[proofplan]
We compare [linear independence](/page/Linear%20Independence) of any fixed set of columns over $\mathbb{Q}$ with linear independence of the same columns after reducing the entries of $A$ to an arbitrary field $F$. Linear independence of a set of $k$ columns is equivalent to the existence of a nonzero $k \times k$ minor in the corresponding column submatrix. Total unimodularity forces every such determinant to be $0$, $1$, or $-1$, and these are nonzero over every field exactly when they are nonzero over $\mathbb{Q}$. Hence every field gives the same independent sets, so the column matroid is representable over every field.
[/proofplan]
[step:Pass the integer matrix to an arbitrary field]
Let $F$ be an arbitrary field. Define the canonical ring homomorphism
\begin{align*}
\pi_F: \mathbb{Z} \to F
\end{align*}
by $\pi_F(k) = k \cdot 1_F$ for every $k \in \mathbb{Z}$. Let $A_F \in F^{m \times n}$ denote the matrix with entries $(A_F)_{ij} := \pi_F(A_{ij})$ for $1 \le i \le m$ and $1 \le j \le n$.
For a subset $I \subset E$, let $A_I \in \mathbb{Z}^{m \times |I|}$ be the submatrix of $A$ consisting of the columns indexed by $I$, and let $(A_F)_I \in F^{m \times |I|}$ be the corresponding column submatrix of $A_F$. We must prove that the columns of $A_I$ are linearly independent over $\mathbb{Q}$ if and only if the columns of $(A_F)_I$ are linearly independent over $F$.
[/step]
[step:Characterize column independence by maximal square minors]
Fix a subset $I \subset E$, and write $k := |I|$. If $k > m$, then no $k$ columns in $F^m$ or in $\mathbb{Q}^m$ can be linearly independent, so both matrices declare $I$ dependent. Suppose therefore that $k \le m$.
For each $k$-element subset $R \subset \{1,\dots,m\}$, let $A_{R,I} \in \mathbb{Z}^{k \times k}$ be the square submatrix of $A$ using rows $R$ and columns $I$. Let $(A_F)_{R,I} \in F^{k \times k}$ be the corresponding submatrix of $A_F$. The columns indexed by $I$ are linearly independent over a field $K$ precisely when at least one $k \times k$ minor of the corresponding $m \times k$ matrix has nonzero determinant in $K$. Applied over $K = \mathbb{Q}$ and $K = F$, this gives
\begin{align*}
I \text{ is independent over } \mathbb{Q} \iff \exists R \subset \{1,\dots,m\},\ |R| = k,\ \det(A_{R,I}) \ne 0 \text{ in } \mathbb{Q}.
\end{align*}
Likewise,
\begin{align*}
I \text{ is independent over } F \iff \exists R \subset \{1,\dots,m\},\ |R| = k,\ \det((A_F)_{R,I}) \ne 0 \text{ in } F.
\end{align*}
[guided]
Fix a subset $I \subset E$ and let $k := |I|$. The question is whether the columns indexed by $I$ have the same dependence relation over every field. If $k > m$, then a family of $k$ vectors in an $m$-dimensional [vector space](/page/Vector%20Space) is linearly dependent over any field, so there is nothing to prove in that case.
Assume $k \le m$. For each $k$-element row set $R \subset \{1,\dots,m\}$, define $A_{R,I} \in \mathbb{Z}^{k \times k}$ to be the square submatrix with rows $R$ and columns $I$. Its image over $F$ is the square matrix $(A_F)_{R,I} \in F^{k \times k}$.
The reason square minors detect independence is the rank criterion for matrices over a field: an $m \times k$ matrix over a field has column rank $k$ if and only if at least one of its $k \times k$ minors has nonzero determinant. Column rank $k$ is exactly linear independence of the $k$ columns. Therefore, over $\mathbb{Q}$,
\begin{align*}
I \text{ is independent over } \mathbb{Q} \iff \exists R \subset \{1,\dots,m\},\ |R| = k,\ \det(A_{R,I}) \ne 0 \text{ in } \mathbb{Q}.
\end{align*}
The same rank criterion applies over the arbitrary field $F$, giving
\begin{align*}
I \text{ is independent over } F \iff \exists R \subset \{1,\dots,m\},\ |R| = k,\ \det((A_F)_{R,I}) \ne 0 \text{ in } F.
\end{align*}
Thus the whole problem has been reduced to comparing whether the same integer minors vanish before and after applying $\pi_F$.
[/guided]
[/step]
[step:Use total unimodularity to preserve nonzero minors over every field]
For each $R \subset \{1,\dots,m\}$ with $|R| = k$, determinant is a polynomial with integer coefficients in the matrix entries, so applying $\pi_F$ entrywise gives
\begin{align*}
\det((A_F)_{R,I}) = \pi_F(\det(A_{R,I})).
\end{align*}
Since $A$ is totally unimodular, $\det(A_{R,I}) \in \{-1,0,1\}$. The element $\pi_F(0)$ is $0_F$, while $\pi_F(1) = 1_F \ne 0_F$ and $\pi_F(-1) = -1_F \ne 0_F$. In characteristic two, $-1_F = 1_F$, but this element is still nonzero. Hence
\begin{align*}
\det(A_{R,I}) \ne 0 \text{ in } \mathbb{Q} \iff \det((A_F)_{R,I}) \ne 0 \text{ in } F.
\end{align*}
By the minor characterization from the previous step, $I$ is independent over $\mathbb{Q}$ if and only if $I$ is independent over $F$.
[/step]
[step:Conclude that the represented matroid is regular]
The field $F$ was arbitrary, and the preceding step proves that every subset $I \subset E$ is independent in the column matroid over $\mathbb{Q}$ if and only if it is independent in the column matroid represented by $A_F$ over $F$. Therefore $A_F$ represents the same matroid $M(A)$ over every field $F$. By the definition of a regular matroid, $M(A)$ is regular.
[/step]
