4
$\begingroup$

I think I am finally beginning to understand tensor products of algebras, and I could use a reality check. If I am understanding correctly, then $k[x]\otimes_k k[x]$ is ring-isomorphic to $k[x,y]$ by the map $x^i\otimes x^j\mapsto x^iy^j$. Is this right?

  • 0
    Related: https://math.stackexchange.com/questions/1758342016-10-01

2 Answers 2

10

Yes, that's right. The tensor product over $k$ is the coproduct in the category of commutative $k$-algebras, and the polynomial ring in $n$ variables over $k$ is the free commutative $k$-algebra on $n$ elements. So the coproduct of the free $k$-algebra on one object and itself can only be the free $k$-algebra on two objects by an examination of the functor it represents.

Geometrically, the tensor product is the product in the category of affine schemes over $k$, and the geometric fact being described here is that the product of $\mathbb{A}^1$ with itself is $\mathbb{A}^2$.

4

I have posted as an answer here of the proof of this isomorphism in the case that $k = \Bbb{C}$. However the proof works I believe for any field and not just for the complex numbers, just replace $\Bbb{C}$ with $k$ in the proof.