The Brown representability theorem can be convenient way to construct a spectrum. But to get a ring spectrum of even a very unstructured form seems to be harder. There's even currently a statement on the nlab to the effect that any multiplicative cohomology theory is represented by a ring spectrum, but my strong suspicion is that this statement was made in error.

**Assumption:** Let $E^\ast: Spaces^{op} \to GrAb$ be a cohomology theory. To avoid subtleties with the unpointed version of Brown's theorem, let's just assume that $E^\ast$ is represented by a spectrum $E$. Now when we impose "multiplicative" structure on $E^\ast$, the question is whether the "multiplication" is also "representable". I can think of at least two possible ways to make this precise:

**Question 1:** Suppose that $E^\ast: Spaces^{op} \to GrAb$ admits the structure of a lax (symmetric) monoidal functor. Then does there exist a map $\mu: E \wedge E \to E$ inducing the (symmetric) monoidal structure of $E^\ast$ in the appropriate sense?

I think that's the cleanest form of the question, but I think it would be more classical to ask it in the following form:

**Question 2:** Suppose that $E^\ast: Spaces^{op} \to GrAb$ admits a lift to a functor $E^\ast: Spaces^{op} \to GrRing$. Then does there exist a map $\mu: E \wedge E \to E$ such that the multiplication $$E^\ast(X) \otimes E^\ast(X) = \pi_{-\ast} F(\Sigma^\infty_+,E) \otimes \pi_{-\ast}F(\Sigma^\infty_+,E) \to \pi_{-\ast}F(\Sigma^\infty_+ X, E) = E^\ast(X)$$ is always given by the formula $$f\otimes g \mapsto \mu_\ast \circ (f \wedge g) \circ \Delta^\ast$$ where ($\Delta: X \to X \times X$ is the diagonal)?

**Remark:** The category "$GrRing$" could be interpreted in multiple ways -- if our rings are associative, we could ask for a homotopy associative $\mu$; if our rings are graded-commutative we could ask for a homotopy commutative $\mu$, etc.

**Note:** The Yoneda lemma does allow us to give an "$E_0$ version" of Brown representability. That is, from the structure of unit elements $u_X \in E^\ast(X)$ we can construct a unit map $u: \mathbb S \to E$, just by the Yoneda lemma. That's why the question is about the multiplication.

**Remark:** Actually -- we might even just as well have our cohomology theory be defined as a functor $E^\ast: Spectra^{op} \to GrAb$, and the question still seems to be interesting.