Let $A$ be a finitely generated $K$-algebra which has no zero divisors. Here $K$ is a field of characteristic $0$. Let $K\subset L$ an algebraic field extension. Now let $f: L\to E$ and $g: \textrm{Quot}(A)\to E$ be two homomorphisms to another field $E$. The universal property of the tensor product gives us a homomorphism $A\otimes_K L \to E$. Is this necessarily injective?
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Is it of any importance that $g$ is $\operatorname{Quot} A → E$ and not simply $A → E$? – k.stm Jun 30 '20 at 09:04
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I also could have required an injective map $A\to E$. – Hans Jun 30 '20 at 09:06
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$ℂ \otimes_ℝ ℂ → ℂ,~x\otimes y ↦ xy$ is not injective, since $1^2 + \mathrm i^2 = 0$.
k.stm
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