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I'm currently stuck on a problem and hope that maybe one of you can shed some light on it.

Here's the situation:

Let $(R_i)_{i\in I}$ be a family of rings. A product of rings $(R_i)_{i\in I}$ is a ring $R$ together with a family of ring homomorphisms $(\phi_i: R \rightarrow R_i)_{i\in I}$ which fulfill the following property:

For every family of ring homomorphisms $(\phi_i: S \rightarrow R_i)_{i\in I}$ there exists a unique ring homomorphism $\phi: S \rightarrow R$ such that $\phi_i = \pi_i \circ \phi$.


The first problem was to show that the direct product of rings $\prod_{i\in I} R_i := \{(x_i)_{i\in I} | x_i \in R_i \;\forall i \in I\} $ together with the family of ring homomorphisms $(\pi_j: \prod_{i\in I } R_i \rightarrow R_j)_{j\in I}$ with $\pi_j((x_i)_{i\in I})=x_j$ (this just gives the j-th component) also fulfills the above property.

That was pretty straight forward. I also gleaned the following information from it:

Since $\phi_i = \pi_i \circ \phi$ has to hold for every family of ring homomorphisms, it has to hold for a family of isomorphisms, too. From this I get:

$\phi_i = \pi_i \circ \phi$ surjective $\Rightarrow$ $\pi_i$ is surjective

$\phi_i = \pi_i \circ \phi$ injective $\Rightarrow$ $\phi$ is injective $\;\forall i\in I$.


The next step was to show that when any two rings with their respective families of ring homomorphisms $(R, (\pi_i)_{i\in I})$ and $(R',(\pi'_i)_{i\in I})$ fulfill the property, then there exists a unique ring homomorphism $\phi: R \rightarrow R'$ with $\pi_i = \pi'_i \circ \phi \quad \forall i \in I$.

To show this, I just plugged in $R$ and $R'$ for $S$.


Now, I ended up with this:

I have unique ring homomorphisms $\phi: R \rightarrow R'$ and $\phi': R' \rightarrow R$, which by the above inferences are injective.

And also ring homomorphisms $(\pi_i: R \rightarrow R_i)_{i\in I}$ and $(\pi'_i: R' \rightarrow R_i)_{i\in I}$, which are surjective $\; \forall i \in I$.


The last thing I need to do is show that $\phi: R \rightarrow R'$ is an isomorphism and I have no idea how.

I could either show that $(\pi_i: R \rightarrow R_i)_{i\in I}$ or $(\pi'_i: R' \rightarrow R_i)_{i\in I}$ are injective or that $\phi: R \rightarrow R'$ is surjective, but I don't know how.

Do any of you see a way?

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Note that $\phi'\circ\phi : R \to R$ is a ring homomorphism satisfying

$$\pi_i\circ(\phi'\circ\phi) = (\pi_i\circ\phi')\circ\phi = \pi_i'\circ\phi = \pi_i.$$

We also have the identity ring homomorphism $\operatorname{id}_R : R \to R$ which satisfies $\pi_i \circ\operatorname{id}_R = \pi_i$. By uniqueness, we must have $\phi'\circ\phi = \operatorname{id}_R$. A similar argument shows that $\phi\circ\phi' = \operatorname{id}_{R'}$. Therefore, $\phi : R \to R'$ and $\phi' : R' \to R$ are isomorphisms. Furthermore, they are inverses.

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  • $\begingroup$ Thanks, Michael. I was so focused on trying to work the injectivity/surjectivity angle that the most basic algebraic notion of an identity slipped my mind. $\endgroup$ – Miriam Nov 17 '15 at 15:33

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