Consider the polynomial $ \ x^3 - 3\ x^2 +2\ x -1$ How can this polynomial be written in mod 3 ?
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
On
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
Your first thought is correct: we don't change any power. Even though $3$ is in the exponent, we don't take the exponents mod 3, but only the coefficients.
So, taking a polynomial mod 3 just means taking each coefficient mod 3 (and leaving the exponents). For example, $6x^3 + 4x^2 - 2 \mod 3$ would be $0x^3 + x^2 + 1$, which is the same as $x^2 + 1$ since we can drop $0$s. This is because $6 \mod 3$ is $0$, $4 \mod 3$ is $1$, and $-2 \mod 3$ is $1$.
If $x$ is an integer, then \begin{align*} &x^3-3x^2+2x-1\\[4pt] \equiv\;&x^3-0x^2+2x-1\;(\text{mod}\;3)\\[4pt] \equiv\;&x^3+2x-1\;(\text{mod}\;3)\\[4pt] \equiv\;&x+2x-1\;(\text{mod}\;3) \qquad\text{[since $x^3\equiv x\;(\text{mod}\;3)$]}\\[4pt] \equiv\;&3x-1\;(\text{mod}\;3)\\[4pt] \equiv\;&0x-1\;(\text{mod}\;3)\\[4pt] \equiv\;&-1\;(\text{mod}\;3)\\[4pt] \equiv\;&2\;(\text{mod}\;3)\\[4pt] \end{align*} If $x$ is just an indeterminate, then you can't replace $x^3$ by $x$, but you can replace coefficients by their residues, mod $3$, hence, with that interpretation, you can get \begin{align*} &x^3-3x^2+2x-1 \qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\;\, \\[4pt] \equiv\;&x^3-0x^2+2x-1\;(\text{mod}\;3)\\[4pt] \equiv\;&x^3+2x-1\;(\text{mod}\;3)\\[4pt] \equiv\;&x^3+2x+2\;(\text{mod}\;3)\\[4pt] \end{align*}