Let $K$ be a field and let $L/K$ be a finite, normal and seperable field extension over $K$. I need to show that $$L\otimes_K L \cong \prod_{\sigma\in \text{Gal(L/K)}}L^\sigma$$ as an $L$-algebra. Any ideas on how to begin?
I tried showing that the map defined by $l \oplus l_0 \mapsto (\sigma_1(l)l_0, \dots, \sigma_n(l)l_0)$ was injective to no avail. I saw a proof for this in the Galois decent section of Bourbaki's Elements of Mathematics where they tensor $L$ with any $L$-vector space $M$, but I was hoping there was possibly a more elementary proof for the more specific case above.
I think what you are trying to prove is that if $L / K$ is a finite Galois extension, with Galois group $G$, then the map, $$ L \otimes_K L \rightarrow \prod_{\sigma \in G} L, $$ defined by $$ a \otimes b \mapsto (a\sigma(b))_\sigma $$ is an isomorphism of $L$-algebras, where $L \otimes_K L$ is an $L$-algebra via the left factor and $\prod_{\sigma \in G} L$ is an $L$-algebra diagonally.
As indicated in the comments above, one can use the primitive element theorem to prove this.
Let $\alpha \in L$ be a primitive element for $L / K$, and $f(x)$ the minimum polynomial of $\alpha$.
There is an isomorphism $K[x] / f(x) \xrightarrow{\sim} L$ defined by $x \mapsto \alpha$.
If $b \in L$, with $b = \sum_i b_i \alpha_i \in L$, let $p_b(x) = \sum_i b_i x_i \in K[x] / f(x)$ be the corresponding element.
Then by the Chinese Reminder Theorem we have a chain of isomorphisms, $$ L \otimes_K L \xrightarrow{\sim} L \otimes_K ( K[x] / f(x) ) \xrightarrow{\sim} L[x] / f(x) = L[x] / \prod_{\sigma \in G}(x - \sigma(\alpha)) \xrightarrow{\sim} \prod_{\sigma \in G} L $$ If we follow the chain we can see that the composition is exactly the map above: $$ a \otimes b \mapsto a \otimes p_b(x) \mapsto a p_b(x) \mapsto (ap_b(\sigma(\alpha)))_{\sigma} = (a\sigma(p_b(\alpha)))_{\sigma} = (a\sigma(b))_{\sigma}. $$