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Matthieu Romagny
Matthieu Romagny
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There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ devidesdivides a general norm. The norm map extandsextends to prime ideals of the integer ring, and if the norm $N_{E/F}(q)$ is $p^3$ for every such ideal over $p$, then $p$ can not possibly be a norm, since its ideal is even not a norm of a fractional ideal of $\mathcal{O}_E$. This happens exactly for those primes $p$ which are inert in $E$, namely those primes modulo which the defining polynomial of $E$ remains irreducible. Indeed, the norm $N_{E/F}(q)$ coincide the the size of $\mathcal{O}_E/q$ which is $p^3$ for inert $q$, the quotient being $\mathbb{F}_{p^3}$.

In our case, for example, the polynomial $x^3+x^2-2x-1$ reduces mod $2$ to the polynomial $x^3+x^2+1$ which is irreducible, so $2$ is an inert prime and the number $2$ can not be realized as a norm. So you can choose $a=2$.

edit: Apparently a similar answer was put in the comments while I written this, sorry for the double answer.

There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ devides a general norm. The norm map extands to prime ideals of the integer ring, and if the norm $N_{E/F}(q)$ is $p^3$ for every such ideal over $p$, then $p$ can not possibly be a norm, since its ideal is even not a norm of a fractional ideal of $\mathcal{O}_E$. This happens exactly for those primes $p$ which are inert in $E$, namely those primes modulo which the defining polynomial of $E$ remains irreducible. Indeed, the norm $N_{E/F}(q)$ coincide the the size of $\mathcal{O}_E/q$ which is $p^3$ for inert $q$, the quotient being $\mathbb{F}_{p^3}$.

In our case, for example, the polynomial $x^3+x^2-2x-1$ reduces mod $2$ to the polynomial $x^3+x^2+1$ which is irreducible, so $2$ is an inert prime and the number $2$ can not be realized as a norm. So you can choose $a=2$.

edit: Apparently a similar answer was put in the comments while I written this, sorry for the double answer.

There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ divides a general norm. The norm map extends to prime ideals of the integer ring, and if the norm $N_{E/F}(q)$ is $p^3$ for every such ideal over $p$, then $p$ can not possibly be a norm, since its ideal is even not a norm of a fractional ideal of $\mathcal{O}_E$. This happens exactly for those primes $p$ which are inert in $E$, namely those primes modulo which the defining polynomial of $E$ remains irreducible. Indeed, the norm $N_{E/F}(q)$ coincide the the size of $\mathcal{O}_E/q$ which is $p^3$ for inert $q$, the quotient being $\mathbb{F}_{p^3}$.

In our case, for example, the polynomial $x^3+x^2-2x-1$ reduces mod $2$ to the polynomial $x^3+x^2+1$ which is irreducible, so $2$ is an inert prime and the number $2$ can not be realized as a norm. So you can choose $a=2$.

edit: Apparently a similar answer was put in the comments while I written this, sorry for the double answer.

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S. carmeli
S. carmeli
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There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ devides a general norm. The norm map extands to prime ideals of the integer ring, and if the norm $N_{E/F}(q)$ is $p^3$ for every such ideal over $p$, then $p$ can not possibly be a norm, since its ideal is even not a norm of a fractional ideal of $\mathcal{O}_E$. This happens exactly for those primes $p$ which are inert in $E$, namely those primes modulo which the defining polynomial of $E$ remains irreducible. Indeed, the norm $N_{E/F}(q)$ coincide the the size of $\mathcal{O}_E/q$ which is $p^3$ for inert $q$, the quotient being $\mathbb{F}_{p^3}$.

In our case, for example, the polynomial $x^3+x^2-2x-1$ reduces mod $2$ to the polynomial $x^3+x^2+1$ which is irreducible, so $2$ is an inert prime and the number $2$ can not be realized as a norm. So you can choose $a=2$.

edit: Apparently a similar answer was put in the comments while I written this, sorry for the double answer.