Let $R$ be a spectrum. Assume that $R$ is bounded-below. Then we can “even-ify” $R$: cone off some generating set of the first nonvanishing odd-dimensional homotopy group of $R$. Do this repeatedly. In the end you have a spectrum $R^{ev}$ with even-dimensional homotopy. Moreover, you only added even cells, so if $R$ had even cells to begin with, then $R^{ev}$ has even cells.

If $R$ is a ring spectrum, then so is $R^{ev}$, by obstruction theory.

Question 1: If $R$ is a connective ring spectrum with even cells, then is the homotopy of $R^{ev}$ zero-divisor-free?

Notes:

• I’m actually not quite sure how to ensure that $R_\ast$ is a commutative ring, so perhaps the title question — asking if $R_\ast$ is in fact polynomial — doesn’t quite make sense.

• For example, when $R$ is the sphere, I think we get $R^{ev} = MU$. At any rate, if we work $p$-locally and take $R$ to be the $p$-local sphere, then I’m certain I’ve been told that $R^{ev} = BP$. So an abstract proof that even-ification produces polynomial homotopy groups might provide some alternate way to think about the Milnor/Quillen computation of the homotopy of $MU$.

• If we do this unstably, then the possible output spaces were completely classified by Steve Wilson in his PhD thesis. I think the theorem says they are all products of spaces of the form $\Omega^{\infty-n}BP$. At any rate, the list is very finite.

• Note that if we don’t require $R$ to have even cells, then there are Eilenberg-MacLane counterexamples (Just take $HR$ where $R$ is an even-graded ring with zero-divisors). But I’m pretty sure that an Eilenberg-MacLane spectrum never has even cells.

Question 2: Is there a classification of (ring) spectra with even homotopy and even cells analogous to Wilson’s classification of spaces with even homotopy and even cells?

Let $R$ be a spectrum. Assume that $R$ is bounded-below. Then we can “even-ify” $R$: cone off some generating set of the first nonvanishing odd-dimensional homotopy group of $R$. Do this repeatedly. In the end you have a spectrum $R^{ev}$ with even-dimensional homotopy. Moreover, you only added even cells, so if $R$ had even cells to begin with, then $R^{ev}$ has even cells.

If $R$ is a ring spectrum, then so is $R^{ev}$, by obstruction theory.

Question 1: If $R$ is a connective ring spectrum with even cells, then is the homotopy of $R^{ev}$ zero-divisor-free?

Question 2: If $R$ is a ring spectrum with even homotopy and even cells, then is $\pi_\ast R$ zero-divisor-free?

Notes:

• I’m actually not quite sure how to ensure that $R_\ast$ is a commutative ring, so perhaps the title question — asking if $R_\ast$ is in fact polynomial — doesn’t quite make sense.

• For example, when $R$ is the sphere, I think we get $R^{ev} = MU$. At any rate, if we work $p$-locally and take $R$ to be the $p$-local sphere, then I’m certain I’ve been told that $R^{ev} = BP$. So an abstract proof that even-ification produces polynomial homotopy groups might provide some alternate way to think about the Milnor/Quillen computation of the homotopy groups of $MU$.

• If we do this unstably, then the possible output spaces were completely classified by Steve Wilson in his PhD thesis. I think the theorem says they are all products of spaces of the form $\Omega^{\infty-n}BP$. At any rate, the list is very finite.

• Note that if we don’t require $R$ to have even cells, then there are Eilenberg-MacLane counterexamples (Just take $HR$ where $R$ is an even-graded ring with zero-divisors). But I’m pretty sure that an Eilenberg-MacLane spectrum never has even cells.

Question 3: Is there a classification of (ring) spectra with even homotopy and even cells analogous to Wilson’s classification of spaces with even homotopy and even cells?