Universal morphism to forgetful functor Ring->Ab

[grief]

I'm reading a book on category theory and I'm stuck on this problem:

For a given abelian group, what is the universal morphism to the forgetful functor Ring->Ab taking each ring to its underlying additive group?

I think if the functor was Ring -> Monoid taking each ring to the underlying multiplicative monoid, the answer would have been the integral monoid ring. But as it is, I have no idea how to solve it.

 

[micromass]

Take an abelian group M, we wish to put a ring structure on it.
Obviously, M is a $$\mathbb{Z}$$-module. Put $$\mathcal{T}^0(M)=\mathbb{Z}$$ and

$$\mathcal{T}^k=M\otimes_\mathbb{Z}...\otimes_\mathbb{Z} M~\text{(k factors)}$$.

Now we put $$\mathcal{T}(M)=\oplus_{k\geq 0}{\mathcal{T}^k(M)}$$. This is a ring with multiplication given by

$$(m_1\otimes...\otimes m_i)(m^\prime_1\otimes...\otimes m^\prime_j)=(m_1\otimes...\otimes m_i\otimes m^\prime_1\otimes...\otimes m^\prime_j)$$

I'll leave the universal property to you...

 

[grief]

Thanks for the reply!

Though I'm still having trouble with the universality part.

I need to decide two things: the morphism of abelian groups from M to $$\mathcal{T}(M)$$, and, given a morphism f of abelian groups from M to a ring R, a morphism of rings from $$\mathcal{T}(M)$$ to R making the standard diagram commute.

For the former morphism, I thought of injection into the $$\mathcal{T}^1$$ coordinate of $$\mathcal{T}(M)$$. This seems like the natural choice.

Now I'm not sure about the morphism from $$\mathcal{T}(M)$$ to R. In order to make the diagram commute, we must take an m in the $$\mathcal{T}^1$$ coordinate to f(m). But in order to make this a ring homomorphism, that means we must take m1m2 (in the $$\mathcal{T}^2$$ coordinate) to f(m1)f(m2). But this seems to not be additive-- m1m2+m3m4=(m1+m3)(m2+m4) will be taken to f(m1+m3)f(m2+m4), which does not equal f(m1)f(m2)+f(m3)f(m4).

Is there a better choice of morphism I'm not thinking about, or did I misunderstand your construction?

 

[micromass]

You wrote $$m_1m_2+m_3m_4=(m_1+m_3)(m_2+m_4)$$. I don't really see that...

What you do is constructe the following map $$M\times M\rightarrow R:(m_1,m_2,...,m_n)\rightarrow m_1m_2...m_n$$, this is a bilinear map. From the universal property of the tensor product, we get a map $$\mathcal{T}^n(M)\rightarrow R$$.
Taking all those maps together yields a map $$\mathcal{T}(M)\rightarrow R$$. And this is the map you want...

 

[Hurkyl]

It turns out this is a fairly direct calculation.


You're looking for the left adjoint to the forgetful functor. Simply because it's a left adjoint, it must preserve colimits.

Furthermore, every abelian group is the colimit of copies of Z.

So we only need to figure out T(Z) and then compute colimits.



(Try computing it yourself at this point)



To compute T(Z), we hvae the formula:

Hom_Ring(TZ, R) ~ Hom_Ab(Z, R)​

The group on the right is (naturally isomorphic to) the elements of R. So the left set as well -- therefore TZ must be the polynomial ring Z[x].



(Try computing it yourself at this point)



Any presentation of an abelian group A is the same thing as writing the group as a colimit, since it let's you write A as the quotient of a free abelian group by a free abelian group.


(Try computing it yourself at this point)


The coproduct of polynomial rings is the polynomial ring on the disjoint union of the variables. Can you now write down a presentation for T(A)?