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

Question

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 $R$.

Answer 1

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.

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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.

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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!

Answer 2

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 $R$.