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In Hoffman's Linear Algebra, in last part of section 1.5, it says that if an equation system $AX=Y$, where the entries of $A$ and the entries of $Y$ are in the field $F_1$, has a solution with $x_1,\ldots,x_n$ in $F$ and $F_1 \leq F$, then it has a solution $x_1,\ldots, x_n$ in $F_1$. He gives an argument, but I really don't follow. Can someone explain in other terms, why this happens?

Thanks in advance.

HeMan
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    You should probably state what assumptions you make about $A$ and $Y$ here. Without anything specified, this is false: $ix = 1$ has a solution $x = -i$ in $\Bbb Q(i)$, but no solution in $\Bbb Q$ (which is certainly a subfield of $\Bbb Q(i)$). – Stahl Aug 27 '17 at 02:18

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You're missing the condition: "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 $F$."

Whether we think of the augmented matrix having elements of $F$ or of $F_1 \subseteq F$, the sequence of elementary row operations all take place in that field. At no time do field operations applied to elements of $F_1$ produce any elements outside of $F_1$, particularly, they do not produce elements of $F \smallsetminus F_1$. (Because fields are closed under their operations.) Consequently, if we start with all elements of $F_1$ and reduce to a solution, that solution only contains elements of $F_1$.

Eric Towers
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  • Yes you're right, I omitted those conditions, but then, given what you say here, then how do I get the implication, if $AX=Y$ has a solution in F_1, then it has a solution in F? I really can't see it. – HeMan Aug 27 '17 at 02:32
  • I can see that with elementary row operations, we stay in the field F_1, but how can I use that in order to prove that if AX=Y has a solution in F, then it has a solution in F_1? – HeMan Aug 27 '17 at 02:33
  • If you want to solve $AX = Y$, then you take the augmented matrix $M$ whose columns are the columns of $A$ in the same order followed by the vector $Y$ as the last column. Then $AX = Y$ has a solution if and only if the row reduced echelon form of $M$ has no rows of the form $[0,0,0,\dots,0,c]$, where $c$ is a nonzero constant. So, given that a solution exists over some field, you know that $M$ will not have a row of this form when converted to RREF. – Stahl Aug 27 '17 at 02:41
  • Then you can read off the solutions to $AX = Y$ by looking at $M$ in RREF: some variables will be free and others will be determined by the free variables. For example, if you have a row $[1,0,1,2,0,4]$ in RREF, then $x_1 = 4 - x_3 - 2x_4$, and $x_3$ and $x_4$ here are free. So all your variables are either free or determined by the free variables, and the coefficients of the determining relationship must lie in your subfield $F_1$ by the answer. – Stahl Aug 27 '17 at 02:44
  • Choosing values of the free variables which lie in $F_1$ and computing the dependent variables will give a sum of products of elements in $F_1$, which is again in $F_1$. – Stahl Aug 27 '17 at 02:44
  • @HeMan : In your first comment, you ask about "if $AX = Y$ has a solution in $F_1$, then it has a solution in $F$." But $F_1$ is a subfield of $F$; every element of $F_1$ is also an element of $F$. Consequently, any solution in $F_1$ is a solution in $F$. – Eric Towers Aug 27 '17 at 20:07
  • @HeMan : In your second comment, you ask about the implication in the other direction. Precisely as I answered: $AX = Y$ only has element of $F_1$ in $A$ and $Y$. This is a linear system in $F_1$. We solve it by reduction to a diagonal system using elementary row operations. Elementary row operations take the $F_1$ elements of a row and multipley by $F_1$ constants and add to other $F_1$ rows, producing a result in $F_1$. That is, if all the numbers start in $F_1$, no reduction step produces a number not in $F_1$. – Eric Towers Aug 27 '17 at 20:10