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This proof is different from the one in Denis Serre's book.

As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$

and

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$

In general, modules given by matrices of relations are isomorphic if and only if the relation matrices are equivalent.

Ths $A$ and $B$ are similar if and only if $A-tI$ and $B-tI$ are equivalent.

This proof is different from the one in Denis Serre's book.

As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$

and

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$

In general, modules given by matrices of relations are isomorphic if and only if the relation matrices are equivalent.

Thus $A$ and $B$ are similar if and only if $A-tI$ and $B-tI$ are equivalent.

I do not have Denis Serre's book, so I do not know if this answer is the same as the one in his book. It is certainly very simple. It is not particularly geometric, and may alas be pretty close to proofs in the literature.

As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$

and

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$

In general, modules over Euclidean domains given by matrices of relations are isomorphic if and only if the matrices have the same Smith normal form. Applying this to the present instance gives the condition that $A-tI$ and $B-tI$ are equivalent.

This proof is different from the one in Denis Serre's book.

As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$

and

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$

In general, modules given by matrices of relations are isomorphic if and only if the relation matrices are equivalent.

Ths $A$ and $B$ are similar if and only if $A-tI$ and $B-tI$ are equivalent.

I do not have Denis Serre's book, so I do not know if this answer is the same as the one in his book. It is certainly very simple. It is not particularly geometric, and may alas be pretty close to proofs in the literature.

As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$

and

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$

In general, modules over Euclidean domains given by matrices of relations are isomorphic if and only if the matrices have the same Smith canonical form. Applying this to the present instance gives the condition that $A-tI$ and $B-tI$ are equivalent.

I do not have Denis Serre's book, so I do not know if this answer is the same as the one in his book. It is certainly very simple. It is not particularly geometric, and may alas be pretty close to proofs in the literature.

As usual, take $M^A$ and $M^B$ to be the $k[t]$-modules with underlying space $k^n$, where $t$ acts by $A$ and $B$ respectively. Then $A$ is similar to $B$ if and only if $M^A$ and $M^B$ are isomorphic as $k[t]$-modules. As $k[t]$ modules $M^A$ and $M^B$ are both generated by the coordinate vectors $e_1,\dotsc,e_n$, and given by relations (in matrix form)

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(A-It)=0$

and

$\begin{pmatrix} e_1 & e_2 & \cdots & e_n \end{pmatrix}(B-It)=0$

In general, modules over Euclidean domains given by matrices of relations are isomorphic if and only if the matrices have the same Smith normal form. Applying this to the present instance gives the condition that $A-tI$ and $B-tI$ are equivalent.