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We know that in the category of (unitary) rings, $\mathbb{Z}$ is the itinial object, i.e. it is the only ring such that for each ring $A,$ there exists a unique ring homomorphism $f:\mathbb{Z} \to A$.

This means, in particular, that $\mathbb{R}$ does not satisfy this property, so for a certain ring $B$, we can construct $g:\mathbb{R} \to B$ and $h:\mathbb{R} \to B$ such that $g \ne h$.

So far, I have proven that $B$ cannot be an ordered ring and cannot be $\mathbb{R}^n \ (n\in\mathbb{N})$. Can you help me finding such a ring $B$?

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Actually I think you can take $B=\mathbb{R}$. $\mathbb{R}$ should have some non-trivial ring endomorphisms. I'm not sure we can explicitly construct one though. – Arnaud D. Mar 18 at 20:28
@HenningMakholm Yes probably. I had in mind something like extending $\sqrt{2}\mapsto -\sqrt{2}$ to $\mathbb{R}$ but I wasn't completely sure about this. – Arnaud D. Mar 18 at 20:32
Notice that if you want to disprove that $\mathbb{R}$ is initial you can also attack existence of uniqueness. That's much easier : you can't find an homomorphism $\mathbb{R}\to \mathbb{Q}$, for example. – Arnaud D. Mar 18 at 20:37
$\mathbb{R}$ has no nontrivial ring endomorphisms (even with choice)! Any ring morphism $f\colon \mathbb{R}\to\mathbb{R}$ must be the identity on $\mathbb{Q}$, and it must preserve the ordering ($a < b$ if and only if $a-b>0$ if and only if there exists $c$ such that $c^2 = a - b$, and this condition is preserved by ring homomorphisms). Since $\mathbb{Q}$ is dense in $\mathbb{R}$, $f$ must be the identity. – Alex Kruckman Mar 18 at 20:40
Non-existence can also be directly argued with characteristic in case of $f\colon\mathbb R\to\mathbb Z/p\mathbb Z$. – Ennar Mar 18 at 20:42

You're wrong in negating the statement of being an initial object.

A ring $R$ is an initial object if and only if, for every ring $S$ there exists a ring homomorphism $R\to S$ and, if $f,g\colon R\to S$ are ring homomorphisms, then $f=g$.

Negating this becomes

$R$ is not an initial object if and only if there exists a ring $S$ such that there is no ring homomorphism $R\to S$ or there are two distinct homomorphisms $R\to S$.

For instance, the ring $\mathbb{F}_2$ has the property that, if a ring homomorphism $f\colon\mathbb{F}_2\to S$ exists, then it is unique. The reason is obvious: $f(0)=0$ and $f(1)=1$, plus the fact that $\mathbb{F}_2=\{0,1\}$.

So there is no ring $S$ with two distinct ring homomorphisms $f,g\colon\mathbb{F}_2\to S$, but $\mathbb{F}_2$ is not an initial object; indeed, there is no ring homomorphism $\mathbb{F}_2\to\mathbb{Z}$.

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Ok that's right thanks! So do you know if such a $B$ exists or it does not exist? – Étienne Tétreault Mar 18 at 21:01
Well said! And just for completeness, the polynomial ring $\Bbb{Z}[x]$ fails on the ground that the homomorphism required by the universal property of an initial object always exists, but is never unique unless $A$ is the trivial ring with 1 element. – Rob Arthan Mar 18 at 21:04
@ÉtienneTétreault Slade's example is good. – egreg Mar 18 at 21:10

Many useful and (sometimes surprisingly) true things have been said in other answers; let me just make some simple remarks.

To begin, you could have consulted the proof of what you implicitly used, namely that if $\def\Z{\Bbb Z}\Z$ is an initial object, then it is automatically unique, up to canonical isomorphism (indeed, for once, up to unique isomorphism). The (very standard) proof goes: suppose $R$ is another initial object in the category, then by assumption there are unique morphisms $f:\Z\to R$ and $g:R\to\Z$. Then $g\circ f:\Z\to\Z$ is a morphism and $\Z$ being initial it is unique and therefore equal to the identity of$~\Z$; similarly $f\circ g:R\to R$ is a morphism and $R$ being initial it is unique and therefore equal to the identity of$~R$, from which it follows that $f$ and $g$ are inverse morphisms, so $R$ is isomorphic to $Z$.

Now taking $\def\R{\Bbb R}R=\R$ it is clear that $f:\Z\to\R$ fails to be surjective so cannot have an inverse; by (the contrapositive of) the above argument this shows that $\R$ cannot be initial, and it does not admit a morphism $g:\R\to\Z$ at all. [That is a small lie. The argument uses the hypothesis that $R$ is initial twice: for the existence of $g$, and for equating $f\circ g$ to the identity of$~R$; one or the other must fail. But it is fairly easy to see that for $R=\R$ the former fails, and indeed one does have the property that any morphism $\R\to\R$ equals the identity, though it takes a bit of works to show that.]

