I have a number of questions which seem linked to me, about basic (?) linear algebra:


Given a field (possibly skew) $K$, and an superfield $L$, one can do linear matrix algebra with coefficients in $L$ (any reference for basic facts about that ?)


So given a square matrix $M\in M\_{n\times n}(L)$, it is left invertible if and only if it is right invertible. What are the known relations between the **left kernel** of $M$ in $L^n$ and its **right kernel** ? 


By left kernel, I mean the space (left-vector space) of rows $\lambda\in L\_{1\times n}$ (one row $n$ columns) such that $\lambda M=(0)$ and by right kernel, I mean the set of $\lambda \in L\_{n\times 1}$ (one column n rows) such that $M\lambda=(0)$.



> **Question 1.** For instance, given $M\in M\_{n\times n}(L)$, if there is a **non-trivial point** of $K^n$ in the left-kernel of $M$, is there also a non-trivial point of $K^n$ in its right kernel?


If $K$ is commutative, I am trying to understand why the notion of dimension of a $K$-vector does not vary if one enlarges $K$. I suppose it is because the fact that a familly of vectors being linearly dependent is expressible by the determinant of a matrix being non-zero, which does not depend on larger $K$ (this must correspond to some kind of $\exists$-quantifier elimination I suppose).



> **Question 2.** What happens if $K$ is skew ? In that case, there is no notion of determinant of a matrix (I mean, is there a notion of skew **determinant** ?). Does the notion of **$K$-dimension** of the $K$-vector $L$ space depends on possible enlargements of $K$ ?


Finally:



> **Question 3.** Why does the **Zariski dimension** of an algebraic subset of $K^n$ ($K$ commutative, non necessarily algebraically closed) does not depend on possible enlargements of $K$ ?



Elementary linear algebra over a (possibly skew) field $K$
Re: question 2, the rank of a free module over any ring $R$ with IBN is well-defined, and invariant under arbitrary extensions of scalars $f : R \to S$ where $S$ also has IBN, because of the straightforward isomorphism


$$R^n \otimes\_R S \cong S^n.$$


In particular, division rings have IBN, as do commutative rings. 


A different and inequivalent question is why the dimension of the kernel of a linear transformation is unchanged by extension of scalars (for a map $f : R\to S$ of division rings). The reason is that a division ring will always be flat as a module over a division ring. 


Re: question 3, it depends on what you mean by "enlarging" a Zariski closed subset of affine space via an extension of scalars $k \to K$. One thing you might mean is taking some finitely generated $k$-algebra and considering first its set of $k$-points and then its set of $K$-points. Then it is *not* true that the dimension is invariant because e.g. the set of $k$-points could be empty while the set of $K$-points is nonempty (consider the variety $x^2 + y^2 = -1$ over $\mathbb{R}$ and then over $\mathbb{C}$). 


