How tho show that $ \mathbb{F}_2[x_1] \otimes \mathbb{F}_2[x_2] \otimes \cdots \otimes \mathbb{F}_2[x_k] \cong \mathbb{F}_2[x_1, \ldots, x_k]$ My question is:
(i) Show that $$ \mathbb{F}_2[x_1] \otimes  \mathbb{F}_2[x_2] \otimes \cdots \otimes  \mathbb{F}_2[x_k] \cong  \mathbb{F}_2[x_1, \ldots, x_k]$$
(ii) If $ \mathbb{F}_2[x_1] \otimes  \mathbb{F}_2[x_2] \cong \mathbb{F}_2[x_1, x_2]$ then I know $$f_1(x_1) \otimes f_2(x_2) = f_1(x_1)f_2(x_2)$$
because  $f_1(x_1) \otimes f_2(x_2) \neq f_2(x_2) \otimes f_1(x_1),$ so $f_1(x_1)f_2(x_2) \neq f_2(x_2)f_1(x_1)\quad (*)$
but $\mathbb{F}_2[x_1, x_2]$ is a commutative algebra then $f_1(x_1)f_2(x_2) = f_2(x_2)f_1(x_1),$ this contradiction with $(*)??$
I don't understand. Can you explain for me? Thank you very much!
 A: I think your issue is that you are assuming that the product in $\mathbb{F}_2[x_1]\otimes\cdots\otimes\mathbb{F}_2[x_k]$ is 'tensor'. Else, I don't see how you arrive at your desired contradiction?
To prove your actual claim you can proceed in several ways. But, the one you mentioned is probably the most correct. Namely, $\mathbb{F}_2[x_1,x_2]$ is the free (commutative) $\mathbb{F}_2$-algebra on two generators, and so to define a map $\mathbb{F}_2[x,y]\to A$, where $A$ is an $\mathbb{F}_2$-algebra, is equivalent to saying where $x$ and $y$ go. So, let $\mathbb{F}_2[x,y]\to\mathbb{F}_2[x]\otimes\mathbb{F}_2[y]$ be defined by sending $x\mapsto x_1\otimes 1$ and $y\mapsto 1\otimes x_2$. Check that this defines an algebra map. It's evidently injective for precisely the reason that you claim. Namely, if $f(x,y)\mapsto 0$, let's say with $\displaystyle f=\sum_{i,j}a_{ij}x^i y^j$ then you'd have that $\displaystyle \sum_{i,j}a_{i,j}(x_1^i\otimes x_2^j)=0$. But, the set $\{x^i\otimes y^j:(i,j)\in\mathbb{N}^2\}$ is linearly independent over $\mathbb{F}_2$ (why?) and thus $a_{ij}=0$. To see it's surjective it suffices to note that you hit monomials (why?) and you evidently do (why?).
PS, I didn't harp on it here, but you shoudl really specify what you're tensoring over. There are only two real natural choices here: $\mathbb{F}_2$ and $\mathbb{Z}$? Does it make a difference which you tensor over?
A: Note that in the tensor product $A\otimes B$ for commutative rings $A, B$, then $a\otimes b \neq b\otimes a$ because the latter doesn't make sense for $a \in A, b \in B$. In the same way, $f_2(x_2)\otimes f_1(x_1)$ doesn't make sense as an element of $\Bbb F_2[x_1]\otimes \Bbb F_2[x_2]$. We do have $\Bbb F_2[x_1]\otimes \Bbb F_2[x_2] \cong \Bbb F_2[x_1, x_2]$, because we can set up an explicit isomorphism.
Let the homomorphism $\varphi:\Bbb F_2[x_1, x_2]\to \Bbb F_2[x_1]\otimes \Bbb F_2[x_2]$ he defined the following way: for  any monomial $x_1^mx_2^n \in \Bbb F_2[x_1, x_2]$ we set $\varphi(x_1^mx_2^n) = x_1^m\otimes x_2^n$. For any polynomial, we extend by linearity.
Why is this an isomorphism? Well, we have an inverse, given by $\psi:\Bbb F_2[x_1]\otimes \Bbb F_2[x_2]\to  F_2[x_1, x_2]$, but before I define it, let's review what an element of $\Bbb F_2[x_1]\otimes \Bbb F_2[x_2]$ looks like. It is a finite sum of the form
$$
f_1(x_1)\otimes g_1(x_2) + f_2(x_1)\otimes g_2(x_2) + \cdots + f_k(x_1) \otimes g_k(x_2)
$$
Because of the bilinearity of the tensor product, we may assume that all the $f_i$ and $g_i$ are monomials. Thus the general element may take a form that looks like this instead:
$$
x_1^{m_1}\otimes x_2^{n_1} + \cdots + x_1^{m_k}\otimes x_2^{n_k}
$$
And at this stage I hope it's pretty clear what $\psi$ ought to be: We define $\psi(x_1^m\otimes x_2^n) = x_1^mx_2^n$, and extend by linearity on the general form above.
Note that the tensor product of commutative rings is a commutative ring, it's just that you checked it on the wrong elements. You shouldn't compare $f(x_1)\otimes g(x_2)$ to $g(x_2)\otimes f(x_1)$. You should compare $$(f_1(x_1)\otimes g_1(x_2))\cdot (f_2(x_1)\otimes g_2(x_2))$$ to $$(f_2(x_1)\otimes g_2(x_2))\cdot (f_1(x_1)\otimes g_1(x_2))$$Or, more generally, insert one of the general forms in each of the brackets above. But in the end, that reduces to this single comparison.
