A Condition for a Vector Space to be a Field

Question

Let $\mathbb{E}$ be a nonempty set with two associated operations, addition, $+ : \mathbb{E} \times \mathbb{E} \to \mathbb{E}$, and multiplication, $\cdot : \mathbb{E} \times \mathbb{E} \to \mathbb{E}$. Suppose $\mathbb{F}$ is a subset of $\mathbb{E}$ such that $(\mathbb{F}, +, \cdot)$ is a field and further suppose that there exists a positive integer $n$ and a set $\{b_i\}_{i = 1}^n \subset \mathbb{E}$ such that every element $e$ of $\mathbb{E}$ can be written as

\begin{equation} e = \sum_{i = 1}^n f_i \cdot b_i \end{equation}

with unique coefficients $\{f_i\}_{i = 1}^n \subset \mathbb{F}$. Must $\mathbb{E}$ be a field?

Currently I have not yet been able to establish a fruitful approach to tackle this question, but I will update it in accordance with my progress.

Answer 1

An easy counterexample to this is to take any field $\mathbb F$, and form the rings $\mathbb F^n$ or $M_n(\mathbb F)$. Both have divisors of zero, so they cannot be fields.

Answer 2

However, with some assumptions, one has the following well-known lemma:

Let $A\longrightarrow B$ be an injective ring homomorphism. Suppose that $A$, $B$ are integral domains, and that $B$ is integral over $A$ (this is the case in particular if $B$ is a finitely generated $A$-module). Then $A$ is a field if and only if $B$ is a field.