# Revision history [back]

sage: Qp=pAdicField(7)
sage: R.<x>=Qp[]
sage: K.<a>=Qp.extension(x^2-5)
sage: OK=K.integer_ring()
sage: OK
7-adic Unramified Extension Ring in a defined by x^2 - 5


Note that 15 is a square in $\mathbb{Q}_p$, so $\mathbb{Q}_p(\sqrt{15})$ is just $\mathbb{Q}_p$.

sage: Qp=pAdicField(7)
sage: R.<x>=Qp[]
sage: K.<a>=Qp.extension(x^2-5)
K.=Qp.extension(x^2-5)
sage: OK=K.integer_ring()
sage: OK
7-adic Unramified Extension Ring in a defined by x^2 - 5

5

Note that 15 is a square in $\mathbb{Q}_p$, $\mathbb{Q}_7$, so $\mathbb{Q}_p(\sqrt{15})$ $\mathbb{Q}_7(\sqrt{15})$ is just $\mathbb{Q}_p$.$\mathbb{Q}_7$.

Try:

sage: Qp=pAdicField(7)
sage: R.<x>=Qp[]
sage: K.=Qp.extension(x^2-5)
K.<a>=Qp.extension(x^2-5)
sage: OK=K.integer_ring()
sage: OK
7-adic Unramified Extension Ring in a defined by x^2 - 55


Note that 15 is a square in $\mathbb{Q}_7$, so $\mathbb{Q}_7(\sqrt{15})$ is just $\mathbb{Q}_7$.

Try:

sage: Qp=pAdicField(7)
sage: R.<x>=Qp[]
sage: K.<a>=Qp.extension(x^2-5)
sage: OK=K.integer_ring()
sage: OK
7-adic Unramified Extension Ring in a defined by x^2 - 5


Note that 15 is a square in $\mathbb{Q}_7$, so $\mathbb{Q}_7(\sqrt{15})$ $\mathbb{Q}_7(\sqrt{3}.\sqrt{5})$ is just $\mathbb{Q}_7$.$\mathbb{Q}_7(\sqrt{5})$.

Try:

sage: Qp=pAdicField(7)
sage: R.<x>=Qp[]
sage: K.<a>=Qp.extension(x^2-5)
sage: OK=K.integer_ring()
sage: OK
7-adic Unramified Extension Ring in a defined by x^2 - 5


Note that 15 is a square in $\mathbb{Q}_7$, so $\mathbb{Q}_7(\sqrt{3}.\sqrt{5})$ $\mathbb{Q}_7(\sqrt{3},\sqrt{5})$ is just $\mathbb{Q}_7(\sqrt{5})$.