 + 3x + 2$ will factor to $(x + 1)(x+2)$ which has zeros at $x = 5, 4$ and further inspection will yield $1, 2$ as the field is relatively small.

Answer 2

Your work and understanding in both cases is correct.

(1) You found all of, and only, the zeros of $x^2+3x+2 \in \mathbb Z_6$

(2) Yes, you should say just $x^2 + 1$ has no zeros (hence no roots) in $\mathbb Z_6$.

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As @Hagen points out in his comment above, this should not be entirely surprising, since it is also true that, in $\mathbb R$, $\,x^2 + 1$ has no zeros (or roots).

The fact that $x^2 + 1$ has zeros in $\mathbb C$ does not change the fact that there are no zeros in $\mathbb R$, and clearly, no zeros in $\mathbb Z_6$.