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The spectrum of the ring of integers $\mathbb{Z}$ consists of the prime ideals, i.e. $Spec(\mathbb{Z}) = \cup_{p \space prime}p\mathbb{Z} \cup (0)$.

i1: S = Spec(ZZ)
i2: nZ = ZZ.ideal(6)
i3: S(nZ)

o3: Point on Spectrum of Integer Ring defined by the Principal ideal (6) of Integer Ring

Obviously, nZ is not a prime ideal, as 6 is composite.

i4: nZ.is_prime()
o4: False

So what does "Point on Spectrum" means exactly in Sage?

Thanks

The spectrum of the ring of integers $\mathbb{Z}$ consists of the prime ideals, i.e. $Spec(\mathbb{Z}) = \cup_{p \space prime}p\mathbb{Z} \cup (0)$.

i1: S = Spec(ZZ)
i2: nZ = ZZ.ideal(6)
i3: S(nZ)

o3: Point on Spectrum of Integer Ring defined by the Principal ideal (6) of Integer Ring

Ring

Obviously, nZ is not a prime ideal, as 6 is composite.

i4: nZ.is_prime()
o4: False

composite. Hence by definition, it is not in $Spec(\mathbb{Z})$. So what does "Point on Spectrum" means exactly in Sage?

Thanks

The spectrum of the ring of integers $\mathbb{Z}$ consists of the prime ideals, i.e. $Spec(\mathbb{Z}) = \cup_{p \space prime}p\mathbb{Z} \cup (0)$.

i1: S = Spec(ZZ)
i2: nZ = ZZ.ideal(6)
i3: S(nZ)
o3: Point on Spectrum of Integer Ring defined by the Principal ideal (6) of Integer Ring

i4: nZ.is_prime()
o4: False

Obviously, nZ is not a prime ideal, as 6 is composite. Hence by definition, it is not in $Spec(\mathbb{Z})$. So what does "Point on Spectrum" means exactly in Sage?

Thanks

The spectrum of the ring of integers $\mathbb{Z}$ consists of the prime ideals, i.e. $Spec(\mathbb{Z}) = \cup_{p \space prime}p\mathbb{Z} \cup (0)$.

i1: S = Spec(ZZ)
i2: nZ = ZZ.ideal(6)
i3: S(nZ)
o3: Point on Spectrum of Integer Ring defined by the Principal ideal (6) of Integer Ring

i4: nZ.is_prime()
o4: False

Obviously, nZ is not a prime ideal, as 6 is composite. Hence by definition, it is not in $Spec(\mathbb{Z})$. So what does "Point on Spectrum" means exactly in Sage?

Thanks