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The Artin-Wedderburn Theorem tells us that if $R$ is a semisimple ring, $R\cong M_{n_1}(D_1)\times \dots\times M_{n_r}(D_r)$ for some $n_1,\dots,n_r\in \mathbb{Z}^+$ and division rings $D_1,\dots,D_r$.

A corollary is that a complete (up to isomorphism) list of irreducible $R$-modules is $(D_1^{n_1}\times \{0\} \times \dots \times\{0\}),(\{0\}\times D_2^{n_2}\times\{0\}\times \dots\times\{0\}),\dots,(\{0\}\times\dots\times\{0\}\times D_r^{n_r}) $, i.e. zeros in each position and where $D_i^{n_i}$ is column vectors with coefficients in $D_i$.

If $\mathbb{F}$ is a field, then this says the irreducible $(\underset{n \text{ times}}{\mathbb{F}\times \dots \times \mathbb{F}})$-modules are of the form $V_i = \underset{\text{$i$-th position}}{\{0\}\times\dots\times \mathbb{F} \times\dots\times \{0\}}$.

I must be missing something--aren't these $V_i$ each isomorphic to $\mathbb{F}$ (and hence there is only one irreducible $(\underset{n \text{ times}}{\mathbb{F}\times \dots \times \mathbb{F}})$-module up to isomorphism? Am I mixing up $\mathbb{F}$-modules and $(\underset{n \text{ times}}{\mathbb{F}\times \dots \times \mathbb{F}})$-modules? So the irreducible $(\underset{n \text{ times}}{\mathbb{F}\times \dots \times \mathbb{F}})$-modules are each isomorphic to $\mathbb{F}$ as $\mathbb{F}$-modules, but not as $(\underset{n \text{ times}}{\mathbb{F}\times \dots \times \mathbb{F}})$-modules?

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They are different as modules because the multiplication is different. For example take $n=2$, then

$(1,0)(a,0)=(a,0)$ but $(1,0)(0,a)=(0,0)$ so the element $(1,0)$ acts as a unit in one module, but acts like multiplication by zero in another module.

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