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Yes, it's possible. Let $R = \mathbb{Q}[[x^{\mathbb{R}_{\ge 0}}]]$ be the ring of formal power series $\sum c_r x^r$ with non-negative real exponents and such that $\{ r : c_r \neq 0 \}$ is well-ordered (this guarantees that multiplication is well-defined); these are known as Hahn series. For $f \in R$ write $\nu(f)$ for the smallest exponent of a nonzero term in $f$. You can see this thread for a proof that the ideals of $R$ have the form

$$I_r = \{ f : \nu(f) \ge r \}$$ $$J_r = \{ f : \nu(f) > r \}$$

and that the only maximal ideals are $I_0$, which is the unit ideal, and $J_0$. Any other ideal strictly contains $J_0$ and there are uncountably many ideals strictly in between, e.g. $I_{r/2}$ for $I_r$ and $J_{r/2}$ for $J_r$.

Yes, it's possible. Let $R = \mathbb{Q}[[x^{\mathbb{R}_{\ge 0}}]]$ be the ring of formal power series $\sum c_r x^r$ with non-negative real exponents and such that $\{ r : c_r \neq 0 \}$ is well-ordered (this guarantees that multiplication is well-defined); these are known as Hahn series. For $f \in R$ write $\nu(f)$ for the smallest exponent of a nonzero term in $f$. You can see this thread for a proof that the ideals of $R$ have the form

$$I_r = \{ f : \nu(f) \ge r \}$$ $$J_r = \{ f : \nu(f) > r \}$$

and that the only maximal ideal is $J_0$. Any other ideal strictly contains $J_0$ and there are uncountably many ideals strictly in between, e.g. $I_{r/2}$ for $I_r$ and $J_{r/2}$ for $J_r$.

Yes, it's possible. Let $R = \mathbb{Q}[[x^{\mathbb{R}_{\ge 0}}]]$ be the ring of formal power series $\sum c_r x^r$ with non-negative real exponents and such that $\{ r : c_r \neq 0 \}$ is well-ordered (this guarantees that multiplication is well-defined). For $f \in R$ write $\nu(f)$ for the smallest exponent of a nonzero term in $f$. You can see this thread for a proof that the ideals of $R$ have the form

$$I_r = \{ f : \nu(f) \ge r \}$$ $$J_r = \{ f : \nu(f) > r \}$$

and that the only maximal ideals are $I_0$, which is the unit ideal, and $J_0$. Any other ideal strictly contains $J_0$ and there are uncountably many ideals strictly in between, e.g. $I_{r/2}$ for $I_r$ and $J_{r/2}$ for $J_r$.

Yes, it's possible. Let $R = \mathbb{Q}[[x^{\mathbb{R}_{\ge 0}}]]$ be the ring of formal power series $\sum c_r x^r$ with non-negative real exponents and such that $\{ r : c_r \neq 0 \}$ is well-ordered (this guarantees that multiplication is well-defined); these are known as Hahn series. For $f \in R$ write $\nu(f)$ for the smallest exponent of a nonzero term in $f$. You can see this thread for a proof that the ideals of $R$ have the form

$$I_r = \{ f : \nu(f) \ge r \}$$ $$J_r = \{ f : \nu(f) > r \}$$

and that the only maximal ideals are $I_0$, which is the unit ideal, and $J_0$. Any other ideal strictly contains $J_0$ and there are uncountably many ideals strictly in between, e.g. $I_{r/2}$ for $I_r$ and $J_{r/2}$ for $J_r$.

Yes, it's possible. Let $k$ be a field of characteristic $0$ and let $R = k[[x^{\mathbb{R}_{\ge 0}}]]$ be the ring of formal power series $\sum c_r x^r$ with non-negative real exponents and such that $\{ r : c_r \neq 0 \}$ is well-ordered (this guarantees that multiplication is well-defined). For $f \in R$ write $\nu(f)$ for the smallest exponent of a nonzero term in $f$. You can see this thread for a proof that the ideals of $R$ have the form

$$I_r = \{ f : \nu(f) \ge r \}$$ $$J_r = \{ f : \nu(f) > r \}$$

and that the only maximal ideals are $I_0$, which is the unit ideal, and $J_0$. Any other ideal strictly contains $J_0$ and there are uncountably many ideals strictly in between, e.g. $I_{r/2}$ for $I_r$ and $J_{r/2}$ for $J_r$.

Yes, it's possible. Let $R = \mathbb{Q}[[x^{\mathbb{R}_{\ge 0}}]]$ be the ring of formal power series $\sum c_r x^r$ with non-negative real exponents and such that $\{ r : c_r \neq 0 \}$ is well-ordered (this guarantees that multiplication is well-defined). For $f \in R$ write $\nu(f)$ for the smallest exponent of a nonzero term in $f$. You can see this thread for a proof that the ideals of $R$ have the form

$$I_r = \{ f : \nu(f) \ge r \}$$ $$J_r = \{ f : \nu(f) > r \}$$

and that the only maximal ideals are $I_0$, which is the unit ideal, and $J_0$. Any other ideal strictly contains $J_0$ and there are uncountably many ideals strictly in between, e.g. $I_{r/2}$ for $I_r$ and $J_{r/2}$ for $J_r$.