Tangent space of a product of algebraic group. I need help with this problem from Shafarevich Basic Algebraic Geometry. 

Let $G$ be an algebraic group and $\Psi:G\times G\rightarrow G$ the
  regular map defined by the group law. Let $T_{e}G$ and be
  $T_{e'}(G\times G)$ the tangent spaces to $G$ and $G\times G$ at their
  respective identity elements. Prove that $T_{e'}(G\times G)=T_{e}G\oplus T_{e}G$ and that $d_e \Psi:T_{e}G\oplus
 T_{e}G\rightarrow T_{e}G$ is given by addition of vectors.

I tried this way but I'm not acquainted with the tensor product. Let $T_{e'}(G\times G)=\left( \frac{M_{e'}}{M_{e'}^2} \right)^{*}$ where $M_{e'}=\{ f\in k[G\times G]\mid f(e')=0 \}$. But $k[G\times G]=k[G]\otimes_k k[G]$ so $M_{e'}=M_{e}\otimes M_{e}$.
Then
$$T_{e'}(G\times G)=\left( \frac{M_{e'}}{M_{e'}^2} \right)^{*} = \left(\frac{M_{e}}{M_{e}^2} \otimes \frac{M_{e}}{M_{e}^2}\right)^{*}=T_e G\oplus T_e G$$.
I'm not sure about the second and the third equalities. Could you help me, please? Thanks!
 A: Using Kahler differentials you may prove a more general lemma valid for arbitrary group schemes. There is a formula for the Kahler differentials of a tensor product: If $A,B$ are commutative unital $k$-algebras it follows there is a canonical isomorphism of $A\otimes_k B$-modules
$$F1.\text{       } \Omega^1_{A\otimes_k B/k}\cong \Omega^1_{A/k}\otimes_k B \oplus A\otimes_k \Omega^1_{B/k}.$$
This is Exercise II.8.3 in Hartshorne's book. Formula $F1$ relies on some basic properties of Kahler differentials that may be found in Hartshorne's book or in the Mastumura book "Commutative ring theory".
If $u: A\rightarrow A_1$and $v: B \rightarrow B_1$ are maps of commutative unital rings and $M$ is an $A$-module, $N$ a $B$-module there is the following formula:
$$M\otimes_k N\otimes_{A\otimes_k B} A_1\otimes_k B_1 \cong$$
$$ (M\otimes_A A_1)\otimes_k (N\otimes_B B_1).$$
If $A_1:=A/I, B_1:=B/J$ we get the formula
$$ M\otimes_k N \otimes_{A\otimes_k B} A/I\otimes_k B/J \cong$$
$$ (M/IM)\otimes_k (N/JN).$$
Hence
$$\Omega^1_{A\otimes_k B/k}\otimes_{A\otimes_k B}(A/I\otimes_k B/J) \cong$$
$$ A/I\otimes_k (\Omega^1_{B/k}\otimes_B B/J)\oplus (\Omega^1_{A/k}\otimes_A A/I)\otimes_k B/J.$$
If $I,J$ are ideals with $A/I\cong B/J\cong k$ we get
$$F2.\text{    }\Omega^1_{A\otimes_k B/k}\otimes_{A\otimes_k B}(A/I\otimes_k B/J) \cong$$
$$\Omega^1_{A/k}\otimes_A k \oplus \Omega^1_{B/k}\otimes_B k.$$
Example 1. If $G:=Spec(A)$ where $A$ is a finitely generated $k$-algebra with $k$ a field and $\mathfrak{m}$ is the maximal ideal corresponding to the identity element $e\in G$, it follows the residue field $\kappa(e)$ satisfies
$$\kappa(e) \cong A/\mathfrak{m} \cong k,$$
since $e$ is a $k$-rational point. Let $B=A$ and $I=J=\mathfrak{m}$ in F2. It follows  there is an isomorphism of $k$-vector spaces
$$\Omega^1_{A\otimes_k A/k}(e,e) \cong \Omega^1_{A/k}(e) \oplus \Omega^1_{A/k}(e).$$
Here $(e,e) \in G\times_k G$ is again a $k$-rational point. Dualizing we get an isomorphism
$$T_{(e,e)}(G\times_k G):= Hom_k(\Omega^1_{A\otimes_k A/k}(e,e) , k) \cong$$
$$ Hom_k(\Omega^1_{A/k}(e) \oplus \Omega^1_{A/k}(e),k)    \cong T_e(G) \oplus T_e(G).$$
Note that in F2 $k$ can be an arbitrary commutative unital ring.
