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A positive answer to the first question for $S_{n}$ follows from the work of Alfred Young. He proved that every irreducible representation of $S_{n}$ over $\mathbb{C}$ is afforded by an explicit $\mathbb{Q}G$-module. Another way of saying this is that any irreducible $\mathbb{Q}S_{n}$-module $V$ is absolutely irreducible, and we have ${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$ This last property is already enough to answer the question (positively) for symmetric groups for the field $k = \mathbb{Q}$, and then for every characteristic zero field $k$, since the prime subfield of $k$ is isomorphic to $\mathbb{Q}.$ This answers 2 positively.

The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible $kG_{i}$-module is isomorphic to $k$. This certainly holds for any algebraically closed field $k$ of characteristic zero.

I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of $S_{n}$ (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).

Later edit: Perhaps a more succinct way to say all this is that $\mathbb{Q}$ is a splitting field for $S_{n}$, and consequently, any field $k$ of characteristic zero is a splitting field for $S_{n}$. On the other hand, if field $k$ of characteristic zero is a splitting field for each finite group $G_{i}: 1 \leq i \leq n$, then $k$ is a splitting field for $G_{1} \times G_{2} \ldots \times G_{n}$, and any irreducible $kG$-module is the tensor product of $n$ irreducible modules, one for each $G_{i}$. This is a consequence of Scur's Lemma and Clifford's Theorem.

A positive answer to the first question for $S_{n}$ follows from the work of Alfred Young. He proved that every irreducible representation of $S_{n}$ over $\mathbb{C}$ is afforded by an explicit $\mathbb{Q}G$-module. Another way of saying this is that any irreducible $\mathbb{Q}S_{n}$-module $V$ is absolutely irreducible, and we have ${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$ This last property is already enough to answer the question (positively) for symmetric groups for the field $k = \mathbb{Q}$, and then for every characteristic zero field $k$, since the prime subfield of $k$ is isomorphic to $\mathbb{Q}.$ This answers 2 positively.

The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible $kG_{i}$-module is isomorphic to $k$. This certainly holds for any algebraically closed field $k$ of characteristic zero.

I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of $S_{n}$ (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).

Later edit: Perhaps a more succinct way to say all this is that $\mathbb{Q}$ is a splitting field for $S_{n}$, and consequently, any field $k$ of characteristic zero is a splitting field for $S_{n}$. On the other hand, if field $k$ of characteristic zero is a splitting field for each finite group $G_{i}: 1 \leq i \leq n$, then $k$ is a splitting field for $G_{1} \times G_{2} \times \ldots \times G_{n}$, and any irreducible $kG$-module is the tensor product of $n$ irreducible modules, one for each $G_{i}$. This is a consequence of Schur's Lemma and Clifford's Theorem.

A positive answer to the first question for $S_{n}$ follows from the work of Alfred Young. He proved that every irreducible representation of $S_{n}$ over $\mathbb{C}$ is afforded by an explicit $\mathbb{Q}G$-module. Another way of saying this is that any irreducible $\mathbb{Q}S_{n}$-module $V$ is absolutely irreducible, and we have ${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$ This last property is already enough to answer the question (positively) for symmetric groups for the field $k = \mathbb{Q}$, and then for every characteristic zero field $k$, since the prime subfield of $k$ is isomorphic to $\mathbb{Q}.$ This answers 2 positively.

The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible $kG_{i}$-module is isomorphic to $k$. This certainly holds for any algebraically closed field $k$ of characteristic zero.

I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of $S_{n}$ (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).

A positive answer to the first question for $S_{n}$ follows from the work of Alfred Young. He proved that every irreducible representation of $S_{n}$ over $\mathbb{C}$ is afforded by an explicit $\mathbb{Q}G$-module. Another way of saying this is that any irreducible $\mathbb{Q}S_{n}$-module $V$ is absolutely irreducible, and we have ${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$ This last property is already enough to answer the question (positively) for symmetric groups for the field $k = \mathbb{Q}$, and then for every characteristic zero field $k$, since the prime subfield of $k$ is isomorphic to $\mathbb{Q}.$ This answers 2 positively.

The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible $kG_{i}$-module is isomorphic to $k$. This certainly holds for any algebraically closed field $k$ of characteristic zero.

I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of $S_{n}$ (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).

Later edit: Perhaps a more succinct way to say all this is that $\mathbb{Q}$ is a splitting field for $S_{n}$, and consequently, any field $k$ of characteristic zero is a splitting field for $S_{n}$. On the other hand, if field $k$ of characteristic zero is a splitting field for each finite group $G_{i}: 1 \leq i \leq n$, then $k$ is a splitting field for $G_{1} \times G_{2} \ldots \times G_{n}$, and any irreducible $kG$-module is the tensor product of $n$ irreducible modules, one for each $G_{i}$. This is a consequence of Scur's Lemma and Clifford's Theorem.

A positive answer to the first question for $S_{n}$ follows from the work of Alfred Young. He proved that every irreducible representation of $S_{n}$ over $\mathbb{C}$ is afforded by an explicit $\mathbb{Q}G$-module. Another way of saying this is that any irreducible $\mathbb{Q}S_{n}$-module $V$ is absolutely irreducible, and we have ${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$ This last property is already enough to answer the question (positively) for symmetric groups for the field $k = \mathbb{Q}$, and then for every characteristic zero field $k$, since the prime subfield of $k$ is isomorphic to $\mathbb{Q}.$ This answers 2 positively.

The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible $k\mathbb{G_{i}}$-module is isomorphic to $k$. This certainly holds for any algebraically closed field $k$ of characteristic zero.

I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of $S_{n}$ (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).

A positive answer to the first question for $S_{n}$ follows from the work of Alfred Young. He proved that every irreducible representation of $S_{n}$ over $\mathbb{C}$ is afforded by an explicit $\mathbb{Q}G$-module. Another way of saying this is that any irreducible $\mathbb{Q}S_{n}$-module $V$ is absolutely irreducible, and we have ${\rm End}_{\mathbb{Q}S_{n}}(V) \cong \mathbb{Q}.$ This last property is already enough to answer the question (positively) for symmetric groups for the field $k = \mathbb{Q}$, and then for every characteristic zero field $k$, since the prime subfield of $k$ is isomorphic to $\mathbb{Q}.$ This answers 2 positively.

The tensor decomposition for irreducible modules of general direct products is a consequence of Schur's Lemma: more precisely, it holds whenever the endomorphism algebra of each irreducible $kG_{i}$-module is isomorphic to $k$. This certainly holds for any algebraically closed field $k$ of characteristic zero.

I think that all this should be covered in Curtis and Reiner's book (1962). If not, an explanation of Young's constructions can be found in almost any book on the representation theory of $S_{n}$ (eg that of G. de B. Robinson, or more recent books of James, James and Kerber, or Fulton and Harris).