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The first part of this answer is merely an elaboration on Marc Kegel and T. Amdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. This wikipedia page may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (the infinite cyclic group), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$. Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to this question. Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.

The first part of this answer is merely an elaboration on Marc Kegel and T. Amdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. This wikipedia page may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (and all infinite cyclic groups are isomorphic to $\mathbb{Z}$), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$. Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to this question. Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.

The first part of this answer is merely an elaboration on Marc Kegel and T. Amdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. This wikipedia page may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (the infinite cyclic group), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$. Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to this question. Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.

The first part of this answer is merely an elaboration on Marc Kegel and T. Amdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. This wikipedia page may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (the infinite cyclic group), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$. Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to this question. Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.

The first part of this answer is merely an elaboration on Marc Kegel and T. Ambdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. This wikipedia page may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (the infinite cyclic group), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$. Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to this question. Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.

The first part of this answer is merely an elaboration on Marc Kegel and T. Amdeberhan's comments, and addresses the question before the edit. As they say, $T^3$ is prime by virtue of being irreducible (a stronger condition) -- since $\pi_2(T^3)$ is trivial, every embedded 2-sphere is null-homotopic in $T^3$ and it follows that it bounds a ball. This wikipedia page may be helpful for you.

While it is true that $S^1\times S^2$ is the only prime closed 3-manifold with fundamental group $\mathbb{Z}$ (the infinite cyclic group), this does not imply that $T^3$ is not prime. (On the off chance this helps, note that $\mathbb{Z}^3$ is not a cyclic group).

In the language of Hatcher's paper, $T^3$ is a prime 3-manifold of type III: it is $K(\mathbb{Z}^3,1)$. Note also that $T^3$ is a Seifert manifold, thus the following statement in Hatcher's paper is consistent with this:

The only Seifert manifold that is not prime is $\mathbb{RP}^3\#\mathbb{RP}^3$, the sum of two copies of real projective 3-space.

I believe the edit of your question is addressed in the answers to this question. Namely, torus bundles are irreducible, so they admit no nontrivial connected sum decomposition.