Filling in some details left out of the comments by @Michael Hardy, @zyx, and myself, suppose $\vec{X} = (X_1, \ldots, X_N)$ where the $X_i, 1 \leq i \leq N,$ are independent $N(\mu, v)$ random variables with known mean $\mu$ and unknown variance $v$. The joint density function is $f_{\vec{X}}(\vec{x}) = (2\pi v)^{-N/2}\exp(-a/v) ~\text{where}~ \vec{x} = (x_1, \ldots, x_N)~ \text{and}~ a = \frac{1}{2} \sum_{i=1}^N (x_i - \mu)^2.$ If $\vec{X}$ is observed to have value $\vec{x}$, the likelihood function is $L(v) = (2\pi v)^{-N/2}\exp(-a/v)$ and it is easy to show that $L(v)$ attains maximum value at $v = \frac{2a}{N} = \frac{1}{N}\sum_{i=1}^N (x_i - \mu)^2$ and so the maximum likelihood estimator (MLE) of $v$ is $\frac{1}{N}\sum_{i=1}^N (x_i - \mu)^2$. We have $ E\left [ \frac{1}{N}\sum_{i=1}^N (X_i - \mu)^2 \right ] = \frac{1}{N}\sum_{i=1}^N E[(X_i - \mu)^2] = \frac{1}{N}\sum_{i=1}^N v = v, $ and thus, contrary to any alleged claims in an unspecified textbook in the possession of Nikos, the MLE for $v$ is unbiased in this instance.
What if $\mu$ is also unknown? The likelihood function is now $ L(\mu, v) = (2\pi v)^{-N/2}\exp\left [ -\frac{1}{2v}\sum_{i=1}^N (x_i - \mu)^2\right ] $ and has a global maximum at $ (\mu, v) = \left (\frac{1}{N}\sum_{i=1}^N x_i, \frac{1}{N}\sum_{i=1}^N \left (x_i -\frac{1}{N}\sum_{i=1}^N x_i \right)^2 \right ) = \left ( \bar{x}, \frac{1}{N}\sum_{i=1}^N (x_i -\bar{x})^2 \right ).$ The MLE for $v$ is thus $\frac{1}{N}\sum_{i=1}^N (x_i -\bar{x})^2$ and is biased since $ E\left [ \frac{1}{N}\sum_{i=1}^N \left (X_i -\frac{1}{N}\sum_{i=1}^N X_i \right)^2 \right ] = \left(\frac{N-1}{N}\right)v $ but, as noted by Nikos's textbook, the MLE for $v$ is asymptotically unbiased in the limit as $N \to \infty$. On the other hand, it should be obvious from the above description that $\frac{1}{N-1}\sum_{i=1}^N (x_i -\bar{x})^2$ is an unbiased estimator for $v$ for all $N \geq 2$.