Don't you want to map from a projective object $P \to M$ and to an injective object $M \to I$?

See e.g. [About Euclidean rings][1] prop.7, sorry I can't find a pdf anywhere on the web. Now if someone knows what the Galois group of the maximal unramified extension is, then I'm interested in a reference. [1]: ams.org/mathscinet/search/…

Will demanding that the determinant of the corresponding matrix (when 2 is invertible) help?

@darij: I edited, so now I'm not allowing those kinds of forms. @Skip: Thanks I'll check that out.

One conceptual difference between $\mathbb{Q}(i)$ and $'\mathbb{Q}( \sqrt{2})$ is that a field in which $-1$ is a square cannot be ordered. Since $'\mathbb{Q}( \sqrt{2})$ has real embeddings it can obviously be ordered. There is a somewhat related invariant called the level of of the field, which if I remember correctly is the least number of summands needed to express -1 as a sum of squares. In the cases above the levels are 1 and $\infty$ respectively. If we had chosen $\sqrt{-2}$ instead the level would have been 2.

Pardon my ignorance, what does "2-cell embedded" mean?

Linear Algebra by Hoffman and Kunze is one of my favorite math books! It was my first introduction to multilinear algebra and I highly recommend it.