$\mathbb{Q}^{alg}[[a,b]] $ is not elementary equivalent to $\mathbb{C}[[a,b]]$, and the same for $\mathbb{Q}^{alg}[a]$ and $\mathbb{C}[a]$?

Question

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] $?

Answer 1

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.