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The sphere $\{p\}\times S^2$ in $S^1\times S^2$ is nonseparating: its complement is $(S^1\setminus \{p\})\times S^2$, which is connected. A sphere bounding a ball separates, so this product is not an irreducible three-manifold.

To show it is prime, suppose a separating sphere cuts it into compact pieces $V,W$ with one spherical boundary each. The Seifert-van Kampen theorem givesA free product of two nontrivial groups is nonabelian: choose nonidentity elements in the two factors; their products in the two orders are distinct reduced words. Hence one factor, say ${\pi }_1(V)$, is trivial. The inclusion of this simply connected compact piece lifts to the universal cover $\mathbb{R}\times S^2\cong {\mathbb{R}}^3\setminus \{0\}$. The lift is an embedding: equality of lifted points would imply equality of their projections in $V$. Its image $\tilde{V}$ is compact with boundary the lifted sphere.

Use the three-dimensional Schoenflies theorem: a locally flat embedded sphere in ${\mathbb{R}}^3$ bounds a ball, and its compact side is that ball. The interior of $\tilde{V}$ is both open and closed in the corresponding complement component, and compactness rules out the unbounded component. Thus $\tilde{V}$ is the bounded ball, and $V$ itself is a ball. Capping $V$ gives $S^3$, so every connected-sum decomposition is trivial. Therefore $S^1\times S^2$ is prime but not irreducible.

Now let a prime closed connected orientable $M$ fail irreducibility. A sphere not bounding a ball cannot separate: if it did, primeness would make one capped side $S^3$, and the three-dimensional Schoenflies theorem would make its uncapped side a ball. Thus there is a nonseparating sphere $S$. Its two-sided neighborhood is $S^2\times [-1,1]$. Since the complement is connected, join its two boundary spheres by an embedded arc in that complement and thicken the arc. The union is a punctured $S^2\times S^1$: the thickened arc joins the two ends of the sphere-cylinder, and the remaining boundary is one sphere. This is the nonseparating sphere gives a sphere-circle summand construction and yieldsThe product is not $S^3$, since its fundamental group is $\mathbb{Z}$. Primeness forces $N\cong S^3$, soThe background results used are van Kampen, the explicit universal cover of the product, and the locally flat three-dimensional Schoenflies theorem; no Poincaré-conjecture argument is needed.