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Example:

Let $ G = \mathbb{Z}_2$. It is more convenient to use the multiplicative notation, so we write $G = \{1, g\}$ where $g^2 = 1$. The group algebra $F[G]$ thus consists of elements of the form $\lambda 1 + \mu g,\ \lambda , \mu \in F $, which are multiplied according to $$ ( \lambda 1 + \mu g ) (\lambda' 1 + \mu'g ) = (\lambda \lambda' + \mu \mu')1 + ( \lambda \mu' + \mu \lambda')g.$$ It is easy to verify that $\lambda 1 + \mu g \longrightarrow (\lambda + \mu , \lambda - \mu)$ is a homomorphism from $F[G]$ into $F \times F$. If $\mathrm{char}(F) \neq 2$, then it is bijective; thus, $F[G] \cong F \times F$ holds in this case. In particular, $F[G]$ is semiprime. This is no longer true if $\mathrm{char}(F) = 2$ since $F(1+g)$ is then a nilpotent ideal of $F[G]$. Note that, in this case, $F[G]$ is isomorphic to the subalgebra of $M_2(F)$ consisting of matrices of the form \begin{pmatrix} \alpha & \beta \\ 0 & \alpha \end{pmatrix} An isomorphism is given by $$ \lambda 1 + \mu g \longrightarrow \begin{pmatrix} \lambda + \mu & \mu \\ 0 & \lambda + \mu \end{pmatrix}$$ Thus, the group algebra $F[\mathbb{Z}_2] $ is semiprime if and only if $\mathrm{char}(F)\ne 2$.

1: In this example was said that $F[G] \cong F \times F$. Is the product of two prime rings semiprime ?

2: How did we find out that $F(1+g)$ is a nilpotent ideal of $F[G]$?

3: Can you explain why $F[G]$ is isomorphic to the subalgebra of $M_2(F)$ consisting of matrices of the form \begin{pmatrix} \alpha & \beta \\ 0 & \alpha \end{pmatrix}

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  • $\begingroup$ 1:In this example we found that $F[G] \cong F \times F$,Is the result of the multiplication of two field( or prime ring), semiprime that we got such a result? I don't understand this question. What do you mean? What does being semiprime have to do with the given map being a ring isomorphism? $\endgroup$ – rschwieb Jan 17 '18 at 17:58
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1:In this example we found that $F[G] \cong F \times F$,Is the result of the multiplication of two field( or prime ring), semiprime that we got such a result?

(Awaiting feedback as to what you mean.) The given map is obviously always onto, and when $2$ is invertible, it is elementary to show it is injective. (Verifying that it's a ring map is straightforward.)

In the first question, I mean : let $F$ be a prime ring, Is $F×F$ semi prime?

Yes. It's obvious that a product of any number of prime rings is semiprime using the definitions you gave here. You could, for example, just apply (i) of the definitions.

2:How How did we find out that $F(1+g)$ is a nilpotent ideal of $F[G]$?

Because $(1+g)^2=1+2g+g^2=1+0+1=0$ in characteristic $2$. And then $(\lambda(1+g))^2=0$ as well. So... $(F(1+g))^2=\{0\}$.

3:Can you explain why $F[G]$ is isomorphic to the subalgebra of $M_2(F)$ consisting of matrices of the form \begin{pmatrix} \alpha & \beta \\ 0 & \alpha \end{pmatrix}

Since $\{1,g\}$ is a basis, $\{1, 1+g\}$ is a basis. Then mapping $1\to \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $1+g\to \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$

extends uniquely to a $F$ linear map of $F[G]$ onto that subring of $M_2(F)$. The map sends $g\to \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$, and respects the operation on $\{1,g\}$, so it is actually a representation of $F[G]$, i.e. a ring homomorphism. By dimensionality considerations, it is an isomorphism.

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  • $\begingroup$ In the first question, I mean : let $F$ be a prime ring, Is $ F \times F $ semi prime? $\endgroup$ – Jak Jan 17 '18 at 18:33
  • $\begingroup$ @Jak This follows immediately from the definitions you wrote down at this post math.stackexchange.com/q/2604015/29335. $\endgroup$ – rschwieb Jan 17 '18 at 18:48

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