I have a question that comes from here (only regarding commutative rings).

Firstly, the theorem: Let $\varphi : R\to S$ be a ring homomorphism. If $S$ is integral, then $\ker\varphi$ is a prime ideal of $R$.

Proof. Preimages of prime ideals are prime under ring homomorphisms which follows directly from the definition. Since $S$ is integral, $(0)$ is prime and thus $\ker\varphi =\varphi^{-1}(0)$ is prime.

Now however we get the contradiction that the trivial homomorphism is not a homomorphism since $R$ can not be a prime ideal. How can this be solved? I mean, clearly $\varphi :R\to S,\ r\to 0$ is a ring homomorphism with $\ker\varphi =R$.

  • 1
    $\begingroup$ A ring homomorphism should send 1 to 1. $\endgroup$ – Ahr Jan 10 '18 at 13:19
  • $\begingroup$ But what if $R$ doesn't have a 1? $\endgroup$ – Buh Jan 10 '18 at 13:20

There are two conventions for what the word "ring" means.

The relevant convention here is the one where the multiplicative unit is part of the ring structure; any ring homomorphism $\varphi : R \to S$ must have $\varphi(1_R) = 1_S$.

  • $\begingroup$ In my lecture we called those "unitary ring homomorphisms". Doesn't this restrict the theory quite a bit? Ring homomorphisms can be implemented on rings without a unit, can't they? Your statement basically implies that there is no ring homomorphism onto the zero ring which - I can imagine - causes a lot of trouble? $\endgroup$ – Buh Jan 10 '18 at 13:24
  • $\begingroup$ @Buh: Actually, the zero ring does have a multiplicative unit. If $Z$ is the zero ring, then $1_Z = 0_Z$. $\endgroup$ – Hurkyl Jan 10 '18 at 13:27
  • $\begingroup$ @Buh: I wouldn't say it restricts the theory, but instead it changes the theory. The study of associative $\mathbb{Z}$-algebras and algebra morphisms simply has a very different flavor from the study of unital rings and unit preserving morphisms. $\endgroup$ – Hurkyl Jan 10 '18 at 13:28
  • $\begingroup$ @Buh: But you can actually pass back and forth between them. You've mentioned one direction; in the other you can actually identify any associative $\mathbb{Z}$-algebra $R$ with its unitalization: the ring whose underlying set is $\mathbb{Z} \times R$ and with operations $(a,b)+(c,d) = (a+c,b+d)$ and $(a,b) \cdot (c,d) = (ac, ad+bc)$. $\endgroup$ – Hurkyl Jan 10 '18 at 13:29
  • $\begingroup$ Okay, you get me there, hehe. But still: $\varphi: \mathbb Z\to \mathbb Z,\ z\mapsto 0$ fullfills all requirements to be a ring homomorphism (aside the unity) so I see no reason to exclude it. What is the reason behind this? $\endgroup$ – Buh Jan 10 '18 at 13:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.