Example 2. If $A:=\mathbb{Z}[x_{ij},t]/(det(x_{ij})t-1)$ and $G:=Spec(A):=GL(n,\mathbb{Z})$, it follows the unit element $I(n)$ is the $\mathbb{Z}$-valued point $\phi: A\rightarrow \mathbb{Z}$ of $G$ defined by $\phi(x_{ij})=0$ if $i\neq j$, $\phi(x_{ii})=1$ and $\phi(t)=1$. Let $\mathfrak{p}:=ker(\phi)$. It follows $I(n) \in G(\mathbb{Z})$ is a $\mathbb{Z}$-valued point (in the language of "representable functors"). F2 gives a formula for the "cotangent space" of the group scheme $G\times_{\mathbb{Z}} G$ at $(I(n),I(n))$ (or the non-closed point $(\mathfrak{p}, \mathfrak{p})$):
$$\Omega^1_{G\times_{\mathbb{Z}} G/\mathbb{Z}}(\mathfrak{p}, \mathfrak{p}) \cong $$
$$\Omega^1_{A/k}\otimes_A \kappa(\mathfrak{p}) \oplus \Omega^1_{A/k}\otimes_A \kappa(\mathfrak{p}).$$
At the level of tangent spaces we get
$$F3.\text{   }T_{(\mathfrak{p}, \mathfrak{p})}(G\times_{\mathbb{Z}} G)\cong T_{\mathfrak{p}}(G) \oplus T_{\mathfrak{p}}(G)$$
Note: The map $\phi:A \rightarrow \mathbb{Z}$ is a surjective map onto an integral domain, hence the kernel $\mathfrak{p}\subseteq A$ is a prime ideal - it is not a maximal ideal. Hence we cannot immediately define $GL(n,\mathbb{Z})$ unless we are willing to work with prime ideals.
You will find a discussion of the "language of representable functors" here:
Equivalence between two definitions of affine algebraic set.
This is a language used when studying more general group schemes such as $GL(n,\mathbb{Z})$ and $GL(n, \mathcal{O}_K)$ with $K$ a number field. You may defined the tangent space of any scheme $X$ at any point $\mathfrak{p}$, closed or non-closed. If you look in the litterature you will find that some authors speak of "distributions" when introducing tangent spaces (Jantzen's book "Representations of algebraic groups"). This is another approach that is much used.
A: You're correct that $k[G\times G]\cong k[G]\otimes_k k[G]$, but the next assertion that you make, $M_{e'}=M_e\otimes M_e$, is not correct (this unfortunately fouls up everything afterwards). For example, if you're working with $\Bbb G_a=\operatorname{Spec} k[t]$, then $k[G\times G]=k[t_1,t_2]$ with maximal ideal at $e'$ $(t_1,t_2)$, while $(t_1)\otimes_k (t_2) = (t_1t_2)$. Instead, you should have $$M_{e'}=(M\otimes k[G]) + (k[G]\otimes M)\subset k[G]\otimes_k k[G].$$
From here, we compute: $M_{e'}^2=M^2\otimes k[G] + M\otimes M + k[G]\otimes M^2$, so applying the third isomorphism theorem to $M\otimes M \subset M_{e'}^2\subset M_{e'}$, we see that $M_{e'}/M_{e'}^2$ is isomorphic to the quotient of $(M\otimes_k k[G]/M) \oplus  (k[G]/M\otimes_k M)$ by $(M^2\otimes_k k[G]/M) \oplus  (k[G]/M\otimes_k M^2)$, where we've upgraded from a sum to a direct sum because the intersection of those submodules is zero. But as $k[G]/M\cong k$ and $M\otimes_k k\cong M$, this is exactly $(M\oplus M)/(M^2\oplus M^2)\cong (M/M^2)\oplus (M/M^2)$. Dualizing, we have the result.
