The following question is a review problem for qualifying material on Jordan normal forms. I am having a little trouble understanding the terminology and using the fact we are given coprime polynomials corresponding to cyclic blocks.
Let $M$ be a block diagonal matrix over a field, consisting of two cyclic blocks whose characteristic polynomials are have a GCD of one.
How do we prove that it is possible to select a basis so the matrix becomes one cyclic block?
On
My guess would be that a cyclic block is one that is (or maybe is similar to) a companion matrix. This means that for the linear operator $\phi$ defined by $M$ your space has a direct sum decomposition $V=W_1+W_2$ into $\phi$-invariant subspaces, where each $W_i$ is cyclic for $\phi$: there is some vector $w_i$ such that its $\phi$-images $\phi^k(w_i)$ (for $0\leq k<\dim(W_i)$) span the subspace $W_i$.
To answer you question, one could take $v=w_1+w_2$. Then for any polynomial $P$ the projections of $P[\phi](v)$ on $W_1$ and $W_2$ are respectively $P[\phi](w_1)$ and $P[\phi](w_2)$. Let $\mu_1,\mu_2$ be the minimal (or characteristic) polynomials of the cyclic factors. Then $P[\phi](v)=0$ if and only if $P[\phi](w_1)=0$ and $P[\phi](w_2)=0$, which means that both $\mu_1$ and $\mu_2$ divide$~P$; since they are given to be coprime, this means that the product $\mu_1\mu_2$ must divide$~P$. In particular this does not happen for any $P\neq0$ with degree less than $\deg(\mu_1\mu_2)=\deg(\mu_1)+\deg(\mu_2)=\dim(W_1)+\dim(W_2)=\dim(V)$. That means that $\phi^k(w_i)$ for $0\leq k<\dim(V)$ are linearly independent, and $v$ is a cyclic vector for$~V$.
The argument is similar to the one showing that two cyclic subgroups of coprime order in an abelian group have a (direct) sum that is again cyclic, where as generator on can take a sum of the generators of the cyclic factors. And all this is related to the Chinese remainder theorem, which has a version for polynomials as well.
On
Let $M$ be an $n$ by $n$ matrix with coefficient in a field $K$; view $K^n$ as a $K[X]$-module, where the indeterminate $X$ acts as $M$; and let $f,g\in K[X]$ be coprime polynomials.
Then the Chinese Remainder Theorem tells us that there is a $K[X]$-algebra isomorphism $$ \frac{K[X]}{(fg)}\ \overset{\sim}{\to}\ \frac{K[X]}{(f)}\times\frac{K[X]}{(g)}\quad. $$
If you are familiar with the Frobenius normal form it is quite easy to prove the statement.
First we compute the Smith form of the characteristic matrix of $A$ (where $A$ is your matrix). Since we have two cyclic blocks this is basically the Smith form of a $2 \times 2$ diagonal matrix which has entries $f_1, f_2$ where those are the characteristic (and minimal) polynomials of your blocks. $f_1$ and $f_2$ are coprime hence your Smith form is $diag(1,...,1,f_1 \cdot f_2)$. Therefore the Frobenius form of your matrix consists of one cyclic block.
This is probably not exactly the desired answer since it does not use the Jordan form but it is a proof in the context of the various normal forms and it is often helpful to consider different forms to prove different statements.