$\Lambda_R [\alpha_1,...,\alpha_n]$ is the graded tensor product over $R$ of $\Lambda_R [\alpha_i]$ where the $\alpha_i$'s have odd dimension.

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In Example 3.16 of Hatcher's Algebraic Topology, it is written that the exterior algebra $\Lambda_R [\alpha_1,...,\alpha_n]$ is the graded tensor product over $R$ of the one-variable exterior algebras $\Lambda_R [\alpha_i]$ where the $\alpha_i$'s have odd dimension.

The definition of the exterior algebra is a given by:

The exterior algebra $\Lambda_R [\alpha_1,...,\alpha_n]$ over a commutative ring $R$ with identity is the free $R$-module with basis the finite products $\alpha_{i_1}...\alpha_{i_k}$ ($i_1<...<i_k)$, with associative, distributive multiplication defined by the rules $\alpha_i \alpha_j = - \alpha_j \alpha_i$ for $i \neq j$ and $\alpha_i^2=0$. The empty product of $\alpha_i$'s are allowed, and provides an identity element $1$ in $\Lambda_R[\alpha_1,...,\alpha_n]$. The exterior algebra becomes a (anti)commutative graded ring by specifying odd dimensions for the generators $\alpha_i$.

I understood the definition of the exterior algebra, but I cannot understand the statement in Example 3.16. What is a "graded tensor product"?

Is it just meaning that $\Lambda_R[\alpha_1,...,\alpha_n]$ is isomorphic as $R$-algebras, to the tensor product $\Lambda_R[\alpha_1] \otimes _R .... \otimes _R \Lambda_R[\alpha_n]$?

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Let $A=\bigoplus_{n=0}^\infty A_n$ and $B=\bigoplus_{n=0}^\infty B_n$ be graded algebras over the ring $R$. We define $C=A\otimes_R B$ as the vector space with grading $$C_n=\bigoplus_{k=0}^n A_k\otimes_R B_{n-k}.$$ There are a couple of ways to make $C$ into a graded $R$ algebra. Here is the less obvious way. Define, for $a\in A_k$, $b\in B_l$, $a'\in A_r$ and $b'\in B_s$, $$(a\otimes b)(a'\otimes b')=(-1)^{lr}(aa')\otimes (bb')\in A_{k+r}\otimes_R B_{l+s}\subseteq C_{k+l+r+s}.$$

If we define graded tensor product this way, then we can extend to products of three or more graded algebras with no difficulty. If we consider $R[\alpha_i]$ where $\alpha_i^2=0$ and $\alpha_i$ is in dimension $1$, then the sign rule enforces the relation $\alpha_i\alpha_j=-\alpha_j\alpha_i$ in the tensor product $R[\alpha_1]\otimes_R\cdots\otimes_RR[\alpha_n]$.