The positive definiteness axiom can be replaced with an axiom asserting the existence of an "orthonormal basis".

Let $R$ be a $*$-ring. We consider a left $*$-module over $R$ which we call $M$. We ask what it means for a Hermitian form $\langle -, - \rangle$ to be an inner product over $M$.

In finite dimensions, you would say that $\langle -, - \rangle$ is an inner product if there exists a finite basis $B=\{b_1,b_2,\dotsc,b_n\}$ of $M$ such that $\langle b_i, b_j \rangle = \delta_{ij}$ where we have used the Kronecker delta symbol.

In possibly infinite dimensions, you would ask that your "orthonormal basis" only be "densely spanning". What I mean by this is that there should exist a subset $B \subset M$ for which $\langle b, b' \rangle=\begin{cases}1, & b = b' \\ 0, & b \neq b' \end{cases}$, and for which every $0 \neq v \in M$ should admit some $b \in B$ for which $\langle b, v \rangle \neq 0$.

For the usual $*$-rings of $\mathbb R$ and $\mathbb C$ (with their usual involutions) this recovers the usual notion of an inner product space. The proof uses Zorn's Lemma.

In foundations not admitting the Axiom of Choice, this definition may be superior to the usual one. It's certainly easier to do maths over an inner product space once you have a basis.

In answer to your question, the appropriate setting is $*$-fields, and not fields. And all $*$-fields can admit inner product spaces.

The positive definiteness axiom can be replaced with an axiom asserting the existence of an "orthonormal basis".

Let $R$ be a $*$-ring. We consider a left $*$-module over $R$ which we call $M$. We ask what it means for a Hermitian form $\langle -, - \rangle$ to be an inner product over $M$.

In finite dimensions, you would say that $\langle -, - \rangle$ is an inner product if there exists a finite basis $B=\{b_1,b_2,\dotsc,b_n\}$ of $M$ such that $\langle b_i, b_j \rangle = \delta_{ij}$ where we have used the Kronecker delta symbol.

In possibly infinite dimensions, you would ask that your "orthonormal basis" only be "densely spanning". What I mean by this is that there should exist a subset $B \subset M$ for which $\langle b, b' \rangle=\begin{cases}1, & b = b' \\ 0, & b \neq b' \end{cases}$, and for which every $0 \neq v \in M$ should admit some $b \in B$ for which $\langle b, v \rangle \neq 0$.

For the usual $*$-rings of $\mathbb R$ and $\mathbb C$ (with their usual involutions) this recovers the usual notion of an inner product space. The proof uses Zorn's Lemma.

In foundations not admitting the Axiom of Choice, this definition may be superior to the usual one. It's certainly easier to do maths over an inner product space once you have an orthonormal basis.

In answer to your question, the appropriate setting is $*$-fields, and not fields. And all $*$-fields can admit inner product spaces.

The positive definiteness axiom can be replaced with an axiom asserting the existence of an "orthonormal basis".

Let $R$ be a $*$-ring. We consider a left $*$-module over $R$ which we call $M$. We ask what it means for a Hermitian form $\langle -, - \rangle$ to be an inner product over $M$.

In finite dimensions, you would say that $\langle -, - \rangle$ is an inner product if there exists a finite basis $B=\{b_1,b_2,\dotsc,b_n\}$ of $M$ such that $\langle b_i, b_j \rangle = \delta_{ij}$ where we have used the Kronecker delta symbol.

In possibly infinite dimensions, you would ask that your "orthonormal basis" only be "densely spanning". What I mean by this is that there should exist a subset $B \subset M$ for which $\langle b, b' \rangle=\begin{cases}1, & b = b' \\ 0, & b \neq b' \end{cases}$, and for which every $0 \neq v \in M$ should admit some $b \in B$ for which $\langle b, v \rangle \neq 0$.

For the usual $*$-rings of $\mathbb R$ and $\mathbb C$ (with their usual involutions) this recovers the usual notion of an inner product space. The proof uses Zorn's Lemma.

In foundations not admitting the Axiom of Choice, this definition may be superior to the usual one. It's certainly easier to do maths over an inner product space once you have a basis.

The positive definiteness axiom can be replaced with an axiom asserting the existence of an "orthonormal basis".

Let $R$ be a $*$-ring. We consider a left $*$-module over $R$ which we call $M$. We ask what it means for a Hermitian form $\langle -, - \rangle$ to be an inner product over $M$.

In finite dimensions, you would say that $\langle -, - \rangle$ is an inner product if there exists a finite basis $B=\{b_1,b_2,\dotsc,b_n\}$ of $M$ such that $\langle b_i, b_j \rangle = \delta_{ij}$ where we have used the Kronecker delta symbol.

In possibly infinite dimensions, you would ask that your "orthonormal basis" only be "densely spanning". What I mean by this is that there should exist a subset $B \subset M$ for which $\langle b, b' \rangle=\begin{cases}1, & b = b' \\ 0, & b \neq b' \end{cases}$, and for which every $0 \neq v \in M$ should admit some $b \in B$ for which $\langle b, v \rangle \neq 0$.

For the usual $*$-rings of $\mathbb R$ and $\mathbb C$ (with their usual involutions) this recovers the usual notion of an inner product space. The proof uses Zorn's Lemma.

In foundations not admitting the Axiom of Choice, this definition may be superior to the usual one. It's certainly easier to do maths over an inner product space once you have a basis.

In answer to your question, the appropriate setting is $*$-fields, and not fields. And all $*$-fields can admit inner product spaces.