Next, and rather unrelated, some words about the tensor product proof that $\R$ does not even have, as the above might suggest, the weaker property that if a morphism $\R\to A$ exists, then it is unique. You can think of the $\otimes$ in the construction of the ring $S\otimes S$ as similar to the fraction bar used to construct $\Bbb Q$, with the following difference: in fractions we decree that one can cancel (or introduce) equal factors ($an/nb=a/b$), but for the tensor product one decrees that one can move a factor across the $\otimes$ symbol in either direction ($an\otimes b=a\otimes nb$). The difference has as consequence that while sums of fractions can always be combined to a single fraction, sum of tensors cannot always.

The factors allowed to be moved are integers (the bare minimum), or possibly some larger class $R$ of scalars, which must then be attached as $\otimes_R$ to indicate this attribute of the construction; it serves as a gatekeeper that allows elements of (the ring) $R$ to cross the $\otimes_R$, but not others. Nonetheless, even for a bare $\otimes_\Z$, one can effectively pass rational factors across: to move $p/q$ from left to right, it suffices to move a factor $p$ from left to right and a factor $q$ from right to left (therefore $\otimes_\Bbb Q$ is not really different from $\otimes_\Z$). However one cannot go beyond rationals: $\sqrt2\otimes_\Z1$ and $1\otimes_\Z\sqrt2$ are different, for if they could be shown equal using a finite number of tensor-equivalences, one would basically have produced a fraction $p/q$ equal to $\sqrt2$, which is impossible. (A formal proof requires work.)

As a concluding remark: the ring $\Bbb Q$ and the finite fields $\Z/p\Z$ with $p$ prime do have the property that morphisms from them are unique if they exist. These are exactly the prime fields: fields without any proper sub-fields.

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Good points (+1), but it seems that the OP was already aware of the argument in your first part. She wanted to see a concrete instance of the definition being false, but negated it incorrectly. – Henning Makholm Mar 19 at 10:49
@HenningMakholm: well yes and no. OP knew that initial objects are unique, and probably has seen a proof of that fact. The point in my first part is that to understand why $\R$ is not initial, it helps to expand the proof for that case; this automatically leads to the proper logical negation of the definition of initial. Good proofs provide understanding, not just conviction. – Marc van Leeuwen Mar 19 at 11:30
Again, I think the OP understands that fine; see her comments to my answer, for example. She already knows that $\mathbb R$ is not initial (exactly because it fails to be isomorphic to $\mathbb Z$ which is), and wants to sanity-check that knowledge by seeing directly that it indeed doesn't satisfy the definition. – Henning Makholm Mar 19 at 11:39
Synthesis of other answers is something I would like to see more of on the site. Thanks for taking the time to write it (+1). – zyx Mar 19 at 16:48

It is easier to show that for a certain ring $B$ there is no ring homomorphism at all $\mathbb R\to B$. In fact we can take $B=\mathbb Z$.

Assume to the contrary that $h:\mathbb R\to\mathbb Z$ is a homomorphism. We seek a contradiction.

If there is any nonzero $a\in\mathbb R$ such that $h(a)=0$, then we're done, because then $h(1)=h(a^{-1}a)=h(a^{-1})h(a)=h(a^{-1})\cdot 0 = 0$, which is not allowed for a unitary ring homomorphism.

Let's consider what $h(\sqrt 2)$ is. As argued above it can't be $0$, so it must be some nonzero integer $n$. Let's assume that $n$ is positive (the other case goes similarly). Then $$ \underbrace{h(\sqrt2/n)+\cdots+h(\sqrt2/n)}_{n\text{ times}} = h(\underbrace{\sqrt2/n+\cdots+\sqrt2/n}_{n\text{ times}}) = h(\sqrt2)=n$$ but the only way this can be true is if $h(\sqrt2/n)=1$, and therefore $$ h(\sqrt2/n-1) = h(\sqrt2/n)-h(1) = 1-1 = 0 $$ However, $a=\sqrt2/n-1$ is irrational and therefore in particular not zero, so this leads to a contradiction as above.

