Putting the scalar field on the lattice was pretty stratight forward. This is not so for fermions even in the simplest case, i.e. free Dirac fields. Let's again start with the Euclidean continuum theory
$$S_E = \int \d^D x\sum_\alpha^{N_f} \bar\psi_\alpha(x)\bigg(\gamma_\mu\partial_\mu+m\bigg)\psi_\alpha(x),$$
where $\alpha$ is a flavour index and Dirac $\gamma$-matrices satisfy the Clifford algebra $\{\gamma_\mu, \gamma_\nu\} = 2\delta_{\mu\nu}$. Then we define a (fromal) mapping to dimensionless lattice variables similar to the scalar field theory through
$$\begin{eqnarray} x &\rightarrow& an, \\ m &\rightarrow& a^{-1} \hat m, \\ \psi_\alpha(x) &\rightarrow& a^{(1-D)/2}\chi_{\alpha, n}, \\ \partial &\rightarrow& a^{-1}\hat\partial, \\ \int\d^D x &\rightarrow& a^D \sum_n\end{eqnarray}$$
where the lattice derivative is defined as $\hat\partial_\mu \chi_n = \frac{1}{2}[\chi_{n+\hat\mu} - \chi_{n-\hat\mu}]$. This results in $S_E^{\mathrm{lattice}} = \sum_{nm}\sum_{\alpha\beta}\bar\chi_{\alpha, n} K_{\alpha\beta, nm} \chi_{\beta, m}$ with
$$K_{\alpha\beta, nm} = \frac{1}{2}\sum_\mu (\gamma_\mu)_{\alpha\beta}[\delta_{m, n + \hat\mu} - \delta_{m, n - \hat\mu}] + \hat m \delta_{\alpha\beta}\delta_{mn}.$$

Passing to the Fourier representation we find
$$K_{\alpha\beta, nm} = \int_{-\pi}^{\pi} \dbar D{\hat k} \sum_\mu (i\gamma_\mu)_{\alpha\beta}\sin(\hat k_\mu) + \hat m\delta_{\alpha\beta} e^{i\hat k\cdot(n -m)}.$$
So, the two-point correlation funstion is given by
$$\langle \chi_{\alpha, n}\bar\chi_{\beta, m}\rangle = (K^{-1})_{\alpha\beta, nm} = \int_{-\pi}^{\pi} \dbar D{\hat k} \frac{[-i\sum_\mu\gamma_\mu\sin(\hat k_\mu)+\hat m]_{\alpha\beta}e^{i\hat k\cdot(n-m)}}{\sum_\mu\sin^2(\hat k_\mu)+\hat m^2}.$$
Taking $a\to 0$ while keeping dimensionful variables on the left-hand side of the lattice mapping constant, we find
$$\langle \psi_\alpha(x)\bar\psi_\beta(y)\rangle = \lim_{a\to 0}\int_{-\frac\pi a}^{\frac\pi a} \dbar D k \frac{[-i\sum_\mu\frac{1}{a}\sin(k_\mu a)\gamma_\mu + m]_{\alpha\beta} e^{ik\cdot(x-y)}}{\sum_\mu\frac{1}{a^2}\sin^2(k_\mu a)+ m ^2}.$$
Now, we could simply replace $\frac{1}{a}\sin(k_\mu a)$ with $k_\mu$ in the limit $a\to 0$ if the only zero of the sine-function ould be at the origin. However, $\sin(k_\mu a)$ also vanishes at the edges of the Brillouin zone, $k_\mu = \pm \pi/a$, that is in the ultraviolet region of the continuum theory. So, for each discrete degrees of freedom that we started with, we get two continuum degrees of freedom, one at low energies and the other at high energies, as we go to the continuum limit. this is the so-called fermion doubling problem.

I'll look at two popular ways of getting around this problem in the second part of this note.