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I am writing up some stuff on polynomials with arbitrarily many variables over/with coefficients in $\mathbb{C}$. While doing this, I vaguely remembered reading something about 'polynomials with uncountably many indeterminates,' or something along those lines, but can't find anything about it now that I look for it.

My question: is it sensible to say a polynomial may have 'uncountably many indeterminates'? What does this...mean? Are we not just handling some element of a polynomial ring $K[x_1,x_2\dots x_n]$, which obviously has countably many indeterminates?

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    $\begingroup$ a polynomial ring is just a free $k$-algebra over a set $X$ (the variables), if you now choose $X$ to be uncountable you get the ring of polynomials in uncountably many variables, but it is highly nonnoetherian, and hence quite ill behaved. $\endgroup$
    – Enkidu
    Dec 9 '19 at 16:18
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    $\begingroup$ Yes, we have $K[x_i\mid i\in \Bbb R]$, why not? $\endgroup$ Dec 9 '19 at 16:18
  • $\begingroup$ @DietrichBurde you mean $i \in \mathbb{R}$, don't you? And sorry, meant "nonnoetherian"! $\endgroup$
    – Enkidu
    Dec 9 '19 at 16:20
  • $\begingroup$ @Enkidu Ah, sorry, you are right. Uncountably many, not infinitely many. $\endgroup$ Dec 9 '19 at 16:23
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The answer to OP is No, a polynomial can't have infinitely many variables.
It is the polynomial ring that has infinitely many variables, not any polynomial.

By definition, a polynomial has only finitely many non-zero coefficients, hence only finitely many variables will appear.

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