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By "pseudo ring homomorphism", I mean a map $f: R \to S$ satisfying all ring homomorphism axioms except for $f(1_R)=f(1_S)$.

Even if we let this last condition drop, there are only two ring pseudo-homomorphisms from $\Bbb Z$ to $\Bbb Z[\omega]$ where $\omega$ is any root of unity, for example. They are the identity homomorphism and the "$0$ pseudo-homomorphism".

I couldn't find a non-zero example. It is easy to conclude that $0=f(a)(1_S-f(1_R))$, so I know that the counterexample will have to be some $S$ with zero-divisors.

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marked as duplicate by Mario Carneiro, USER91500, probablyme, T. Bongers, Claude Leibovici Jan 21 at 6:05

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The embedding of a ring $A_1$ into a direct product of nontrivial rings $A_1 \times A_2$ is an example. – darij grinberg Jan 20 at 23:57
Oh, I had never thought of those embeddings in the right way! Thanks! I wonder if people have thought of other examples too... – Rodrigo Jan 21 at 0:04
Shoiuld $\: f(1_S) \:$ be replaced with $1_S$? $\;\;\;\;$ – Ricky Demer Jan 21 at 5:32
up vote 4 down vote accepted

Every commutative example takes the following form. Suppose $f : R \to S$ is a non-unital ring homomorphism between two commutative rings. Then $f(1_R)$ is some idempotent $m \in S$, as Jendrik Stelzner remarks. $mS$ is a "non-unital" subring of $S$ (it's a subring except that its unit is $m$, not $1_S$), and $f$ is a ring homomorphism in the ordinary sense to this subring. Moreover, $S$ decomposes as a product of rings

$$S \cong mS \times (1 - m)S.$$

So darij's comment essentially exhausts all examples.

Geometrically such a morphism corresponds to a "partially defined" morphism $\text{Spec } S \to \text{Spec } R$ of affine schemes, where "partially defined" means defined on some union of connected components. There are analogous statements one can make about non-unital C*-algebra homomorphisms between commutative C*-algebras.

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Because $f(1_R) = f(1_R^2) = f(1_R)^2$ be know that $1_R$ must be mapped to an idempotent element of $S$. We can use this to construct some counterexamples:

Take for example a commutative ring $R$ and a not-necessarily commutative or unital (but associative) $R$-algebra $A$. For any idempotent $e \in A$ the map $R \to A, r \mapsto re$ does the trick. This map is only unital if $A$ is unital and $e = 1_A$.

One examples of this is a ring (i.e. $\mathbb{Z}$-algebra) $S$ with an idempotent $e \in S$. (If $ne = 0$ for some $n \geq 2$ then this also results a pseudo ring homomorphism $\mathbb{Z}/n \to S$. Matt Samuel’s answer is an example of this with $S = \mathbb{Z}/6$, $e = 3$ and $n = 2$.)

If $k$ is a field then the matrix algebra $\mathrm{M}_n(k)$ gives us lots of idempotents (unless $k$ is finite and $n$ is small, or $n = 1$), as there is a bijection between the idempotents of $\mathrm{M}_n(k)$ and the direct sum decompositions $k^n = U \oplus V$ (where we regard $U \oplus V$ and $V \oplus U$ as two different decompositions).

Another class of examples arises by taking a product of unital rings $\prod_{i \in I} R_i$ and considering the idempotent $e_j = (\delta_{ij})_{i \in I} \in \prod_{i \in I} R_i$; we can use this to construct the pseudo ring homomorphism $R_j \to \prod_{i \in I} R_i$, $r \mapsto r e_j$, which is only unital if $|I| = 1$.

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For your matrix algebra example, if $k$ is say $\Bbb C$, where there are lots of idempotents, and $n=1$ so that $M_1=\Bbb C$, I don't see the bijection between the idempotents of $\Bbb C$ and the direct sum decompositions of $\Bbb C^1$ (of which there is only one) – Rodrigo Jan 21 at 15:08
Every idempotent $E \in M_n(\mathbb{C})$ gives a direct sum decomposition $\mathbb{C}^n = \ker E \oplus \operatorname{im} E$. On the other hand every decomposition $\mathbb{C}^n = U \oplus V$ results in a linear map $\mathbb{C}^n = U \oplus V \to \mathbb{C}^n$ given by $u+v \mapsto u$. These two constructions are inverse to each other, granted that we distinguish between the two decompositions $U \oplus V$ and $V \oplus U$. I also added that $n \geq 2$, because in the special case of $n = 1$ there are only two decompositions $\mathbb{C}$, corresponding to $0, 1 \in M_1(\mathbb{C})$. – Jendrik Stelzner Jan 21 at 15:23

Try the homomorphism $$\mathbb Z_2\to \mathbb Z_6$$ given by $$[1]_2\mapsto [3]_6$$

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In particular, $f(a) = f(a)f(1_R)$ implies $f(1_R) = f(1_R)^2$ so $f(1_R)$ is an idempotent and $\mathbb{Z_6}$ is the smallest ring of the form $\mathbb{Z_n}$ that has a non-trivial idempotent. Idempotency is also critical for the map to be a ("pseudo") homomorphism: $$f(a)f(b) = (3a)(3b) = 3^2 ab = 3 ab = f(ab)$$ – Derek Elkins Jan 21 at 0:36
Note also that $\Bbb Z_6$ factors into the isomorphic ring $\Bbb Z_2\times\Bbb Z_3$ (the elements are $(0,0),(1,1),(0,2),(1,0),(0,1),(1,2)$ in $\Bbb Z_6$ order), so this can also be rewritten in the direct product characterization. – Mario Carneiro Jan 21 at 4:38

The (conceptually) simplest example I know is $f:\Bbb Z\to\Bbb Z\times\Bbb Z$ defined by $f(n)=(n,0)$. The ring identity of $\Bbb Z\times\Bbb Z$ is $(1,1)$ but $1$ is instead mapped to $(1,0)$. The same example shows that a subset of a ring that is a ring is not necessarily a subring if you don't require it to contain the ring unit.

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This example can of course be generalized, as darij and Qiaochu point out, but I think it is better to lead with the simple case. – Mario Carneiro Jan 21 at 4:34

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