# $x\otimes 1\neq 1\otimes x$

In Bourbaki, Algèbre 5, section 5, one has $A$ et $B$ two $K$-algebras in an extension $\Omega$ of $K$. It is said that if the morphism $A\otimes_K B\to \Omega$ is injective then $A\cap B=K$. I see the reason: if not there would exist $x\in A\cap B\setminus K$ so that $x\otimes 1=1\otimes x$ which is false.

But why $1\otimes x\neq x\otimes 1$ for $x\notin K$?

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The short answer: if you tensor over $K$, you are only allowed to transport elements of $K$ to the other side of the $\otimes$. – Fredrik Meyer Jan 25 '13 at 13:18
@Fredik: See my comment to Thomas' answer: This is not a proof, only a mnemonic. – Martin Brandenburg Jan 25 '13 at 14:52
There's a nice argument that $A\otimes_K B \to A \otimes_{A \cap B} B$ is an isomorphism, but alas I couldn't see how to infer from that an isomorphism $K \to A \cap B$. :( – Hurkyl Jan 25 '13 at 15:39

For any $K$-module $T$ there is a canonical isomorphism $$\operatorname{Hom}_K(A\otimes B,T)\simeq\operatorname{Bil}_K(A\times B,T).$$ Thus in order to prove that $x\otimes1\neq1\otimes x$ it is enough to produce a bilinear map $\phi:A\times B\rightarrow T$ such that $\phi(x,1)\neq\phi(1,x)$.
Take $T=K$ and let $\lambda\in A^\ast$, $\mu\in B^\ast$ such that $\lambda(x)=0$, $\lambda(1)=1$ and $\mu(x)=1$. Then $\phi(a,b)=\lambda(a)\mu(b)$ works.
I hope you know that if $\{v_i\}$ is a basis of $V$ and $\{w_j\}$ is a basis of $W$, then $\{v_i\otimes w_j\}$ is a basis of $V\otimes W$. Now since $x\notin K$, we can extend $\{1,x\}$ to a basis of $A$ and $B$, respectively. Now as a corollary of the above claim you have in particular that $1\otimes x$ and $x\otimes 1$ are linearly independent. In particular they are not equal.
No, you need some variants of choice in order to prove this. There are models of ZF with nontrivial $K$-algebras $A,B$ ($K$ a field) but $A \otimes_K B$ trivial. – Martin Brandenburg Jan 25 '13 at 14:48
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