I am reading about nim operations extended to ordinals (developed by Conway).
In Siegel's Combinatorial Game Theory, Lemma 4.4 (page 440) states that if $\gamma$ is a field where every polynomial of degree $\leq n$ has a root in $\alpha$, then polynomials of degree $\leq n$ and coefficients $< \gamma$ are evaluated normally, that is, with the classic sum and product operations of ordinals.
As $\omega$ is a quadratically closed field (this is in fact a classic game theory result), $\omega \otimes 2 = \omega \cdot 2$ However, I wanted to compute $\omega \otimes 2$ with the recursive definition, that is:
$\alpha \otimes \beta = mex \{ \alpha' \otimes \beta : \alpha \otimes \beta' : \alpha' < \alpha,\beta' < \beta\}$
So I have to know the results of $\omega \otimes 1, \omega \otimes 0$, and $n \otimes 2$ for $n\in \mathbb{N}$.
$\omega \otimes 1 = 1, \omega \otimes 0 = 0$; those are easy.
As $\omega$ is a field, in particular every natural number has an inverse in $\omega$. In fact, $\frac{1}{2}=3 \in \mathbb{N}$. Take $k\in\mathbb{N}$. Then taking $n= \frac{k}{2} = k \otimes 3$, we have that $n \otimes 2 = k \otimes 3 \otimes 2 = k$.
So $n \otimes 2$ is suryective in $\omega$ (and stays in $\omega$ as $\omega$ is a field and is closed under nim product).
So $\omega \otimes 2 = mex \{0,\omega,n : n\in \omega\} = mex\ \omega +1 = \omega + 1$. This is different from $\omega \cdot 2$.
I don't understand why! Did I do something wrong?
Your definition of $\alpha\otimes\beta$ is wrong. The correct definition is $$\alpha\otimes\beta=\mathrm{mex}\{\alpha'\otimes\beta\oplus\alpha\otimes\beta'\oplus \alpha'\otimes\beta':\alpha'<\alpha,\beta'<\beta\}.$$