If an identity in the language of rings holds for all fields, does it necessarily hold for all commutative rings?

Question

It is weirdly difficult to find new identities for ring theory (other than commutativity) that make it more like field theory. This motivates my:

Question. If an identity in the language of rings holds for all fields, does it necessarily hold for all commutative rings?

Answer 1

Yes. In fact, we don't even need to know it in all fields – it suffices to know it in, say, the rational function field $\mathbb{Q} (X_1, X_2, \ldots)$.

Indeed, let $f (X_1, \ldots, X_n)$ and $g (X_1, \ldots, X_n)$ be terms in the language of rings, with variables drawn from $X_1, \ldots, X_n$. Then they are provably equal if and only if they denote the same element of $\mathbb{Z} [X_1, \ldots, X_n]$. (Use the universal property of polynomial rings.) Now use the existence of a ring embedding $\mathbb{Z} [X_1, \ldots, X_n] \hookrightarrow \mathbb{Q} (X_1, X_2, \ldots)$.