Grading and tensor product I read that if $B=\bigoplus_k B_k$ is a graded $A$-algebra and $C$ is another $A$-algebra (not graded) then $B\otimes_A C$ is graded by $B_k\otimes_A C$. I can see that 
$ B\otimes_A C=\bigoplus B_k\otimes_A C$ but I don't understand why we can consider the $B_k\otimes_A C$ as subset of $B\otimes_A C$ because we do not necessarily have
$$ B_k\otimes_A C\hookrightarrow B\otimes_A C $$
if $C$ is not flat. 
 A: This is a good question. The key point is that $B_k$ is a direct summand of $B$. In this case, we don't need $C$ to be flat. The only thing we need is that $\otimes $ is an additive functor.

Suppose $0 \to A \to B \to C \to 0$ is a split exact sequence (i.e., we have a splitting map, either $B\to A$ or $C\to B$, so that $B\simeq A\oplus C$). Suppose $T$ is an additive functor. Then $0\to TA \to TB \to TC \to 0$ is also split exact.

Proof.
Let $\iota_A:A\to B$, $\iota_C:C\to B$, $\pi_A:B\to A$, and $\pi_C:B\to C$ be the inclusions and projections. We know $\pi_A\iota_A =1_A$, $\pi_C\iota_C=1_C$, and $\iota_A\pi_A + \iota_C\pi_C = 1_B$. These are the properties required to show that $B\simeq A\oplus C$. (Specifically that the diagram: 
$$ \require{AMScd}
\begin{CD}
0 @>>> A @>\iota_A>> B @>\pi_C>> C @>>> 0 \\
@.@V1_AVV@V(\pi_A,\pi_C)VV@V1_CVV@.\\
0 @>>> A @>>> A\oplus C @>>> C @>>> 0
\end{CD}$$
commutes and the middle vertical arrow is an isomorphism.)
Now apply $T$ to these equations. We have $(T\pi_A)(T\iota_A)=1_{TA}$, $(T\pi_C)(T\iota_C)=1_{TC}$, and 
$$(T\iota_A)(T\pi_A)+(T\iota_C)(T\pi_C)=1_{TB}. \quad \text{(this is where additivity is used)}$$
This implies that the following diagram commutes and the middle vertical arrow is an isomorphism:
$$ \require{AMScd}
\begin{CD}
0 @>>> TA @>T\iota_A>> TB @>T\pi_C>> TC @>>> 0 \\
@.@V1_{TA}VV@V(T\pi_A,T\pi_C)VV@V1_{TC}VV@.\\
0 @>>> TA @>>> TA\oplus TC @>>> TC @>>> 0
\end{CD}$$
Then exactness of the bottom sequence yields exactness of the top sequence.
A: Tensoring is a left-adjoint functor and thus commutes with colimits (direct sum in this case). So, we have
$$B \otimes_A C = \left(\bigoplus B_k \right) \otimes_A C \cong \bigoplus \left(B_k \otimes_A C \right).$$
Now we have a natural inclusion
$$B_k \otimes_A C \hookrightarrow \bigoplus \left(B_k \otimes_A C \right) \cong B \otimes_A C.$$
So, we can identify each summand $B_k \otimes_A C$ with a copy inside $B \otimes_A C$.
