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Since ACF is complete, $\mathbb{Q}^{\text{alg}}$ is elementary equivalent to $\mathbb{C}$, and by Ax-Kochen $\mathbb{Q}^{\text{alg}}[[a]]$ is elementary equivalent to $\mathbb{C}[[a]]$. But how should I show that $\mathbb{Q}^{\text{alg}}[[a,b]] $ is not elementary equivalent to $\mathbb{C}[[a,b]]$ and also $\mathbb{Q}^{\text{alg}}[a]$ is not elementary equivalent to $\mathbb{C}[a] $?

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  • $\begingroup$ Did you really mean to ask about $\mathbb{Q}^{\text{alg}}[a]$ and $\mathbb{C}[a,b]$? Did you mean $\mathbb{C}[a]$? $\endgroup$ – Alex Kruckman Apr 23 at 15:59
  • $\begingroup$ Yes, thanks I didn't know how to write them...so sorry. $\endgroup$ – user297564 Apr 23 at 17:09
  • $\begingroup$ Oh, I wasn't complaining about the symbols (though I'm glad you've learned how to use \mathbb!). I was asking about the number of variables: [a] vs [a,b]. $\endgroup$ – Alex Kruckman Apr 23 at 17:19
  • $\begingroup$ Oh! It was wrong! Edited! $\endgroup$ – user297564 Apr 23 at 19:03
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My go-to reference for these kinds of questions is the book Model Theoretic Algebra by Jensen and Lenzing.

The question of how to prove that $\mathbb{Q}^{\text{alg}}[a]$ and $\mathbb{C}[a]$ are not elementarily equivalent has been asked before on this site. It is Example 3.12 in Jensen and Lenzing, and it is proceeded by a full proof, which I summarized in my answer to the linked question.

Remark 3.39 in Jensen and Lenzing states that $\mathbb{Q}^{\text{alg}}((x_1,\dots,x_n))$ and $\mathbb{C}((x_1,\dots,x_n))$ are not elementarily equivalent for any $n>1$. They do not prove this, but give a reference to the paper Indécidabilité de la théorie des anneaux de séries formelles à plusieurs indéterminées by Françoise Delon. The fact that $\mathbb{Q}^{\text{alg}}[[a,b]]$ and $\mathbb{C}[[a,b]]$ are not elementarily equivalent follows immediately, since the quotient fields $\mathbb{Q}^{\text{alg}}((a,b))$ and $\mathbb{C}((a,b))$ are interpretable in the power series rings.

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