I know this is not a "research Mathematical question", but this is a question that I would ask to my Math colleagues, but no one could give me a readily "yes-no" answer. Unfortunately, this is not a subject my colleagues or me are specialist.

I also posted in MO, but with no answer yet:

I am investigating the relation between polynomial identities and finite-dimensional simple (non-associative) algebras. An interesting question is if the polynomial identities uniquely define the simple finite-dimensional algebra, up to isomorphism.

Limiting to simple-associative algebras over algebraically closed fields, this question is rather trivial, and can be solved using basic PI-theory (Amitsur–Levitzki theorem and the structure of simple finite-dimensional associative algebras). This question was investigated in some context - simple Lie, simple Jordan, simple non-associative algebras, and more recently, simple graded associative algebras.

But Razmyslov, in his book (1994), gave a surprisingly general answer for the question, in the context of algebra with an arbitrary signature. Checking every step of the proof, Razmyslov uses a result from field theory, which for his purposes, can be stated as follows:

Lemma. Let $K$ be any infinite field (of any characteristic) and suppose $L\mid K$ an arbitrary field extension (not necessarily finite). Let $K_0=K(x_i\mid i\in\mathbb{N})$ be the field of fractions of the polynomial ring over $K$ with countable infinite number of variables. Suppose $\delta_1:L\to K_0$ and $\delta_2:L\to K_0$ two $K$-monomorphisms. Then there exist a field extension $K_0\subset M$ and a $K$-automorphism $\sigma$ of $M$ such that $\sigma\circ\delta_1=\delta_2$.

Since I want to use Razmyslov's Theorem in my research, I was advised to check its proof. The Lemma above is really essential in the arguments. So my question is: does anyone knows an exact reference to prove the validity of the statement? Or a counter-example?

Thanks in advance!


In order to avoid having a question without answer in the forum, Eric Wofsey posted an answer here.


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