Is $\Bbb{Q}[F]$ a field whenever $F$ is a field?

Question

Let $F = $ a finite field for example. Let $\Bbb{Q}[F]$ be a field ring, which is similar to a group ring except you have both operations extended homomorphically to the whole space, so that distributivity occurs naturally and you have a ring.

But we chose $\Bbb{Q}$ as our coefficient ring. Is $\Bbb{Q}[F]$ a field? It's sort of like a finite extension when $F$ is finite! Interesting. Since we're not declaring a parent space.

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What both operations extended homorphically to all of $\Bbb{Z}[F]$ means is this:

$a = \sum_{f \in F} a_f f, \ b = \sum_{f \in F} b_f f$

Then since $\cdot$ and $+$ are homomorphic in each argument (via extending ring-homomorphically), we have:

$$ a + b = \sum_{f \in F} (a_f + b_f) f \\ a\cdot b = \sum_{f \in F} \sum_{gh = f} (a_gb_h) f $$

Similar to standard group ring stuff.

Answer 1

Suppose $F$ is a finite field of cardinality $p$. Assume that $\mathbb Q[F]$ is a field. Then $\mathbb Q[F]$ contains all $(p-1)$th roots of unity.

Also, $\mathbb Q$ embeds in $\mathbb Q[F]$, by the map $q \mapsto q \cdot 1_F$. Hence it is a number field of degree $p$. But it contains $\zeta_{p-1}$, so it contains a field of degree $\varphi(p-1)$. Hence $\varphi(p-1) \mid p$. This is not possible for $p \geq 5$.

Answer 2

The definition of $\mathbb Q[F]$ does not depend on the addition of $F$, so it is the monoid ring $\mathbb Q[F]$ where $F$ is considered as a multiplicative monoid.

The element $0_F \in \mathbb Q[F]$ is nonzero, and it has the property that $0_F^2 = 0_F$.

Likewise, the element $1_F$ is nonzero and satisfies $1_F^2 = 1_F$.

But a field cannot have two nonzero idempotents.

Answer 3

So the OP wanted to define, with $M$ the multiplicative monoid of $\Bbb{F}_q$, the monoid ring $$k[M]=k[ \Bbb{F}_q^*\cup 0]\cong k[x,z]/(x^{q-1}-1,z^2-z, xz-z)$$ where $x$ is thought as a generator of $\Bbb{F}_q^\times$ and $z$ is thought as $0\in \Bbb{F}_q$.

It is never an integral domain because it contains $k[x]/(x^{q-1}-1)$ and $(x-1)(\sum_{j=0}^{q-2}x^j-1)=0$, and for $q=2$ it is because it contains the non-zero idempotent element $z$ which is not $1$ because $xz\ne x$.