# Isomorphism of tensor product+field extension

Let $k$ be a field, $f(x)\in k[x]$ be an irreducible polynomial over $k$, and $\alpha$ be a root of $f$. If $L$ is a field extension to $k$, what does $k(\alpha)\otimes_k L$ isomorphic to?

I'm trying to use the fact that $k(\alpha)\otimes_k L$ isomorphic to $k[x]/(f(x))\otimes_k L$ to drive some formulas, but I get struck at the beginning. Any help?

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 I think you should have $k[X]/(f)\otimes_K L \simeq L[X]/(f)$, right? In this case, what this ring looks like will depend on how $f$ factors over $L$. In particular, if all of $f$'s roots are in $L$, you'll have a direct sum of copies of $L$. – vgty6h7uij May 23 '12 at 11:30

I'll give you a hint. So, you have the exact sequence

$$0\to (f(x))\xrightarrow{i} k[x]\xrightarrow{\pi} k(\alpha)\to0$$

Tensoring you get the exact sequence

$$L\otimes_k (f(x))\xrightarrow{1\otimes i} L\otimes_k k[x]\xrightarrow{1\otimes \pi} L\otimes_k k(\alpha)\to0$$

Thus, $L\otimes_k k(\alpha)\cong (L\otimes_k k[x])/\ker(1\otimes \pi)$. But, by exactness $\ker(1\otimes \pi)=\text{im}(1\otimes i)$. It's trivial that $L\otimes_k k[x]=L[x]$ and that when you consider this isomorphism one finds that $\text{im}(1\otimes i)=(f(x))$.

I leave the rest to you.

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Following Alex's hint, you can prove a more general statement: let $\phi \colon A \to B$ be a ring homomorphism and let $I \subseteq A[x_1, \dots, x_n]$ be the ideal generated by the polynomials $f_1, \dots, f_r \in A[x_1, \dots, x_r]$. Then $$\frac{A[x_1, \dots, x_n]}{I} \otimes_A B \simeq \frac{B[x_1, \dots, x_n]}{(\bar{f_1}, \dots, \bar{f_r})},$$ where $\bar{f_i} \in B[x_1, \dots, x_n ]$ is the image of $f_i$ after $\phi$.

A once more general statement is the following: if $R \to S$ is a ring homomorphism and $I \subseteq R$ is an ideal, then $$\frac{R}{I} \otimes_R S \simeq \frac{S}{IS}.$$

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