Does there exist an injective ring homomorphism from the ring $\mathbb{R}\times\mathbb{R}$ to the ring of continuous functions over $\mathbb{R}$?

I know that $\mathbb{R}\times\mathbb{R}$ is a field. So, any ring homomrphism should have either identity or the whole field as a kernel. So, does it imply that there is such a homomorphism? I am unable to think of any examples. Any hints. Thanks beforehand.

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    $\begingroup$ $\mathbb R \times \mathbb R$ is not a field. Remember, the direct product of fields is not going to be a field in general, because, for example, $(1,0) \cdot (0,1) = (0,0)$. $\endgroup$ – астон вілла олоф мэллбэрг Dec 10 '18 at 12:10
  • $\begingroup$ @астонвіллаолофмэллбэрг oh, thanks for that fact. So, then, it is not even an integral domain! $\endgroup$ – vidyarthi Dec 10 '18 at 12:11
  • $\begingroup$ By the definition of the direct product, two homomorphisms from $\mathbb R$ to the ring of continuous functions over $\mathbb R$ give rise to a homomorphism from $\mathbb R \times \mathbb R$ to the latter space, with the two maps as components/projections. However, the pair of maps being injective does not imply that the combined map will be injective. $\endgroup$ – астон вілла олоф мэллбэрг Dec 10 '18 at 12:18
  • $\begingroup$ @астонвіллаолофмэллбэрг ok, any good examples? $\endgroup$ – vidyarthi Dec 10 '18 at 12:23

Let us make a general statement, and then apply it to this case.

Definition : Given a ring $R$, an element $e \in R$ is said to be idempotent, if $e^2 = e$. Note that $0,1$ are idempotent. Any other idempotent will be referred to as non-trivial.

Let $R,S$ be rings such that $R$ has a non-trivial idempotent but $S$ does not have any non-trivial idempotent. Then, any homomorphism $\phi : R \to S$ is not injective.

Proof : Let $e \in R$ be a non-trivial idempotent, which is to say that $e^2 = e$, and $e \neq 0,1$. Let us look at $\phi(e)$. Then, $(\phi(e))^2 = \phi(e^2)= \phi(e)$, so $\phi(e)$ is idempotent in $S$, hence equals zero or one. if $\phi(e) = 0$ then $\phi$ is not injective as $e \neq 0$. If $\phi(e) =1$, then $\phi(1-e) = 0$, and again injectivity is contradicted.

You can use the general statement in lots of places. Try to find candidates for $R$ and $S$.

In our case, $S$ , as the ring of continuous functions over $\mathbb R$,contains no non-trivial idempotent, since if $f^2 = f$, then for each $x$ we have $f(x) = 0$ or $1$. However, as $f$ is continuous, $f(\mathbb R)$ is connected, hence an interval, but has to be a subset of $\{0,1\}$, which is discrete. Consequently, $f$ is either identically zero or identically one, hence is a trivial idempotent.

However, $\mathbb R \times \mathbb R$ contains the non-trivial idempotent $(0,1)$. Hence, the general statement gives the result you have desired.

As to non-injective maps, there are many of them. As mentioned earlier, consider $f,g \in S$ not necessarily distinct. Then, $(a,b) \to af + bg$ is a homomorphism. It is not difficult to see that every homomorphism is of this form.

Also, think about what happens if you treat $R$ and $S$ as just abelian groups under addition. In that case, can you find an injective group homomorphism between the two? Note that the concept of idempotent cannot be defined without the multiplication, so this might be an interesting side question.


Suppose there is an injective ring homomorphism. The unit element $ (1,1) $ of $ \mathbb{R} \times \mathbb{R} $ maps to the unit element of $ \mathcal{C} (\mathbb{R} ) $, i.e. the constant function $ 1_{\mathbb{R}} $. If $ f,g $ are the images of $ (1,0) , (0,1) $ respectively, then $ f+g = 1_{\mathbb{R}} $ and $ fg = 0 $. This shows that $ f(1_{\mathbb{R}}-f) = 0 \implies f = f^2 $. Since $ f $ is continuous, $ f $ is identically $ 0 $ or identically $ 1 $. The first case contradicts injectivity and so does the second because then $ g = 0 $. So there does not exist any injective ring map.

  • $\begingroup$ +1: Very nice argument! And from this we readily find that the only ring homomorphisms $\Bbb R\times\Bbb R\to\mathcal C(\Bbb R)$ are $(x,y)\mapsto x\cdot 1_{\Bbb R}$ and $(x,y)\mapsto y\cdot 1_{\Bbb R}.$ $\endgroup$ – Cameron Buie Dec 10 '18 at 18:36

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