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why $h(\sqrt2)$ is an integer? – Tsemo Aristide Mar 18 at 20:40
@TsemoAristide: I'm proving that it is not the case that for every $B$ there is a unique homomorphism $\mathbb R\to B$. For this it is enough to show a single counterexample, and the counterexampe I have selected is $B=\mathbb Z$. It doesn't matter that there might be other possible $B$s that are not counterexamples. One counterexample sufficies to disprove a universal claim. – Henning Makholm Mar 18 at 20:45
Much more simply, a ring homomorphism from a field is injective… – egreg Mar 18 at 20:48
@ÉtienneTétreault: You know that $\mathbb Z$ is an initial object. – Henning Makholm Mar 18 at 20:51
Yeah well that is what I mean, in category theory we already talk up to isomorphism and I know $\mathbb{Z}$ and $\mathbb{R}$ are not isomorphic! And we don't need to make a debate about it I just wanted explicit examples thanks for trying! – Étienne Tétreault Mar 18 at 20:58

In fact, there is a universal ring with two distinct morphisms from $\mathbb{R}$, the ring $\mathbb{R}\otimes_\mathbb{Z}\mathbb{R}$.

I am unsure of whether it has nicely-presented quotients with the same property.

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Very interesting,but I'm not familiar with that notation, can you send me a link that explains that construction? – Étienne Tétreault Mar 18 at 20:48

Here's an answer to the original question which doesn't use tensor products (but does use some field theory).

Let $i$ be the usual inclusion of $\mathbb{R}$ into $\mathbb{C}$, and let $\sigma$ be any automorphism of $\mathbb{C}$ which does not fix $\mathbb{R}$ pointwise. Then $\sigma\circ i\neq i$ are distinct ring homomorphisms $\mathbb{R}\to\mathbb{C}$.

To find an automorphism of $\mathbb{C}$ which does not fix $\mathbb{R}$ pointwise, you could, for example,

  • Extend the automorphism of $\mathbb{Q}[\sqrt{2}]$ given by $\sqrt{2}\mapsto -\sqrt{2}$ to an automorphism of $\mathbb{C}$.
  • Pick a transcendence basis $\mathcal{B}$ of $\mathbb{C}$ over the algebraic closure of $\mathbb{Q}$ which contains $\pi$. Then any permutation of $\mathcal{B}$ which moves $\pi$ extends to an automorphism of $\mathbb{C}$.
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$\mathbb{R}$ has two distinct embeddings into the tensor product $\mathbb{R}\otimes_{\mathbb{Z}} \mathbb{R}$, given by $r\mapsto r\otimes 1$ and $r\mapsto 1\otimes r$.

These are the two inclusions coming from the fact that the tensor product is the coproduct in the category of rings.

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In the category of commutative rings. – egreg Mar 18 at 20:46
There is a unique embedding $\mathbb{F}_2\to\mathbb{F}_2\otimes_{\mathbb{Z}}\mathbb{F}_2$. – egreg Mar 18 at 20:59
Are those embeddings only maps or are they ring homomorphism? I don't know many about module theory! – Étienne Tétreault Mar 18 at 21:06
@AlexKruckman My objection is that just being the coproduct isn't sufficient. – egreg Mar 18 at 21:11
@egreg's objection has some merit, at least if you define the tensor product by its universal property. Then the only way you can know that the two canonical injections into the coproduct must be distinct is if you already know some $B$ with two distinct morphisms $\mathbb R\to B$. – Henning Makholm Mar 18 at 21:17

Let $\phi:\Bbb R\to B$ be a ring morphism. Then $\ker\phi$ is an ideal of $\Bbb R$. But the only ideals of $\Bbb R$ are $\Bbb R$ and $\langle 0\rangle$. Thus $\ker\phi=\Bbb R$ or $\ker\phi=0$.

Suppose $\ker\phi=\Bbb R$. Then $B$ must be the trivial ring since $\phi(1)=1_B$ and the only ring satisfying $1_B=0_B$ is the trivial ring. In this case $\phi:\Bbb R\to B$ is unique.

Suppose $\ker\phi=0$. Then $\phi$ is injective. It follows that $\DeclareMathOperator{image}{image}\image(\phi)$ is a subring of $B$ isomorphic to $\Bbb R$.

This proves that every nontrivial ring $B$ without a subring isomorphic to $\Bbb R$ admits no homomorphisms $\phi:\Bbb R\to B$.

Examples include $B=\Bbb Z$, $B=\Bbb Z_n$, $B=\Bbb Q$.

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... or indeed any countable ring. – Marc Paul Mar 18 at 21:55
This also shows that to have a ring with more than one morphism from $\Bbb R$, one must have a ring with more than one subring isomorphic to $\Bbb R$. That such rings exist (and indeed that $\Bbb C$ is one) should definitely be surprising for beginners in ring theory. Also noteworthy is that they don't exist if one replaces $\Bbb R$ by $\Bbb Q$. – Marc van Leeuwen Mar 19 at 3:11

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