Are nimbers the largest field* of characteristic 2? The nimbers are defined as follows: $$a+b=\operatorname{mex}(\{a+b'\}\cup\{a'+b\})$$ $$ab=\operatorname{mex}(\{ab'+a'b+a'b'\})$$ Where $a'<a,b'<b$, and $a$ and $b$ are arbitrary ordinals. If yes, is there a proof? If not, what field is larger?
*Technically the nimbers don't form a field because a field's domain is a set, while the nimbers are a proper class
Remarkably, this is true, so long as you're willing to stretch the definition of a field to allow its elements to form a proper class. This is briefly mentioned here, and the definition of "largest" I am using is that
First, suppose $\Bbbk$ is an algebraically closed small field of characteristic $2$, then up to isomorphism it is given by the algebraic closure of $\Bbb F_2(X)$ for $X$ some set of independent variables (this is because every algebraically closed field is uniquely determined up to isomorphism by its characteristic and its transcendence degree over its prime field, which in the case of characteristic $2$ is given by $\Bbb F_2$).
Now, since the "Field" $\mathbf{On}_2$ of nimbers forms a proper class, we can find a set $A$ of algebraically independent elements such that $|A|=|X|$ (otherwise, every subset of cardinality $|X|$ would be algebraically dependent, which would force the cardinality of $\mathbf{On}_2$ to be either countably infinite or of size $|X|$, which would force it to be a set). This allows us to embed $\Bbb F_2(X)\hookrightarrow\mathbf{On}_2$. Since $\mathbf{On}_2$ is algebraically closed, this map will factor through its algebraic closure, giving an embedding $\Bbbk\hookrightarrow\mathbf{On}_2$.
If $\Bbbk$ is not algebraically closed, then it factors into its own algebraic closure $\bar{\Bbbk}$. By the above argument, this gives an embedding $\Bbbk\hookrightarrow\bar{\Bbbk}\hookrightarrow\mathbf{On}_2$.
Therefore, in this sense of the word, it is somehow the "largest field" of characteristic two. However, the answer I linked also mentions that it's not unique with this property. The simplest counterexample is $\mathbf{On}_2(x)$ of rational functions with nimber coefficients. Since any field of characteristic $2$ embeds into $\mathbf{On}_2$, it will also embed into $\mathbf{On}_2(x)$. However, this field is not algebraically closed (you can't solve $t^2-x=0$ for $t$).
If you assume global choice (as mentioned by another answer on the MO question I linked), the Field of nimbers is the unique algebraically closed large Field of characteristic $2$, so perhaps in this sense, it is the nicest "largest field" of characteristic $2$.