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The forgetful functor from the category of $\lambda$-rings to that of rings has a right adjoint in the form of the universal $\lambda$ functor $\Lambda$, which is isomorphic to the big Witt vectors functor. But, does the forgetful functor have a left adjoint? Some kind of free $\lambda$-ring functor?

EDIT: I see this as obviously true when I forget all the way down to sets. But, I only want to forget down to the category of commutative rings. Would it just be to take the free ring on all the elements, mod out by the $\lambda$ relations and all the ring relations?

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This is obviously true from the definition of a lambda ring as a commutative ring with operators and certain relations among these operators. – Torsten Ekedahl Sep 8 '11 at 16:53
up vote 5 down vote accepted

The edit suggests to me that you want not only the existence of the free $\lambda$-ring on any given ring (which is immediate from general results about adjoints of forgetful functors between varieties of algebras) but a construction using terms. Given a ring R, first form the set of all formal $\lambda$-ring terms over $R$, i.e., well-formed expressions built from the operations of $\lambda$-rings (the ring operations and the $\lambda$'s) and symbols for all the elements of $R$. Then divide out by the equivalence relation that identifies two such terms if their equality is provable (in equational logic) from the axioms of $\lambda$-rings plus the diagram of $R$. (For the diagram, it suffices to take all those equations that are true in $R$ and have the form $a+b=c$ or $ab=c$, where $a,b,c$ are among the symbols for elements of $R$.)

What I've written here is, of course, not specific to $\lambda$-rings. The left adjoint of any forgetful functor between varieties of algebras can be constructed in an analogous way.

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