I'm reading the book Linear Algebra by Hoffman and Kunze and on page 15 of the 2nd edition, they made an assertion which intrigues me.
Let us observe one final thing about the system $AX = Y$. Suppose the entries of the matrix $A$ and the scalars $y_1,\dots,y_m$ happen to lie in a subfield $F_1$ of the field $F$. If the system of equations $AX = Y$ has a solution with $x_1,\dots,x_n$ in $F$, it has a solution with $x_1,\dots,x_n$ in $F_1$. For, over either field, the condition for the system to have a solution is that certain relations hold between $y_1,\dots,y_m$ in $F_1$ (the relations $z_i = 0$ for $i > r$, above). For example, if $AX =Y$ is a system of linear equations in which the scalars $y_k$ and $A_{ij}$ are real numbers, and if there is a solution in which $x_1,\dots,x_n$ are complex numbers, then there is a solution with $x_1,\dots,x_n$ real numbers.
The converse is obviously true. However, the claim made by the authors seems very mysterious to me (see for instance: $x^2+1=0$ has solutions in $\mathbb C$, but not in $\mathbb R$). Can anyone throw light on this?
To solve the system of linear equations $AX = Y$, we perform a sequence of elementary row operations on the augmented matrix $A' = [A_1\ \cdots\ A_n \ Y]$ to obtain the system $RX = Z$, which is equivalent to the original system.
If $R$ has exactly $r$ non-zero rows ($1 \leq r \leq m$), then in the system $RX = Z$ the latter $m-r$ equations are $$ \begin{align} 0 &= z_{r+1}\\ \vdots &\quad\ \vdots \\ 0 &= z_m. \end{align} $$ The authors have shown that a solution to $AX = Y$ exists if and only if a solution to $RX = Z$ exists if and only if $z_i = 0$ for all $i > r$.
Now, suppose that when we view the system $AX = Y$ over the field $F \supset F_1$ it has a solution, that is, with $x_1,\dots,x_n \in F$. Then, the conditions $z_i = 0$ for all $i > r$ are satisfied, where each $z_i$ is a linear combination of $y_1,\dots,y_m$ with coefficients in $F$. But, since the entries of $A$ and $Y$ lie in the subfield $F_1$ of $F$, each $z_i$ is a linear combination of $y_1,\dots,y_m$ with coefficients in $F_1$. This does not change the fact that each $z_i = 0$, so the condition for a solution of $AX = Y$ to exist over the field $F_1$ is automatically satisfied, namely that $z_i = 0$ for all $i > r$. Hence, if a solution exists over $F$, it also exists over $F_1$.
Regarding your example, it is true that $x^2 + 1 = 0$ has a solution over $\mathbb{C}$ but not over $\mathbb{R}$, but this is not a contradiction of the above conclusion because $x^2 + 1$ is not linear. So, such a relation can never arise among the entries $y_1,\dots,y_m$ in the above problem.
This shows that the assertion that you pointed out is indeed non-trivial.