# $k[x]\otimes_k k[x]\cong k[x,y]$?

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?

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Yes, that is right. a⊗b * c⊗d is defined to be ac⊗bd, which works out just fine, and the rest of the check is just "bilinearity" over and over. Q[i]⊗Q[√2] = Q[i,√2], but Q[i]⊗Q[i] is a little funny since you have two copies of i. I think it is just Q[i]×Q[i], but that is shown by fiddling with polynomial factorizations. – Jack Schmidt Jun 14 '12 at 22:34

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 $\mathbb{A}^2$.
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.