Tensor product of polynomial ring with itself I have two questions with regards to taking tensor products of polynomial rings.
First, when I thought about $$\mathbb{F}[X] \otimes_\mathbb{F} \mathbb{F}[X],$$ I thought, wrongly of course, that it is $$\cong \mathbb{F}[X].$$ I want to intuitively understand why this is wrong. I thought the isomorphism is just $$f(x) \otimes g(x) \mapsto f(x)g(x).$$ The problem here is that you can pull-through not just scalars, but even $x$, i.e. under this map, $xf \otimes g = f \otimes xg$. I am only taking the tensor product over $\mathbb{F}$, so this is wrong. Is my reasoning correct here?
Second, is it true that $$\mathbb{F}[X] \otimes_{\mathbb{F}[X]} \mathbb{F}[X] \cong \mathbb{F}[X]?$$
 A: The issue indeed is that you can't pull non-scalars across the tensor symbol (without some other theorem that might apply in certain scenarios, like what @KCd mentions).
Here's a way to see that $\mathbb{F}[X]\otimes_{\mathbb{F}}\mathbb{F}[X]$ can't possibly be isomorphic to $\mathbb{F}[X]$:

*

*First, let's define an $\mathbb{F}$-bilinear function $\mathbb{F}[X]\times \mathbb{F}[X] \to \mathbb{F}[X_1,X_2]$ with $(X^m,X^n)\mapsto X_1^mX_2^n$ for all $m,n\in\{0,1,2,\dots\}$ (verification of well-definedness and bilinearity left to the reader), and hence there is a corresponding $\mathbb{F}$-linear map $\mathbb{F}[X]\otimes_{\mathbb{F}} \mathbb{F}[X] \to \mathbb{F}[X_1,X_2]$ by the universal property of tensor products.  The map is certainly surjective.  For injectivity, suppose $\sum_{i,j} c_{ij} X^i\otimes X^j$ were in the kernel.  The image is $\sum_{i,j} c_{ij} X_1^iX_2^j$, and since polynomials let us read off coefficients, we see each $c_{ij}$ is $0$.
Hence the map is an isomorphism.


*Second, if there were an isomorphism $\mathbb{F}[X]\otimes_{\mathbb{F}}\mathbb{F}[X]\cong \mathbb{F}[X]$, by composing isomorphisms we would have an isomorphism $\mathbb{F}[X_1,X_2]\cong\mathbb{F}[X]$.  Assuming $\mathbb{F}$ is a field (since you're using that letter) or at least a Noetherian ring, then by either using transcendence degree of the fraction field or Krull dimension, we get a contradiction.
It is true that $\mathbb{F}[X]\otimes_{\mathbb{F}[X]}\mathbb{F}[X]\cong \mathbb{F}[X]$, and more generally $R\otimes_RR\cong R$ for any (unital) ring whatsoever, commutative or noncommutative.  By going through the same universal property of tensor products, we can define an $R$-linear map $R\otimes_R R\to R$ such that $1\otimes 1\mapsto 1$, and it is an isomorphism.  Concretely, the isomorphism has the formula $\sum_{i,j} a_i\otimes b_j \mapsto \sum_{i,j}a_ib_j$.
