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Stream: learning: questions

Topic: Ring homomorphisms between integral domains

Madeleine Birchfield (Feb 21 2024 at 16:40):

Every ring homomorphism between fields is an injection and a ring monomorphism. Is this still true for ring homomorphisms between integral domains?

Josselin Poiret (Feb 21 2024 at 16:56):

isn't $\mathbb{Z} \to \mathbb{Z}/2$ a counter example?

Madeleine Birchfield (Feb 21 2024 at 17:26):

Right, I forgot about that. Since the integers are the initial commutative ring, the unique ring homomorphism from the integers to any integral domain with prime characteristic is not injective.

Madeleine Birchfield (Feb 21 2024 at 17:27):

But I think I am actually interested in ordered integral domains, so is it true that every ring homomorphism between ordered integral domains is an injection and a ring monomorphism? The ordering makes the integral domain have characteristic zero, so excludes the above example.

John Baez (Feb 21 2024 at 18:14):

Is the fraction field of an ordered integral domain an ordered field?

Kevin Arlin (Feb 21 2024 at 18:32):

How about evaluation at $0$ from $\mathbb Z[T]\to\mathbb Z,$ where $T$ is infinitesimally small?

Damiano Mazza (Feb 21 2024 at 19:07):

What order are you putting on $\mathbb Z[T]$?

David Egolf (Feb 21 2024 at 19:13):

I tried to show that one can have a non-injective ring homomorphism between two integral domains with characteristic zero. Let $i: \mathbb{Z} \to \mathbb{Z}[x]$ be the unique ring homomorphism, which sends $a \mapsto a$ for all $a \in \mathbb{Z}$. Note that $\mathbb{Z}[x]$ is an integral domain with characteristic zero.

We are looking for a map $\pi: \mathbb{Z}[x] \to \mathbb{Z}[x]/I$ for some ideal $I$, such that $\pi \circ i$ is still injective. If $\pi \circ i$ is injective, that means that $\mathbb{Z}[x]/I$ has characteristic zero. We also want $\mathbb{Z}[x]/I$ to be an integral domain. We can ensure the quotient is an integral domain by choosing $I$ to be a prime ideal.

David Egolf (Feb 21 2024 at 19:14):

To meet these two conditions simultaneously, we pick a prime ideal of $I$ that doesn't include any integer multiples of 1. We set $I= \langle x+1 \rangle$. This should be a prime ideal as $x+1$ is an irreducible polynomial.

We then consider $\mathbb{Z} \to_i \mathbb{Z}[x] \to_\pi \mathbb{Z}[x]/\langle x+1 \rangle$. Since the image of $i$ is disjoint from the kernel of $\pi$, the composite map is injective. The injectivity of $\pi \circ i$ implies that $\mathbb{Z}[x]/\langle x+1 \rangle$ has characteristic zero. And since $\langle x+1 \rangle$ is a prime ideal, $\mathbb{Z}[x]/\langle x+1 \rangle$ is an integral domain.

Therefore, $\pi: \mathbb{Z}[x] \to \mathbb{Z}[x]/\langle x+1 \rangle$ is a non-injective ring homomorphism between integral domains with characteristic zero.

(Hopefully I didn't make a mistake! I'm also not sure how this relates to the desired result for ordered integral domains).

Kevin Arlin (Feb 21 2024 at 19:14):

@Damiano Mazza I think saying $T$ is infinitesimally small uniquely determines the order: a polynomial is positive if and only if its smallest-degree coefficient is positive.

Kevin Arlin (Feb 21 2024 at 19:19):

@David Egolf, that works (for the non-ordered case), but there's an isomorphism of $\mathbb Z[x]/\langle x+1\rangle$ with a more familiar ring that would make it much clearer!

Damiano Mazza (Feb 21 2024 at 19:24):

Kevin Arlin said:

a polynomial is positive if and only if its smallest-degree coefficient is positive.

Ok, thanks, you're right. Well then yours is definitely a counterexample!

Damiano Mazza (Feb 21 2024 at 19:26):

Ah no, wait, is it? If $T$ is infinitely small doesn't that mean that $\mathbb Z[T]$ is not an integral domain? I'm taking "infinitely small" to mean something like $T^k=0$ for some $k>1$. But that must not be it. What do you mean then?

Kevin Arlin (Feb 21 2024 at 19:31):

No, I mean infinitely small in the sense that $kT<1$ for every integer $k.$

Kevin Arlin (Feb 21 2024 at 19:32):

This kind of construction is a standard way of getting at a "poor man's" approximation to, say, hyperreal numbers: you want a field containing $\mathbb R$ and also some infinitely small $\varepsilon,$ so try $\mathbb R(T)$ and just define $T$ to be infinitely small (but still invertible!)

Damiano Mazza (Feb 21 2024 at 19:35):

Nice!!! Thanks, I had never seen this.

Kevin Arlin (Feb 21 2024 at 19:41):

Sure :)

David Egolf (Feb 21 2024 at 19:46):

Kevin Arlin said:

David Egolf, that works (for the non-ordered case), but there's an isomorphism of $\mathbb Z[x]/\langle x+1\rangle$ with a more familiar ring that would make it much clearer!

Thanks for pointing that out! After a bit of thought, I think the evaluation homomorphism $f: \mathbb{Z}[x] \to \mathbb{Z}$ at $x=-1$ should also have the properties we want. (That is, it is a non-injective ring homomorphism between integral domains with characteristic zero).

The kernel of this map is $\langle x+1 \rangle$ and its image is all of $\mathbb{Z}$, so I think we have an isomorphism $\mathbb{Z}[x]/\langle x+1 \rangle \cong \Z$ given by evaluating any representative at $x=-1$.

Kevin Arlin (Feb 21 2024 at 19:55):

Yep, that's what I was thinking.

Madeleine Birchfield (Feb 22 2024 at 03:22):

Damiano Mazza said:

What order are you putting on $\mathbb Z[T]$?

Kevin Arlin said:

Damiano Mazza I think saying $T$ is infinitesimally small uniquely determines the order: a polynomial is positive if and only if its smallest-degree coefficient is positive.

One can also take $T$ to be any transcendental real number and consider $\mathbb{Z}[T]$ to be the commutative subring of the real numbers generated by $T$; this ring is an integral domain because the real numbers do not have zero divisors, and the order relation is inherited from the order relation of the real numbers.

Kevin Arlin (Feb 22 2024 at 03:43):

I thought of that too but I actually don’t think it works. WLOG let $T=\pi.$ Then the desired homomorphism is antitone when applied to $3$ and $4-\pi,$ for instance.

Kevin Arlin (Feb 22 2024 at 03:44):

In particular these are isomorphic rings but not isomorphic ordered rings.

James Deikun (Feb 22 2024 at 03:46):

The homomorphisms were never required to be monotone, but it doesn't hurt that you found an example where it is.

Kevin Arlin (Feb 22 2024 at 17:37):

Oh, funny, right!