If we consider an isometric immersion $\bar{f}:M^2\to\mathbb{R}^3$ which is minimal, is it true that $$ f:M^2\to\mathbb{R}^4 $$ (defined by $\bar{f}$ on the first three coordinates and $0$ on the last one) is still minimal? I tried to prove this but I couldn't find how the mean curvature vector field changes by extending the immersion in a constant manner on one coordinate.
Is there any literature dealing with this question?
Yes because the inclusion $i:\mathbb{R}^3\hookrightarrow{}\mathbb{R}^4$ is Totally Geodesic (this is obvious by the form of the covariant derivative in $\mathbb{R}^n$).
What holds true is the following:
Proposition: Given $f:M^r\to N^k$ minimal immersion and $g:N^k\to P^l$ totally geodesic, then $g\circ f:M^r\to P^l$ is minimal.
Proof: Split the second fundamental form of $M^r\subset P^l$, in the tangent and normal part of $N(\subset P)$. The normal part vanishes by definition of Totally geodesic submanifold, hence, taking the trace, we obtain the mean curvature of $M^r\subset P^l$ equalling the mean curvature of $M^r\subset N^k$ (Note that the covariant derivative of $N$ is just the tangential part of the covariant derivative of $P$). By minimality, the latest is zero and we conclude. $\blacksquare$
As a reference, I suggest Chapter 6 of Riemannian Geometry by do Carmo (note that exercise 3 has the same flavour).