Smiths normal form is similar to $xI-A.$ I am reading Rational Canonical form from The Abstract Algebra book by Dummit and Foote. I have some doubt in Smith normal form. Smiths normal for says for any $n\times n$ square matrix $A$ over an arbitrary field $F,$ $xI-A$ is equivalent to diagonal matrix in $F[x]$ whose diagonal elements are either $1$ or the invariant factors of the pair $(F^n,A)$. But after looking at other references it seems to me that $xI-A$ is not only equivalent but Similar to such diagonal matrix in $F[x].$  I can't understand how they are similar. I need some help to understand the similarity. 
And also I want to know if there are references for the Canonical form in the modern approach by what I mean using the results of modules over PID.
Thank you. 
 A: I don't want to set up all the machinery related to this problem but only give some basic ideas.
Let $A\in M_n(F)$, where $F$ is a field, $m$ be its minimal polynomial and $\chi_A$ be its characteristic polynomial. If $p\in F[x]$, then $C_p$ denotes the companion matrix of $p$.
Note that $F^n$ can be viewed as a finitely generated module over the PID $F[A]$: $p(A).v=p(A)v$. Then there is the so called "structure theorem" cf. 
https://en.wikipedia.org/wiki/Structure_theorem_for_finitely_generated_modules_over_a_principal_ideal_domain
Step 1. The Frobenius normal form of $A$ over the field $F$. For every $v\in F^n$, there is $q$, a divisor of $m$ of degree $r$ s.t. $q(A)v=0$ and s.t. $v,Av,\cdots,A^{r-1}v$ is linearly independent. By this way, we can construct a basis of $F^n$, using polynomials $p_1,\cdots,p_k=m$ s.t. $p_i$ divides $p_{i+1}$ and $p_1\cdots p_k=\chi_A$; thus $A$ is similar to its Frobenius form $diag(C_{p_1},\cdots,C_{p_k})\in M_n(F)$. Moreover,
(*) the polynomials $(p_i)_i$  uniquely define the similarity class of $A$ over $F$.
Step 2. The smith normal form of $xI_n-A$ over the PID $F[x]$. There are $S,T\in M_n(F[x])$, where $\det(S),\det(T)\in F^*$ s.t. $S(xI-A)T=diag(I_r,p_1,\cdots,p_k)$ with $r=n-k$, the Betti number associated to $xI-A$. Thus $xI-A$ is equivalent to its Smith normal form. The key of the proof is to show that if $A=C_p$, then the Smith normal form of $xI-A$ is $diag(I_{n-1},p)$. cf. for example  
http://www.numbertheory.org/courses/MP274/smith.pdf
Roughly speaking, Frobenius and Smith say pretty much the same thing.
Conclusion. According to (*), 
$xI-A,xI-B\in M_n(F[x])$ have same Smith normal form IFF $A,B$ are similar over $F$.
Note that $xI-A$ is absolutely not similar to its Smith normal form (in general $xI-A$ is not diagonalizable).
