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A similar question like this has been asked here, apologies, but need to clarify something at the end

Our homework question was to show that any ring homomorphism $f:K\rightarrow R $ (where K is a field, R is a ring) is injective.

So by definition of a ring homomorphism, $ker(f)$ is an ideal in K. By K a field $\Rightarrow$ the only ideals are $\{0\}, K$

$ker(f)={0} \Rightarrow f$ is injective, by definition of an injective ring homomorphism.

However I answered saying that $ker(f)=K \Rightarrow$ f is the zero map, $f(k)=0\ \forall k\in K$. So $f$ is either injective, or it is the zero map.

To which my corrector asked, " Is the zero map a ring homomorphism?"

Obviously if you've got something like $f:\Bbb R \rightarrow \Bbb Z_4 $, then $f(1_R)=f(1)=0 \ne 1 $ so this can't be a ring homomorphism

But if you considered $f:\Bbb R \rightarrow \{0\} $ , then $f(1_R)=f(1)=0 = 1_0 $

Does this work as a ring homomorphism? So can the zero map sometimes be a ring homomorphism?

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  • $\begingroup$ Yeah, it's a ring homomorphism, if you consider $\{0\}$ to be ring. $\endgroup$ – Karl Kroningfeld Dec 21 '13 at 23:22
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    $\begingroup$ Depends you whether you require a ring homomorphism to send $1$ to $1$. But even you do, then a morphism to the ring $\{0\}$ preserves the unity and is nevertheless the zero map. $\endgroup$ – Stefan Hamcke Dec 21 '13 at 23:24
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    $\begingroup$ The instructor's objection is valid only if you require: (a) your rings to have $1\ne0$; (b) ring homomorphisms map $1$ to $1$. $\endgroup$ – egreg Dec 21 '13 at 23:28
  • $\begingroup$ Actually you want {0} to be a commutative ring with 1 to have a terminal object in that category, I'd say. At least from the definitions I'm used to, OP's point of view makes sense. $\endgroup$ – Louis Dec 22 '13 at 1:26
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We have to distinguish between two algebraic structures with appropriate homomorphisms between them (in fact, they constitute categories)

$\bullet$ Rings (which I always assume to be unital) with homomorphisms of rings.

As always, a homomorphism of blupp preserves the whole structure of blupp, so in particular for blupp=Ring the unit is preserved by definition. This definition is (or should be ...) universally accepted.

$\bullet$ Rngs or non-unital rings with homomorphisms of rngs

Following the general principle, a homomorphism of rngs preserves the addition and the multiplication (and then also additive inverses and the zero), but not the unit because there is none.

There is a so-called forgetful functor from Rings to Rngs which forgets the unit. Notice that if $R$ is a ring, the underlying rng $|R|$ is a different object (and similarly for homomorphisms). This is usually ignored in textbooks and lectures, leading to utter confusions. Similar problems arise with other forgetful functors. These problems will end immediately when we take forgetful functors seriously. Unfortunately, some authors consider rings with unit but consider also non-unital homomorphisms between them, which doesn't make sense at all because this actually ignores the general principles of universal algebra, secretly applies the forgetful functor all the time, and doesn't take rings seriously. And some authors even don't consider $0$ as a ring because "it has no unit", sic! Of course the zero ring is a ring. It's quite cold these days, so books following this approach would make a warm fire ...

Anyway, now the question is easily answered:

Let $f : R \to R'$ be a homomorphism of rings (thus preserving the unit by definition). Then $f=0$ iff $R'=0$. It doesn't matter if $R$ is a field or not, the condition doesn't depend on $R$ at all.

On the other hand, there is always a zero homomorphism of the underlying rngs $0 : |R| \to |R'|$. Notice, again, that these are different objects than the rings themselves!

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    $\begingroup$ I don't think the paragraph here which promises to answer the question actually does that -- in the question the field is the domain of the arrow, but your paragraph speaks about the codomain being a field. (Also, your explanation seems to ignore the fact the zero ring is in fact a unital ring, according to most definitions, and the zero map $R\to 0$ is always a ring homomorphism). $\endgroup$ – hmakholm left over Monica Dec 22 '13 at 1:02
  • $\begingroup$ Read carefully. Actually I've answered more than just the question. But OK I've edited it so that everyone understands ... $\endgroup$ – Martin Brandenburg Dec 22 '13 at 1:22
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If we assume that your instructor is competent, then when he asks "is the zero map a homomorphism?", he's not necessarily claiming that it never is, but merely pointing out that your answer is incomplete until you include some discussion of the circumstances in which it is a homomorphism -- namely that it is a (unital) homomorphism exactly when the codomain is the zero ring.

Of course the homework problem is flawed when it asks you to prove something that isn't strictly true. That just means that the perfect solution will point out the flaw and find a way to repair it with complete rigor. It sounds like it may have been unclear from your solution whether you were wrong (and thought the zero map is always a solution), or you were right (and were contemplating the zero ring) but just failed to write your correct thoughts down completely.

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