# Do subrings contain 0, the additive identity because $1-1=0$ in subrings as in subfields?

Algebra by Michael Artin Ch3, Ch11

Artin has different definitions of rings particularly that his rings are commutative in both addition and multiplication. Based on his definitions, (*) I believe that $$0$$ is in subrings for the same reason $$0$$ is in subfields: (**) $$1-1=0$$

Am I mistaken?

(*)

Definition of a subring of the ring of complex numbers $$\mathbb C$$ (and then I guess this is extended to a subring of a ring $$R$$)

Definition of a ring

(**)

Earlier, subfields of the field $$\mathbb C$$, fields and subfields of fields were defined similarly.

Definition of a subfields of the field of complex numbers $$\mathbb C$$ (and then I guess this is extended to a subfield of a field $$F$$)

Definition of a field

• By every definition I've ever heard, a subring contains $0$ (the additive identity). You're correct in saying that it is implied by closure under subtraction and inclusion of the element $1$, but alternatively we know that subrings are rings in their own right, and every ring has an underlying additive abelian group (which requires the existence of an identity element, $0$) Commented Oct 23, 2018 at 2:43
• @D.Beec Thanks! So, we could actually say to Artin that 'that subfields are fields in their own right' is an alternative proof than '$1-1=0$' to show $0$ is in every subfield? Hmmm....I mean, that $0$ is in every subfield sounds like part of the proof of showing that every subfield is a field. So, how would you show that every subfield is a field without using $1-1=0$ to show $0$ is in every subfield?
– BCLC
Commented Oct 23, 2018 at 2:48
• Here's an alternative way to consider it: Let $F$ be a field with the given operations of $+$ and $\times$. We call $A$ a subfield of $F$ if $A\subseteq F$ and if $A$ is a field with respect to the same operations $+$ and $\times$. Now, in 3.2.1 Artin is trying to provide a minimal set of properties that if true, show that $A$ is indeed a subfield of $F$. In this list, he doesn't explicitly state "$A$ must contain the additive identity", but that property is implied by this list. He's trying to keep the list short so if you ever have to check for a subfield, you can do it quickly Commented Oct 23, 2018 at 3:02

Subrings contain $$0$$ because they are, in particular, groups (written additively). Recall that a ring is a group written additively with a mutliplicative structure. So, not all rings have 1 but all rings have 0 since that is the identity element in the underlying group.
• It is slightly more subtle than that. Consider that if we don't explicitly require a subring to contain $1$, then the subring may still contain a multiplicative identity that works in that subring. For example $\mathbb Z/6\mathbb Z$ could have $\{0,3\}$ as a subring, with $3$ filling the role of the multiplicative identity. This situation cannot arise for additive identities because a ring is an additive group. In a group the equation $a+a=a$ characterizes the identity uniquely, and therefore the subring's $0$ must necessarily equal the $0$ of the larger ring. Commented Oct 23, 2018 at 2:54
• Thanks RhythmInk and @Henning Makholm! Turns out I might be pretty screwed for the GRE because Artin defines rings to include $1$.