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In a proof that I am studying, the author makes a substitution/ change of variables.

and claims $\mathbb C [x,y]= \mathbb C[x+iy,x-iy]$.

But how can one rigorously show this? I have a few problems with it.

First, is it really the case that these rings are equal, or shouldn't we be more careful and say isomorphic?

Second, when we define a polynomial ring, we don't many any mention of distributing over a variable in the definition. so if we have a variable of the form $x+iy$ there is nothing in our axioms that says we can take an element of the ring, call it $r$, and distribute it over the variable to get $r(x+iy)=rx+iry$.

What is really going on here?

Also, when we define a polynomial ring with some unknowns, we don't specify where the unknowns "live." The unknowns in the ring $\mathbb C[x,y]$ could be anything, but in the ring $\mathbb C[x+iy,x-iy]$ it is almost like we are forcing them to be complex numbers?

violet
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1 Answers1

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The subring of $\mathbb{C}[x, y]$ that is generated by $\mathbb{C}$ and both $x+iy$ and $x-iy$ is all of $\mathbb{C}[x,y]$.

That is: we have a ring $\mathbb{C}[x,y]$ and we ask "what is the smallest subring that contains $\mathbb{C}$ and both $x+iy$ and $x-iy$?". The answer is that the only such ring is all of $\mathbb{C}[x,y]$.

There is another interpretation in terms of ring homomorphisms. Consider the homomorphism $\mathbb{C}[u,v] \rightarrow \mathbb{C}[x,y]$ that sends $u \rightarrow x+iy, v \rightarrow x-iy$. The image is a subring that contains $\mathbb{C}$ and both $x+iy$ and $x-iy$. Therefore, the ring homomorphism is surjective.

David Lui
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