This is crossposted from MSE. The question:

Find the Wedderburn decomposition of $D_{5},$ the dihedral group of order 10, over the field $\mathbb{F}_{3}.$

I have shown that the irreducible representations of $D_{5}$ over $\mathbb{F}_{3}$ are the two trivial 1-dimensional representations, and two 4-dimensional representations which derive from considering the action of $D_{5}$ on the group of $5^{\text{th}}$ roots of unity in $\mathbb{F}_{3}.$

Now, we can decompose the group ring $\mathbb{F}_{3}(D_{5})$ into a direct sum of two copies of $\mathbb{F}_{3}$ and two copies of $\mathbb{F}_{3}^{4}$. However, it seems that the two copies of $\mathbb{F}_{3}^{4}$ combine to form the matrix ring $M_{2\times 2}(\mathbb{F}_{3^{2}})$, giving the final Wedderburn decomposition into matrix rings.

Is there a nice explanation for this last step? I would appreciate any help in understanding this.

This is crossposted from MSE. The question:

Find the Wedderburn decomposition of $D_{5},$ the dihedral group of order 10, over the field $\mathbb{F}_{3}.$

I have shown that the irreducible representations of $D_{5}$ over $\mathbb{F}_{3}$ are the two trivial 1-dimensional representations, and two 4-dimensional representations which derive from considering the action of $D_{5}$ on the group of $5^{\text{th}}$ roots of unity in $\mathbb{F}_{3}.$

Now, we can decompose the group ring $\mathbb{F}_{3}(D_{5})$ into a direct sum of two copies of $\mathbb{F}_{3}$ and two copies of $\mathbb{F}_{3}^{4}$. However, it seems that the two copies of $\mathbb{F}_{3}^{4}$ combine to form the matrix ring $M_{2\times 2}(\mathbb{F}_{3^{2}})$, giving the final Wedderburn decomposition into matrix rings.

Is there a nice explanation for this last step? I would appreciate any help in understanding this.

This is crossposted from MSE. The question:

Find the Wedderburn decomposition of $D_{5},$ the dihedral group of order 10, over the field $\mathbb{F}_{3}.$

I have shown that the irreducible representations of $D_{5}$ over $\mathbb{F}_{3}$ are the two trivial 1-dimensional representations, and two 4-dimensional representations which derive from considering the action of $D_{5}$ on the group of $5^{\text{th}}$ roots of unity in $\mathbb{F}_{3}.$

Now, we can decompose the group ring $\mathbb{F}_{3}(D_{5})$ into a direct sum of two copies of $\mathbb{F}_{3}$ and two copies of $\mathbb{F}_{3}^{4}$. However, it seems that the two copies of $\mathbb{F}_{3}^{4}$ combine to form the matrix ring $M_{2\times 2}(\mathbb{F}_{3}^{2})$, giving the final Wedderburn decomposition into matrix rings.

Is there a nice explanation for this last step? I would appreciate any help in understanding this.

This is crossposted from MSE. The question:

Find the Wedderburn decomposition of $D_{5},$ the dihedral group of order 10, over the field $\mathbb{F}_{3}.$

I have shown that the irreducible representations of $D_{5}$ over $\mathbb{F}_{3}$ are the two trivial 1-dimensional representations, and two 4-dimensional representations which derive from considering the action of $D_{5}$ on the group of $5^{\text{th}}$ roots of unity in $\mathbb{F}_{3}.$

Now, we can decompose the group ring $\mathbb{F}_{3}(D_{5})$ into a direct sum of two copies of $\mathbb{F}_{3}$ and two copies of $\mathbb{F}_{3}^{4}$. However, it seems that the two copies of $\mathbb{F}_{3}^{4}$ combine to form the matrix ring $M_{2\times 2}(\mathbb{F}_{3^{2}})$, giving the final Wedderburn decomposition into matrix rings.

Is there a nice explanation for this last step? I would appreciate any help in understanding this.