Return to Answer

A bit too long for a comment, so I write it as an answer.

$(1)$ $P$ is the sum of the terms in the expansion of the determinant $\det (\sigma_i a_j)$ corresponding to even permutations, end hence in $K$. Also -$N$ is the sum of the terms corresponding to odd permutations, hence also in $K$.

$(2)$ Since either $\sigma(P)=P$ and $\sigma(N)=N$ or $\sigma(P)=N$ and $\sigma(N)=P$, it always follows that $\sigma(P+N)=P+N$ and $\sigma(PN)=PN$, for $\sigma\in Gal(L/\mathbb{Q})$, where $L$ is the Galois closure of $K$ over $\mathbb{Q}$.

$(3)$ So $P+N$ and $PN$ are fixed under elements of $Gal(L/\mathbb{Q})$, hence in $\mathbb{Q}$ by definition.

EDIT: Look here for more details (page 12): web.mit.edu/~holden1/www/math/ant.pdf

A bit too long for a comment, so I write it as an answer.

$(1)$ Let $L$ be the Galois closure of $K$ over $\mathbb{Q}$. We can apply $\sigma\in Gal(L,\mathbb{Q})$ on $P+N$ and $PN$, which lie in $L$.

$(2)$ Since either $\sigma(P)=P$ and $\sigma(N)=N$ or $\sigma(P)=N$ and $\sigma(N)=P$, it always follows that $\sigma(P+N)=P+N$ and $\sigma(PN)=PN$, for $\sigma\in Gal(L/\mathbb{Q})$, where $L$ is the Galois closure of $K$ over $\mathbb{Q}$.

$(3)$ So $P+N$ and $PN$ are fixed under elements of $Gal(L/\mathbb{Q})$, hence in $\mathbb{Q}$ by definition.

EDIT: Look here for more details (page 12): web.mit.edu/~holden1/www/math/ant.pdf

A bit too long for a comment, so I write it as an answer.

$(1)$ $P$ is the sum of the terms in the expansion of the determinant $\det (\sigma_i a_j)$ corresponding to even permutations, end hence in $K$. Also -$N$ is the sum of the terms corresponding to odd permutations, hence also in $K$.

$(2)$ Since either $\sigma(P)=P$ and $\sigma(N)=N$ or $\sigma(P)=N$ and $\sigma(N)=P$, it always follows that $\sigma(P+N)=P+N$ and $\sigma(PN)=PN$, for $\sigma\in Gal(L/\mathbb{Q})$, where $L$ is the Galois closure of $K$ over $\mathbb{Q}$.

$(3)$ So $P+N$ and $PN$ are fixed under elements of $Gal(L/\mathbb{Q})$, hence in $\mathbb{Q}$ by definition.

A bit too long for a comment, so I write it as an answer.

$(1)$ $P$ is the sum of the terms in the expansion of the determinant $\det (\sigma_i a_j)$ corresponding to even permutations, end hence in $K$. Also -$N$ is the sum of the terms corresponding to odd permutations, hence also in $K$.

$(2)$ Since either $\sigma(P)=P$ and $\sigma(N)=N$ or $\sigma(P)=N$ and $\sigma(N)=P$, it always follows that $\sigma(P+N)=P+N$ and $\sigma(PN)=PN$, for $\sigma\in Gal(L/\mathbb{Q})$, where $L$ is the Galois closure of $K$ over $\mathbb{Q}$.

$(3)$ So $P+N$ and $PN$ are fixed under elements of $Gal(L/\mathbb{Q})$, hence in $\mathbb{Q}$ by definition.

EDIT: Look here for more details (page 12): web.mit.edu/~holden1/www/math/ant.pdf

A bit too long for a comment, so I write it as an answer.

$(1)$ $P$ is the sum of the terms in the expansion of the determinant $\det (\sigma_i a_j)$ corresponding to even permutations, end hence in $K$. Also -$N$ is the sum of the terms corresponding to odd permutations, hence also in $K$.

$(2)$ Since either $\sigma(P)=P$ and $\sigma(N)=N$ or $\sigma(P)=N$ and $\sigma(N)=P$, it always follows that $\sigma(P+N)=P+N$ and $\sigma(PN)=PN$.

$(3)$ So $P+N$ and $PN$ are fixed under elements of $Gal(K/\mathbb{Q})$, hence in $\mathbb{Q}$ by definition.

A bit too long for a comment, so I write it as an answer.

$(1)$ $P$ is the sum of the terms in the expansion of the determinant $\det (\sigma_i a_j)$ corresponding to even permutations, end hence in $K$. Also -$N$ is the sum of the terms corresponding to odd permutations, hence also in $K$.

$(2)$ Since either $\sigma(P)=P$ and $\sigma(N)=N$ or $\sigma(P)=N$ and $\sigma(N)=P$, it always follows that $\sigma(P+N)=P+N$ and $\sigma(PN)=PN$, for $\sigma\in Gal(L/\mathbb{Q})$, where $L$ is the Galois closure of $K$ over $\mathbb{Q}$.

$(3)$ So $P+N$ and $PN$ are fixed under elements of $Gal(L/\mathbb{Q})$, hence in $\mathbb{Q}$ by definition.