A generic 4-torus${}=\frac{C^2}{\Lambda}$ is simple for $\Lambda$ a rank $4$ lattice in $C^2$. Let $\Lambda$ be a generic rank 4 lattice in $C^2$ where $C$ is the complex number. Then $A=\frac{C^2}{\Lambda}$ is generically non-isogenous to a product of elliptic curves.
$\textbf{Q:}$ How do I see generic lattice $\Lambda$ cannot be split into $\Lambda_1\oplus\Lambda_2?$ (In other words, the generic $4$-torus is not isogenous to the product of $2-$tori as an abelian variety.) 
 A: A generic $\mathbb C^2 / \Lambda$ is not even isomorphic to a product $(\mathbb C / \Lambda_1)\times(\mathbb C / \Lambda_2)$ as a complex manifold (let alone being isogenous as a torus with group action).
One way to see that these spaces are not isomorphic as complex manifolds is as follows. Let $X$ be the relevant complex manifold, and consider the second cohomology group $H^2(X, \mathbb C)$. By Hodge decomposition, this decomposes as
$$ H^2(X, \mathbb C)\cong H^{2,0}(X) \oplus H^{1,1}(X) \oplus H^{0,2}(X).$$
Furtherfore, the complex vector space $H^2(X, \mathbb C)$ contains within it the integral lattice $$H^2(X, \mathbb Z) \subset H^2(X, \mathbb C)$$
which is spanned by the Poincare duals of the 2-cycles in $X$.
Now consider the intersection
$$ H^2(X, \mathbb Z) \cap H^{1,1}(X) \subset H^2(X, \mathbb C).$$
This intersection $H^2(X, \mathbb Z) \cap H^{1,1}(X)$ is spanned by the Poincare duals of the 2-cycles with the special property that they can be represented as holomorphic submanifolds of $X$ - see this article on the Lefshetz (1,1) theorem, for example.
We can use the rank of $H^2(X, \mathbb Z) \cap H^{1,1}(X) $ to distinguish a generic complex-two-torus from a product of elliptic curves:


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*For $X$ equal to a generic $\mathbb C^2 / \Lambda$, $H^2(X, \mathbb Z) \cap H^{1,1}(X)$ is trivial. It's easy to see this explicitly using coordinates. Indeed, if $(x_1, x_2, x_3, x_4)$ are real coordinates chosen so that $\Lambda$ is generated by the unit vectors in these coordinates, then the Poincare duals of 2-cycles in $X$ are simply the $dx_i \wedge dx_j$'s; $H^2(X, \mathbb Z)$ is the $\mathbb Z$-module generated by the $dx_i \wedge dx_j$. Now let's go to holomorphic coordinates, $z = \sum_{i=1}^4 a_i x_i$ and $w = \sum_{i=1}^4 b_i x_i$, where the coefficients $a_i, b_i \in \mathbb C$ are chosen generically (because $X = \mathbb C^2 / \Lambda$ is supposed to be generic). Then $H^{1,1}(X, \mathbb C)$ is the $\mathbb C$-vector space spanned by $dz \wedge d\bar z$, $dz \wedge d\bar w$, $dw \wedge d\bar z$, $dw \wedge d\bar w$. The intersection $H^{2}(X,\mathbb Z) \cap H^{1,1}(X)$ is then the intersection between the rank-six $\mathbb Z$-lattice $H^2(X, \mathbb Z)$ and the complex four-plane $H^{1,1}(X)$ within the six-dimensional complex vector space $H^2(X, \mathbb C)$. With the $a_i$ and $b_i$ chosen generically, the four-plane $H^{1,1}(X)$ will not intersect any of the lattice points in $H^2(X, \mathbb Z)$ other than the zero point.

*For $X$ equal to $(\mathbb C / \Lambda_1) \oplus (\mathbb C / \Lambda_2)$, the intersection $H^{2}(X,\mathbb Z) \cap H^{1,1}(X)$ is a $\mathbb Z$-module of at least rank two: it contains the Poincare duals of $\{ 0 \} \times (\mathbb C / \Lambda_2 )$and $(\mathbb C / \Lambda_1 ) \times \{ 0 \}$, which are clearly holomorphic submanifolds of $X$. If you wish, you can write out the generators explicitly in coordinates as follows. If $x_1, x_2$ (resp. $y_1, y_2$) are the real coordinates on $\mathbb C / \Lambda_1$ (resp. $\mathbb C / \Lambda_2$) with respect to which the lattice generators are given by unit vectors, then $dx_1 \wedge dx_2$, $dy_1 \wedge dy_2$ and the $dx_i \wedge dy_j$'s are the integral generators of $H^2(X, \mathbb Z)$. Meanwhile, if $z = a_1 x_1 + a_2 x_2$ and $w = b_1 y_1 + b_2 y_2$ are the complex coordinates, then it's clear that $dx_1 \wedge dx_2 $ is in $H^{1,1}(X)$ (being a multiple of $dz \wedge d\bar z$), and $dy_1 \wedge dy_2$ is in $H^{1,1}(X)$ (being a multiple of $dw \wedge d\bar w$). So $H^{2}(X,\mathbb Z) \cap H^{1,1}(X)$ is non-trivial, as we have shown that it contains $dx_1 \wedge dx_2$ and $dy_1 \wedge dy_2$.
