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The inclusion $\mathbb{R}_{\ge 0} \rightarrow \mathbb{R}$ is a semi-ring morphism. So, every $\mathbb{R}$-vector space becomes a $\mathbb{R}_{\ge 0}$-module by restriction of scalars, in particular $\mathbb{R}^n$. Now, the convex cones in $\mathbb{R}^n$ are exactly the sub-$\mathbb{R}_{\ge 0}$-modules of $\mathbb{R}^n$. And the polyhedral cones are exactly the convex cones which are finitely generated as a $\mathbb{R}_{\ge 0}$-module.
But note that I suppose that a convex cone is a subset $X$ of $\mathbb{R}^n$ such that $\alpha.x+\beta.y \in X$ if $\alpha,\beta \in \mathbb{R}_{\ge 0}$ and $x,y \in X$, so they are convex cones $X$ with $0 \in X$.
You would also have the same if you replace $\mathbb{R}_{\ge 0}$ by $\mathbb{R}_{\gt 0}$, but in this case you should note require a zero in the definition of semi-ring and module over a semi-ring.
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