Let $R$ be a commutative ring with unity, $M,N$ are $R$-modules, $T(M),T(N)$ their tensor algebras over $R$. In page $571$, D. Eisenbud, Commutative Algebra with a View Toward Algebraic Geometry, the author states the following:
$T(M \oplus N) = T(M) \otimes T(N) \otimes T(M) \otimes T(N) \otimes...$, the set of finite sums of tensor products of elements all but finitely many of which are $1 \in R = T_0(M) = T_0(N)$.
I have tried and could not explicit the equality, I believe that it should be just deduced from the commutativity of tensor product and direct sum. Moreover, in case of graded modules, $M = \oplus M_i$, we can grade the tensor algebra by: $$T(M)_i = \oplus_{j_1+...+j_n = i}M_{j_1} \otimes... \otimes M_{j_n},$$ and the symmetric algebra $S(M)$ is the quotient algebra defined by the relations: $$ab = (-1)^{\left|a \right|\left|b \right|}ba, \ a^2 = 0 \ \text{if} \ \left|a \right| \ \text{odd}.$$ Then from the equality $T(M \oplus N) = T(M) \otimes T(N) \otimes T(M) \otimes T(N) \otimes...$, the author again deduce that $S(M \oplus N) = S(M) \otimes S(N)$. I am not really satisfied with his argument so perhaps it is better if someone could help me write down the isomorphism.