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Stream: deprecated: mathematics

Topic: Algebra over a ring in two different ways

Jean-Baptiste Vienney (Dec 09 2022 at 14:41):

Let $S$ be a commutative ring. Let $R$ be an algebra over $S$ ie. a commutative ring with a morphism $S \rightarrow R$. Can $S$ be an algebra over $R$ in two different ways, ie. can we have two morphisms of rings $i:R \rightarrow S$ and $j:R \rightarrow S$ such that $i \neq j$?

I'm pretty sure the answer is yes and it should be easy but I don't have an example. It would be surprising that there can't be more than one morphism between two rings :big_smile:. Do you have an easy example or a proof that it is false? Maybe it depends of the ring $R,S$, if they are field or not, of the characteristics...

All I know is that for a $\mathbb{Z}$-algebra $S$, it is an algebra in a unique way, because $\mathbb{Z}$ is the initial object in the category of commutative rings...

Oscar Cunningham (Dec 09 2022 at 14:46):

For example the ring of integer polynomials $\Bbb{Z}[X]$ can be mapped to $\Bbb{Z}$ by evaluating the polynomials by setting $X$ to a particular integer. So there is one morphism $\Bbb{Z}[X]\to\Bbb{Z}$ for each integer.

Jean-Baptiste Vienney (Dec 09 2022 at 14:46):

Oh nice, thanks!

Oscar Cunningham (Dec 09 2022 at 14:48):

Another simple example would be that complex conjugation is a morphism $\Bbb{C}\to\Bbb{C}$, distinct from the identity.

Oscar Cunningham (Dec 09 2022 at 14:49):

But it would feel weird to phrase that as '$\Bbb{C}$ is a $\Bbb{C}$-algebra in two different ways'.

Jean-Baptiste Vienney (Dec 09 2022 at 14:52):

Maybe saying that $\mathbb{C}$ is a $\mathbb{C}$-algebra in a unique way makes sense because ring morphisms from a field are always injective so that the image of a field in an algebra over a field is unique up to isomorphism but for rings that's not necessarily the case...

Jean-Baptiste Vienney (Dec 09 2022 at 14:54):

So I think that your previous example really shows that $\mathbb{Z}$ is a $\mathbb{Z}[X]$-algebra in two different ways because if you change the value you set $X$ two, you change the image of the morphism.

Oscar Cunningham (Dec 09 2022 at 14:54):

I think you wrote 'bijective' when you meant 'injective'. Morphisms out of fields are not necessarily bijective, for example $\Bbb{R}\to\Bbb{C}$.

Jean-Baptiste Vienney (Dec 09 2022 at 14:55):

Oh yes thanks

Oscar Cunningham (Dec 09 2022 at 14:58):

Also, I think the image of $\Bbb{Z}[X]\to\Bbb{Z}$ is always $\Bbb{Z}$, because polynomials include constants.

Oscar Cunningham (Dec 09 2022 at 14:59):

But yes, there's no automorphism $f$ of $\Bbb{Z}$ such that $f\circ \mathrm{eval}_2 = \mathrm{eval}_3$.

Jean-Baptiste Vienney (Dec 09 2022 at 15:03):

So, do you have an example of two ring morphisms $f,g:R \rightarrow S$ such that $im(f) \not\cong im(g)$?

Oscar Cunningham (Dec 09 2022 at 15:06):

Again, $\Bbb{Z}[X]$ is useful because you can set $X$ to be whatever you want. So for example in the ring $\Bbb{Z}[Y,Z]/Y^2$ there's no automorphism taking $Y$ to $Z$, because $Y$ squares to 0 and $Z$ doesn't. So then if we consider the maps $\Bbb{Z}[X]\to\Bbb{Z}[Y,Z]/Y^2$, one of which sends $X$ to $Y$ and the other of which sends $X$ to $Z$, their images will look completely different.

Morgan Rogers (he/him) (Dec 09 2022 at 15:07):

For an example with fields, consider $\mathbb{Q}[X]/(X^3 - 1)$. This has three homomorphisms to $\mathbb{C}$, mapping $X$ to the respective roots of unity. One of them has image contained in the reals, the other two do not.

Jean-Baptiste Vienney (Dec 09 2022 at 15:08):

Okay, so it even exists with fields.

Tobias Fritz (Dec 09 2022 at 15:11):

Recently the question whether $\mathbb{R}$-algebra structure is unique in this sense came up on MathOverflow and nobody seemed to know the answer (although that was in a comment, so not many people may have seen the question). Perhaps someone here can make progress on this? Note that $\mathbb{R}$ has no field automorphisms.

Jean-Baptiste Vienney (Dec 12 2022 at 10:49):

I think that maybe for augmented $\mathbb{R}$-algebras it could be the case that they are $\mathbb{R}$-algebras in a unique way. An augmented $\mathbb{R}$-algebra is a $\mathbb{R}$-algebra $i:\mathbb{R} \rightarrow A$ together with an augmentation $g:A \rightarrow \mathbb{R}$ ie. a ring homomorphism such that $i;g=Id$.

