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

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 $$\mathbb{C}[a]$$?

• Did you really mean to ask about $\mathbb{Q}^{\text{alg}}[a]$ and $\mathbb{C}[a,b]$? Did you mean $\mathbb{C}[a]$? – Alex Kruckman Apr 23 at 15:59
• Yes, thanks I didn't know how to write them...so sorry. – user297564 Apr 23 at 17:09
• 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]. – Alex Kruckman Apr 23 at 17:19
• Oh! It was wrong! Edited! – user297564 Apr 23 at 19:03

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.