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Is the kernel of a ring homomorphism a subring or an ideal?

Dummit & Foote (Abstract Algebra, $3^{rd}$ ed., 2004.) state in Proposition 7.3.5 (2), page 239,

Proposition 5. Let $R$ and $S$ be rings and let $\varphi: R\rightarrow S$ be a homomorphism.

(1) The image of $\varphi$ is a subring of $S$.

(2) The kernel of $\varphi$ is a subring of $R$. Furthermore, if $\alpha \in \operatorname{ker} \varphi$ then $r \alpha$ and $\alpha r \in\operatorname{ker} \varphi$ for every $r \in R$, i.e., $\operatorname{ker} \varphi$ is closed under multiplication by elements from $R$.

Four pages later, page 243, in the First Isomorphism Theorem, D&F state,

Theorem 7.

(1) (The First Isomorphism Theorem for Rings) If $\varphi: R \rightarrow S$ is a homomorphism of rings, then the kernel of $\varphi$ is an ideal of $R$, the image of $\varphi$ is a subring of $S$ and $R / \operatorname{ker} \varphi$ is isomorphic as a ring to $\varphi(R)$.

(2) (omitted)

What about the distinction between and the (possibly different) roles played by subrings and ideals? Part of the issue is whether a particular ring has an identity $1_R$. And, for the ring homomorphism, $\varphi: R \rightarrow S$, does $\varphi(1_R) = 1_S$? This seems to be a matter of definition,

  1. whether a ring R has an identity, $1_R$, and
  2. whether a ring homomorphism $\varphi: R \rightarrow S$ should map $1_R \mapsto 1_S$.

P. Aluffi (Algebra: Chapter 0, 2009.) discusses subrings and ideals on page 139, stating, "Ideals are close to being subrings . . ." He concludes that ideals are more important, claiming in a footnote that, "ideals are precisely the submodules of a ring $R$.

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    $\begingroup$ Some people require rings to have a unity, and morphisms to respect the unity. In that setting, you would also necessarily require subrings to share the unity of the whole ring; Aluffi seems to be in that group. If you require a subring to have a unity and in fact share the unity of the overring, then only the ideal $(1)$ is also a subring. On the other hand, others do not require a unity (or not require morphisms to send the unity to the unity), in which case a subring would just be a subgroup that is closed under multiplication, making ideals a subring. It's a matter of convention. $\endgroup$ Jul 31, 2023 at 17:57
  • $\begingroup$ Kernel is an ideal. It cannot be a subring, as you want $1_R$ to be in your ring, and thus the ring homomorphism would need to be the zero map. $\endgroup$ Jul 31, 2023 at 17:57
  • $\begingroup$ Thank you all for the clarification! After a closer reading of D&F, they only require a "ring homomorphism" to preserve multiplication, in addition to being a group homomorphism. Also, the D&F definition of subring, in addition to being a group, is that it is closed under multiplication. Proposition 5 (2), states the ker 𝜑 is a subring. But, the "Furthermore" clause goes on to say that ker 𝜑 is closed by multiplication of elements from R. This makes ker 𝜑 an ideal. $\endgroup$
    – Cliff
    Jul 31, 2023 at 20:23
  • $\begingroup$ This is an interesting question. There is (obviously) some overlap between subrings and ideals. A subring, being closed under multiplication, must contain the identity element, 1. If an ideal contained the identity element, (1), then it would equal to the whole ring, R. But, an ideal I, is closed under multiplication by elements of the parent ring, R, not necessarily contained in the ideal I. $\endgroup$
    – Cliff
    Sep 14, 2023 at 21:23

2 Answers 2

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The kernel of a ring homomorphism is always an ideal, as noted in both propositions. Whether it is a subring is a matter of definition.


If rings are assumed to have unity, then it is not a subring, except in the case of the "$0$" homomorphism (assuming you do not require $\phi(1) = 1$; if you do, then it is never a subring).

If rings are NOT assumed to have unity, then ideals are subrings, and, since kernels are ideals, they are also subrings.


Ring theory, just like other branches of mathematics, has a bit of a terminology issue, in which different authors will use the same name for different structures. I'd say "ring" is one of the terms with the most distinct definitions - commutativity, associativity and unity are all included or excluded depending of the context!

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  • $\begingroup$ Thank you all! After a closer reading of D&F, they only require a "ring homomorphism" to preserve multiplication, in addition to being a group homomorphism. Also, the D&F definition of subring, in addition to being a group, is that it is closed under multiplication. $\endgroup$
    – Cliff
    Jul 31, 2023 at 20:15
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Let $\phi:R \to S$ be a ring homomorphism. I claim ker$\phi$ is an ideal of $R$. To show this, we must show for every $x,y \in$ ker$\phi$, we have $x-y \in$ ker$\phi$ and for every $x \in$ ker$\phi$ and all $r \in R$ one has $rx \in$ ker$\phi$. Let $x,y \in$ ker$\phi$, Then to check if $x-y \in$ ker$\phi$, we check if $\phi(x-y)=0$, note that

\begin{align} \phi(x-y)&=\phi(x)-\phi(y) && \text{as $\phi$ is a ring homomorphism}\\ &=0 - 0 && \text{as $x,y \in$ ker$\phi$}\\ &=0. \end{align} Thus $x-y \in$ ker$\phi$.

Next let $x \in$ ker$\phi$ and $r \in R$, then

\begin{align} \phi(rx)&=\phi(r)\phi(x) && \text{as $\phi$ is a ring homomorphism}\\ &=\phi(r) \cdot 0 && \text{as $x \in$ ker$\phi$}\\ &=0 && \text{as $0 \cdot s=0$ for every $s \in S$} \end{align}

This forces $rx \in$ ker$\phi$, thus ker$\phi$ is an ideal of $R$.

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