MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

3 typo

Suppose that the field extension $L/K$ is separable and that $L$ K$is infinite. Let$A$,$B\in M_n(K)$and suppose there exists a matrix$Q\in M_n(L)$is such that$QA=BQ$and which has rank$Q=r$. We want to show there is a matrix$Q'\in M_n(K)$such that$Q'A=BQ'$and which has rank at least$r$. First, by replacing$L$by the subfield of$L$generated over$K$by the coefficients of$Q$if we need to, we can suppose that$L/K$is a finitely generated extension. By using a maximal purely trascendental extension of$K$contained in$L$as an intermediate step, we see that it is enough to consider separately the cases in which (i)$L/K$is purely trascendental or (ii)$L/K$is finite. In case (i), let$S$be a trascendence basis of$L/K$. Since the matrix$Q$has rank$r$, it has an$r\times r$minor$M$with non-zero determinant. As the entries of$Q$are finitely many rational functions in a finite number of elements of the indeterminates$S$, and since the field$K$is infinite, we assign values from$K$to the indeterminates which appear in$Q$in such a way that we obtain a matrix$Q'\in M_n(K)$(ie, we avoid zeroes in denominators) and such that the minor of$Q'$corresponding to$M$still has non-zero determinant. It is clear that$Q'A=BQ'$and that the rank of$Q'$is at least$r$, so we are done in this case. Let us now consider case (ii). Up to enlarging$L$, we can assume that$L/K$is Galois, with Galois group$G$. As before, the matrix$Q$has an$r\times r$minor$M$with non-zero determinant. Suppose the elements of$G$are$g_1=1_G,g_2,\dots,g_n$, and consider the polynomial$f(X_1,\dots,X_n)=\det_M\left(\sum_{i=1}^ng_i(Q)X_i\right)\in L[X_1,\dots,X_n]$; here the elements of$G$act on the matrix$Q$in the obvious way, and$\det_M$denotes the determinant of the minor of its argument corresponding to$M$. Notice that$f$is not the zero polynomial, because the coefficient of$X_1^r$is precisely$\det_MQ\neq0$. Since$L$is infinite and the elements of$G$are algebraically independent (Lang, Algebra, VI, \S12, Theorem 12.2), the map $$x \in L \mapsto f(g_1(x),\dots,g_n(x))\in L$$ is not identically identically zero. It follows that there exists a$\xi\in L$such that the matrix$Q'=\sum_{i=1}^ng_i(\xi)g_i(Q)$has$\det_MQ'\neq0$; in particular, the rank of$Q'$is at least$r$. Since the extension$L/K$is Galois and$Q'$is fixed by all elements in$G$, we see that$Q'\in M_n(K)$. Finally, since the matrices$A$and$B$have their coefficients in$K$,$Q'A=BQ'$. 2 added 9 characters in body Suppose that the field extension$L/K$is separable and that$L$is infinite. Let$A$,$B\in M_n(K)$and suppose there exists a matrix$Q\in M_n(L)$is such that$QA=BQ$and which has rank$Q=r$. We want to show there is a matrix$Q'\in M_n(K)$such that$Q'A=BQ'$and which has rank at least$r$. First, by replacing$L$by the subfield of$L$generated over$K$by the coefficients of$Q$if we need to, we can suppose that$L/K$is a finitely generated extension. By using a maximal purely trascendental extension of$K$contained in$L$as an intermediate step, we see that it is enough to consider separately the cases in which (i)$L/K$is purely trascendental or (ii)$L/K$is finite. In case (i), let$S$be a trascendence basis of$L/K$. Since the matrix$Q$has rank$r$, it has an$r\times r$minor$M$with non-zero determinant. As the entries of$Q$are finitely many rational functions in a finite number of elements of the indeterminates$S$, and since the field$K$is infinite, we assign values from$K$to the indeterminates which appear in$Q$in such a way that we obtain a matrix$Q'\in M_n(K)$(ie, we avoid zeroes in denominators) and such that the minor of$Q'$corresponding to$M$still has non-zero determinant. It is clear that$Q'A=BQ'$and that the rank of$Q'$is at least$r$, so we are done in this case. Let us now consider case (ii). Up to enlarging$L$, we can assume that$L/K$is Galois, with Galois group$G$. As before, the matrix$Q$has an$r\times r$minor$M$with non-zero determinant. Suppose the elements of$G$are$g_1,\dots,g_n$, g_1=1_G,g_2,\dots,g_n$, and consider the polynomial $f(X_1,\dots,X_n)=\det_M\left(\sum_{i=1}^ng_i(Q)X_i\right)\in L[X_1,\dots,X_n]$; here the elements of $G$ act on the matrix $Q$ in the obvious way, and $\det_M$ denotes the determinant of the minor of its argument corresponding to $M$. Notice that $f$ is not the zero polynomial, because the coefficient of $X_1^r$ is precisely $\det_MQ\neq0$.