Jean-Baptiste Vienney (Dec 12 2022 at 10:54):

It would be the case that two $\mathbb{R}$-algebras $A,B$ whose underlying rings are isomorphic are also isomorphic as $\mathbb{R}$-algebras if either $A$ or $B$ is augmented, if the unique ring epimorphism $\mathbb{R} \rightarrow \mathbb{R}$ was the identity (edit: in fact, I don't know).

Jean-Baptiste Vienney (Dec 12 2022 at 11:05):

And I also suspect that every non-zero $\mathbb{R}$-algebra is augmented.

Morgan Rogers (he/him) (Dec 12 2022 at 11:19):

Beware of your intuition here. There is at most one ring homomorphism from $\mathbb{Q}$, and since this is dense in $\R$, a continuous ring homomorphism from $\R$ is unique if it exists. Any counterexample would therefore be discontinuous, and one may be constructible using the axiom of choice, for example by constructing a sufficiently "nice" $\mathbb{Q}$-basis of $\R$ and mapping the basis elements to different places.

Morgan Rogers (he/him) (Dec 12 2022 at 11:20):

Jean-Baptiste Vienney said:

It would be the case that two $\mathbb{R}$-algebras $A,B$ whose underlying rings are isomorphic are also isomorphic as $\mathbb{R}$-algebras if either $A$ or $B$ is augmented, if the unique ring epimorphism $\mathbb{R} \rightarrow \mathbb{R}$ was the identity.

I don't understand the argument here, could you explain?

Jean-Baptiste Vienney (Dec 12 2022 at 11:25):

You are right, it doesn't work, I wanted to write:

"Firstly, given two $\mathbb{R}$-algebras $i:\mathbb{R} \rightarrow A$, $j:\mathbb{R} \rightarrow B$, a ring homomorphism $f:A \rightarrow B$ is a $\mathbb{R}$-algebra homomorphism iff $i;f=j$.

Jean-Baptiste Vienney (Dec 12 2022 at 11:26):

If $A$ is augmented, you have a ring homomorphism $g:A \rightarrow \mathbb{R}$ such that $i;g=Id$.

Jean-Baptiste Vienney (Dec 12 2022 at 11:27):

Then $g$ is a ring epimorphism.

Jean-Baptiste Vienney (Dec 12 2022 at 11:28):

If $f:A \rightarrow B$ is a ring isomorphism, then $j;f^{-1};g:\mathbb{R} \rightarrow \mathbb{R}$ is a ring epimorphism.

Jean-Baptiste Vienney (Dec 12 2022 at 11:30):

So, if the unique ring epimporphism $\mathbb{R} \rightarrow \mathbb{R}$ is the identity, then $j;f^{-1};g = Id$."

Jean-Baptiste Vienney (Dec 12 2022 at 11:38):

(If you had $g;i=Id$, it would give $j;f^{-1}=i$ and then $f^{-1}$ would be a $\mathbb{R}$-algebra homomorphism, but in this case $A \cong B \cong \mathbb{R}$...)

Jean-Baptiste Vienney (Dec 12 2022 at 11:40):

Maybe it could be true for connected graded $\mathbb{R}$-algebras ie. the ones of the form $\underset{n \ge 0}{\bigoplus}A_{n}$ where $A_{0} \cong \mathbb{R}$ and $A_{i}A_{j}\subseteq A_{i+j}$ (still if the unique epimorphism $\mathbb{R} \rightarrow \mathbb{R}$ is the identity.)

Jean-Baptiste Vienney (Dec 12 2022 at 12:34):

I'm gonna try to see if it works in these conditions...

Jean-Baptiste Vienney (Dec 12 2022 at 13:33):

Anyway, I have absolutely no intuition for the discontinuous ring homorphisms. So, I hope someone is gonna solve your problem. As to me, I learned a lot about ring and algebra homomorphisms in this thread.

Tobias Fritz (Dec 12 2022 at 14:17):

Good thinking in any case, thanks for sharing!

John Baez (Dec 12 2022 at 14:34):

Here's something vaguely, not precisely, related to the question of whether there can be more than one ring homomorphism $f: \mathbb{R} \to R$ for some ring $R$.

John Baez (Dec 12 2022 at 14:37):

There's a concept of [[real closed field]]. Besides the real numbers $\mathbb{R}$ we can find another real closed field $\mathbb{R}'$, not isomorphic to $\mathbb{R}$, with the same cardinality. We can build the complex numbers $\mathbb{C}$ using pairs of real numbers. We can use the exact same formula to build a field $\mathbb{C}'$ using pairs of guys in $\mathbb{R}'$. But it’s easy to check that this funny field $\mathbb{C}'$ is algebraically closed and of characteristic zero. Since it has the same cardinality as $\mathbb{C}$, it must be isomorphic to $\mathbb{C}$.

John Baez (Dec 12 2022 at 14:38):

So, $\mathbb{C}$ contains real closed fields other than $\mathbb{R}$.

Jean-Baptiste Vienney (Dec 12 2022 at 15:28):

And what is this $\mathbb{R}'$? Does it have a name?

John Baez (Dec 12 2022 at 21:17):

I believe there are many real closed fields with same cardinality as the reals. You can use any one as $\mathbb{R}'$. For example, you can use the field of Puiseaux series with real coefficients. You can also use the field of hyperreal numbers.