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I am writing some strict algebra notes and am stuck with the most basic stuff, as is the First Isomorphism theorem for Rings. I am working over commutative rings with unity. The problematic part of the formulation of the First Isomorphism theorem for Rings says that if $f: R \to S$ is a ring homomorphism, then the image of $f$ is a subring of $S$. By definition, the subring of $S$ has to contain $1$ (neutral for multiplication in $S$). This is not correct: $f(x) = 0$ for all $x \in R$ is a ring homomorphism (why not, $f(a+b) = f(a)+f(b)$ and $f(ab) = f(a)f(b)$, that is how it is defined for all rings). Hence, here $0$ is not a subring of $S$! What am I missing? Is homomorphism of rings with unity defined with $f(1) = 1$ to escape such ridiculous situations? As even if we put that $f$ is not a zero homomorphism, we still do not get that $f(1) = 1$ (i.e. why would $f(1)*a = a$ for $a \notin \Im(f)$ ($\Im(f)$ is my notation for the image of $f$)). We need that $f$ is surjective to prove that $\Im(f)$ is a subring of $S$... Once again, what am I missing?

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  • $\begingroup$ $f(x) = 0$ is a ring homomorphism. $\endgroup$
    – Afntu
    Commented Apr 23 at 7:04
  • $\begingroup$ $f(x)=0$ is a rng homomorphism, but it is only a ring homomorphism if the codomain is the zero ring. $\endgroup$ Commented Apr 23 at 8:01
  • $\begingroup$ also, a note on typesetting: the command \Im (i.e. $\Im$) is supposed to be used for the imaginary part of a complex number. Compare with \Re (i.e. $\Re$). $\endgroup$ Commented Apr 23 at 8:02
  • $\begingroup$ Beware that many common results can fail if ring homs / subrings don't preserve $1$, e.g. being coprime $(a,b)=(1)$ need not be preserved - see the counterexample I gave here. $\ \ $ $\endgroup$ Commented Apr 23 at 15:20

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If you are working over rings with unity, then usually the morphisms are unital morphisms, i.e. the condition $f(1)=1$ is part of the definition of a morphism.

There's another notion, approppiate for "rings that don't necessarily have unity", nowadays sometimes called rngs to avoid confusion (you drop the i because there's no multiplicative identity). Since rngs may not have multiplicative identity, a rng morphism is just required to preserve addition and multiplication. Since addition forms a group, you get $f(0)=0$ automatically, but again there's no $f(1)=1$ requirement ($1$ may not even exist in either the domain or the codomain rng).

Of course, all rings are also rngs, so if $R,S$ are actually rings (they have identity) you can consider either rng morphisms or ring morphisms between them. The image of a rng morphism $f:R\to S$ is a subrng of $S$ but not necessarily a subring. However, if $f$ is a ring morphism (i.e. $f(1)=1$) then $f(R)$ is a subring of $S$, i.e. contains the identity of $S$.

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    $\begingroup$ rng and ring is one of my more favourite instances of math humour :) $\endgroup$
    – peek-a-boo
    Commented Apr 23 at 7:49
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$f : R \to S$ is a ring homomorphism, now if $R$ is a commutative ring with unity, then $R$ contains the unity, but $S$, may not contain the unity, for example, $f: R \to S$ by $f(x) = 0$ is a ring homomorphism, and $f(R) = \{0\} \subseteq S$. You are correct that the subring of a commutative ring also contains unity, but in this example, $S$ is not a subring of $R$, so your claim is incorrect. In this particular example $f(1) = 1.0 =0$.

Actually, If $S \neq \{0\}$ and $f$ is onto map then we can say $f(1)$ is the unity of $S$. But in this case, the unity of $R$ is mapped to the unity of $S$.

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To answer your first question: $f(x) = 0$ is a ring homomorphism into the zero ring, in which $1=0$ (you can easily check that this satisfies all the ring axioms). The fact that for a homomorphism $f: R\rightarrow S$ of rings with unity we have $f(1_R) = 1_S$ is actually not a definition but immediately follows from the fact that $f$ preserves the multiplicative structure: Let $0\neq r\in R$. Then $f(r) = f(1_Rr) = f(1_R)f(r)$ from which it follows that $f(1_R) = 1_S$, but again, it should be stressed that in $S$ we could have $1=0$.

EDIT: apparently, it is indeed a definition. See the comment below.

Second, it does not matter if $a$ lies within the image of $f$. You can multiply an element of a subring with an element not in the subring. Choose for example the embedding homomorphism $g:\Bbb Z \rightarrow \Bbb Q$, $x\mapsto x/1$. Then, for coprime integers $a$, $b$, certainly $a/b$ does not lie within the image of $g$, but we can still write $g(1_\Bbb Z)\cdot a/b = 1_\Bbb Q\cdot a/b = a/b$.

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    $\begingroup$ Consider $f:\mathbb Z\to\mathbb Z^2$ given by $f(n)=(0,n)$. It preserves addition and multiplication but $f(1)\neq (1,1)$. Your argument works if $f$ is surjective, though. $\endgroup$ Commented Apr 23 at 7:52
  • $\begingroup$ that is an interesting case: we then would have $\mathrm{im} f \cong 0\times\Bbb Z$, and indeed all $(a,1)$ would act as an identity on $\mathrm{im} f$. I will have to consider this further. $\endgroup$
    – paulina
    Commented Apr 23 at 7:59
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It depends on what you consider as part of the data of a ring. If you formally think of a ring $R$ as a tuple $(|R|, +, 0, \cdot, 1)$ satisfying certain axioms, where $|R|$ is a set, $+$ and $\cdot$ are maps $|R| \times |R| \to |R|$, and $0,1$ are distinguished elements of $|R|$, then a morphism between rings should preserve all of this data.

I.e. if $S = (|S|, \oplus, 0', \times, 1')$ is another ring under this definition, then formally a morphism $\alpha: R \to S$ is a tuple

$$\alpha = (f, R, S) $$

where $f: |R| \to |S|$ is a function between the underlying sets that preserves all of the data of $R$ and $S$, which includes the multiplicative identities $1$ and $1'$.

Why should the elements $0$ and $1$ of a ring should be considered as parts of its data, since additive and multiplicative identities are unique if they exist? A potential answer is that you can imagine a class $\mathcal{A}$ of algebraic structures that have distinguished elements (e.g. an element $x$ satisfying $x^2 = x$) which are not uniquely determined by the axioms, so must be included as part of the data.

So a good definition of an algebraic structure is: a tuple consisting of a set, some operations on the set, and some distinguished elements, along with a list of axioms/laws. A morphism between two algebraic structures of the same type should then preserve both the operations and distinguished elements. This is the subject of universal algebra (see also these notes).

Coming back to rings, a typical definition says something like

there exists an element $1 \in R$ such that $x1 = 1x$ for all $x \in R$.

Pretending we don't know that multiplicative identities are unique, it seems reasonable to think that this element $1 \in R$ is pretty important, and should be encoded within the ring.

At the end of the day, what you consider as part of the data/structure of a mathematical object is a choice, and your choice will affect what a morphism between objects should be. Category theory is a good language for this stuff.

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