Since $L$ is infinite and the elements of $G$ are algebraically independent (Lang, Algebra, VI, \S12, Theorem 12.2), the map $$x \in L \mapsto f(g_1(x),\dots,g_n(x))\in L$$ is not identically identically zero. It follows that there exists a $\xi\in L$ such that the matrix $Q'=\sum_{i=1}^ng_i(\xi)g_i(Q)$ has $\det_MQ'\neq0$; in particular, the rank of $Q'$ is at least $r$. Since the extension $L/K$ is Galois and $Q'$ is fixed by all elements in $G$, we see that $Q'\in M_n(K)$. Finally, since the matrices $A$ and $B$ have their coefficients in $K$, $Q'A=BQ'$.

1

Suppose that the field extension $L/K$ is separable and that $L$ is infinite.

Let $A$, $B\in M_n(K)$ and suppose there exists a matrix $Q\in M_n(L)$ is such that $QA=BQ$ and which has rank $Q=r$. We want to show there is a matrix $Q'\in M_n(K)$ such that $Q'A=BQ'$ and which has rank at least $r$.

First, by replacing $L$ by the subfield of $L$ generated over $K$ by the coefficients of $Q$ if we need to, we can suppose that $L/K$ is a finitely generated extension. By using a maximal purely trascendental extension of $K$ contained in $L$ as an intermediate step, we see that it is enough to consider separately the cases in which (i) $L/K$ is purely trascendental or (ii) $L/K$ is finite.

In case (i), let $S$ be a trascendence basis of $L/K$. Since the matrix $Q$ has rank $r$, it has an $r\times r$ minor $M$ with non-zero determinant. As the entries of $Q$ are finitely many rational functions in a finite number of elements of the indeterminates $S$, and since the field $K$ is infinite, we assign values from $K$ to the indeterminates which appear in $Q$ in such a way that we obtain a matrix $Q'\in M_n(K)$ (ie, we avoid zeroes in denominators) and such that the minor of $Q'$ corresponding to $M$ still has non-zero determinant. It is clear that $Q'A=BQ'$ and that the rank of $Q'$ is at least $r$, so we are done in this case.

Let us now consider case (ii). Up to enlarging $L$, we can assume that $L/K$ is Galois, with Galois group $G$. As before, the matrix $Q$ has an $r\times r$ minor $M$ with non-zero determinant. Suppose the elements of $G$ are $g_1,\dots,g_n$, and consider the polynomial $f(X_1,\dots,X_n)=\det_M\left(\sum_{i=1}^ng_i(Q)X_i\right)\in L[X_1,\dots,X_n]$; here the elements of $G$ act on the matrix $Q$ in the obvious way, and $\det_M$ denotes the determinant of the minor of its argument corresponding to $M$. Notice that $f$ is not the zero polynomial, because the coefficient of $X_1^r$ is precisely $\det_MQ\neq0$.

Since $L$ is infinite and the elements of $G$ are algebraically independent (Lang, Algebra, VI, \S12, Theorem 12.2), the map $$x \in L \mapsto f(g_1(x),\dots,g_n(x))\in L$$ is not identically identically zero. It follows that there exists a $\xi\in L$ such that the matrix $Q'=\sum_{i=1}^ng_i(\xi)g_i(Q)$ has $\det_MQ'\neq0$; in particular, the rank of $Q'$ is at least $r$. Since the extension $L/K$ is Galois and $Q'$ is fixed by all elements in $G$, we see that $Q'\in M_n(K)$. Finally, since the matrices $A$ and $B$ have their coefficients in $K$, $Q'A=BQ'$.