Are matrices always diagonalizable in the complex field? As the title says, considering the roots of the characteristic polynomial are the eigenvalues of the matrix, and in the complex field every polynomial has $n$ different roots where $n$ is the polynomial degree, this means every eigenvalue on the complex field is a simple eigenvalue, so every matrix in $\mathbb C$ will be diagonalizable. Am I wrong?
 A: No, not every matrix over $\Bbb C$ is diagonalizable. Indeed, the standard example $\begin{pmatrix} 0&1\\0&0 \end{pmatrix}$ remains non-diagonalizable over the complex numbers.
Here's where your argument breaks down. You've correctly argued that every $n\times n$ matrix over $\Bbb C$ has $n$ eigenvalues counting multiplicity. In other words, the algebraic multiplicities of the eigenvalues add to $n$. However, the geometric multiplicities are not necessarily the same as the algebraic multiplicities (when the algebraic multiplicity is $k$ then the geometric multiplicity can be any integer from $1$ to $k$). And diagonalizability is equivalent to the sum of the geometric multiplicities being $n$, not the sum of the algebraic multiplicities.
Looking at your exact wording, you have said "every polynomials have $n$ different roots"; that's a mistake—complex polynomials of degree $n$ have $n$ roots counting multiplicity, but nothing stops those roots from coinciding sometimes—consider $(x-i)^n$ for example.
