You might be interested also in the theory of generic polynomials. See, for instance, the book "Generic polynomials: Constructive Aspects of the Inverse Galois Problem" (Jensen–Ledet–Yui, Cambridge University Press, 2003).

I certainly agree with your first paragraph, but perhaps less with your second, and in particular with your final sentence. There are groundbreaking papers that do not "solve a problem" but create new mathematical tools and theories. Of course, you clearly stated that this is a barometer that you use for yourself, so it's hard to disagree with that ;-)

If the division rings are finite-dimensional over their center, then of course the left and right dimensions of $\ell$ over $k$ coincide. But indeed, for all finite $m,n > 1$, there exist examples of skew field extensions $\ell$ over $k$ such that the left dimension is $m$ and the right dimension is $n$. (See mathoverflow.net/questions/320526/…)

Simply because it is $2$-dimensional over $k = Z(A)$... (Write $B = k.1 + k.b$.)

Rather than defining $-y$ by a non-equality as you do, it might be easier to define it using the equality $-y := i(y) - 1 = i(y).i(i(y)^{-1})$ (unless $y=1$).

Related: mathoverflow.net/questions/59884/…

@semisimpleton That's very unlikely, these other primes tend to behave very differently. Here is a possible (old) reference for the classical groups: jstor.org/stable/2033424

The first book that comes to mind is the book "Gradings on simple Lie algebras" by Elduque and Kochetov (bookstore.ams.org/surv-189). It doesn't consider Lie superalgebras, though.

What exactly is it that you find unconvincing about his proof? It looks completely accurate to me: just pass to the projectivization of the affine plane. There is indeed the tricky aspect that for non-Desarguesian projective planes, the removal of different lines could result in non-isomorphic affine planes, but this is not relevant for this proof.

@YCor Thanks, I have modified the definition of $G$ (this is indeed what I meant to say).