whether a kind of surgery can go on infinitely many steps? let $M$ be a closed ortientable irreducible 3-mfd, let $T$ be a non-separating torus in $M$, we cut $M$ along $T$ and glue two solid tori along the two boundary tori, we get a new closed 3-mfd $M_1$ (we need $M_1$ also is irreducible). Now If there exists nonseperable torus $T_1$ in $M_1$, we go on the above process, we get a new closed 3-mfd $M_2$ ...
My question is, whether you can find a $M$, choose suitable $T_i$, glue solid tori suitablely, the process will go infinitely?
or can you prove that it is impossible to find such an example? (for example, from $M$---> $M_1$, some invariant of 3-mfds decrease strictly).

After Kevin's example, I added the condition "$M_1$ also is irreducible". This condition is natural
in the original field (for this question): 3-mfd with Anosov flow. 
 A: The answer is "no", although it seems that a homological argument is not enough as Kevin's and Bin's examples show. I describe an argument which uses geometrization.
There is a quantity which decreases strictly at each operation. It is crucial to suppose that both $M$ and $M_1$ are irreducible. The quantity for an irreducible manifold $M$ is the triple $(\|M\|, n(M), s(M))$ of real numbers, where 

*
$\|M\|$ is Gromov's norm (i.e. the sum of the volumes of the hyperbolic pieces in the JSJ decomposition of $M$)

* $n(M)$ is the number of tori in the JSJ decomposition of $M$

* $s(M)$ is the sum of the $-\chi(S)$ for each Seifert piece of the decomposition with some base surface $S$.
Triples $(\|M\|, n(M), s(M))$ are ordered lexicographically. I prove below that the surgery you describe (cut along a non-separating torus and glue two solid tori) indeed strictly decreases this quantity, supposing that the resulting manifold $M_1$ is still irreducible (this hypothesis is important). The result then follows because Gromov norms of 3-manifolds form a well-ordered set.
Let $T$ be the torus you cut. If it is adjacent to at least one hyerbolic piece, the filled manifold $M_2$ has strictly smaller Gromov norm thanks to Thurston's Dehn filling theorem. It remains to consider the case $T$ is adjacent to two (possibly coinciding) Seifert pieces and Gromov norm does not decrease. If $T$ is a torus of the JSJ then $n(M)$ decreases. If $T$ is a torus inside a Seifert piece, then $s(M)$ decreases. You use here the following fact: the hypothesis that $M_2$ is irreducible ensures that the meridian of your Dehn filled tori are not fiber-parallel, and hence only add some (possibly non-singular) fiber at the adjacent Seifert pieces. Therefore the JSJ of the new manifold is easily controlled by the JSJ of the old manifold.
It is possible that this argument extends to the relaxed case where you only suppose $T$ to be incompressible (and $M$, $M_1$ are any manifolds).
A: Start with $S^2\times \{1\}$ inside $S^2\times S^1$.  Increase the genus of this non-separating 2-sphere to obtain a non-separating torus $T\subset S^2\times S^1$.  Cutting along $T$ yields $S^2\times I$ with a 1-handle attached to each boundary component.  It is possible to glue solid tori to the two boundary components to obtain $S^2\times S^1 \# S^2\times S^1$.  This process can be continued indefinitely, always increasing first Betti number.
Perhaps you want to require that $T$ is incompressible?
