Isomorphism of $F$-algebras Let $K/F$ be a field extension, and consider $R=F[x]/(g(x))$ for some $g\in F[x]$. Then $R$ is an $F$-vector space and an $F$-algebra. Further, $K\otimes_F R$ is an $F$-algebra. I'd like to show that $K\otimes_F R$ is isomorphic to $K[x]/(g(x))$ as an $F$-algebra.
To do this, I start by using the universal property for tensor products. Consider the map $\varphi:K\times R\rightarrow K[x]/(g(x))$ given by $\varphi(k,f(x)+(g(x)))=kf(x)+(g(x))$. This is an $F$-bilinear map, and by the universal property, this induces an $F$-module homomorphism $\Phi:K\otimes_F R\rightarrow K[x]/(g(x))$, where $\varphi=\Phi\circ\iota$, and where $\iota :K\times R\rightarrow K\otimes_F R$ is the natural map. 
Now, it is not hard to see that this map is surjective: since the image of $\varphi$ generates $K[x]/(g(x))$ as an abelian group, we get that $\varphi$ is surjective and so $\Phi$ is  surjective as well.
My problem lies in trying to show that $\Phi$ is an injective $F$-module homomorphism, and that it also preserves multiplication. How do I go about doing this? Do I need to construct a map from $K[x]/(g(x))$ to $K\otimes_F R$? Is this even possible?  
 A: There are many possible ways to approach the problem.
One could use the fact the morphism $\Phi$ is $K$-linear and prove that the $K$-spaces $K \otimes_F R$ and $K[x]/(g(x))$ have the same dimensions.
Nevertheless I prefer a more direct approach that proves an inverse for $\Phi$. 
We start by building a mapping $\iota' \colon K[x] \to K \otimes_F R$ defined by 
$$\iota'(\sum_n a_n x^n) = \sum_n a_n \otimes (x^n + (g(x)))$$ 
which can be proven to be a $F$-linear map (indeed it is a $K$-linear map) whose kernel contain $g(x)$. 
Hence by the universal property of quotients it follows that this map factors through a unique $\iota \colon K[x]/(g(x)) \to K \otimes_F R$, which is the map such that 
$$\iota(\sum_n a_n x^n + (g(x)))=\sum_n a_n \otimes (x^n+(g(x)))\ .$$
An easy computation shows us that 
$$\iota\circ\Phi(\lambda \otimes (x^n+(g(x))))=\iota(\lambda x^n +(g(x)))=\lambda \otimes (x^n+(g(x)))$$
that is $\iota\circ\Phi$ coincides with the identity on a set of generators for $K\otimes_F R$. From this it follows that $\iota\circ\Phi=\text{id}$, hence the injectivity of $\Phi$ (similarly one can prove that $\Phi \circ \iota = \text{id}$).
Hope this helps. 
