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Any homomorphism $φ$ between the rings $\mathbb{Z}_{18}$ and $\mathbb{Z}_{15}$ is completely defined by $φ(1)$. So from

$$0 = φ(0) = φ(18) = φ(18 \cdot 1) = 18 \cdot φ(1) = 15 \cdot φ(1) + 3 \cdot φ(1) = 3 \cdot φ(1)$$

we get that $φ(1)$ is either $5$ or $10$. But how can I prove or disprove that these two are valid homomorphisms?

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Ring homomorphisms must fix the identity. – user126 Jul 21 '10 at 10:12
In this case we want to relax that requirement – Tomer Vromen Jul 21 '10 at 10:49
any homomorphism ( ring and group ) must fix identity. – Surya Nov 17 '13 at 23:52
up vote 19 down vote accepted

If one has a homomorphism of two rings $R, S$, and $R~$ has an identity, then the identity must be mapped to an idempotent element of $S$, because the equation $x^2=x$ is preserved under homomorphisms. Now $5$ is not an idempotent element in $\Bbb Z_{15}$, so the map generated by $1 \to 5$ is not a homomorphism.

However, 10 is an idempotent element of $\Bbb Z_{15}$. In particular, the subring $T \subset \Bbb Z_{15}$ generated by 10 has unit 10. Since it is annihilated by 3, and consequently by 18, there is a unital homomorphism $\Bbb Z_{18} \to T$ (i.e., mapping $1$ to $10$). So your second map is a legitimate homomorphism of rings (composing with the injection $T \to \Bbb Z_{15}$).

Basically, the point of this answer is to check that one of your maps preserves the relations of the two rings, while the other doesn't.

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What do you mean by $T \subset Z_{10}$? In fact, what do you mean by $T$? – Casebash Jul 21 '10 at 13:11
Ack, I meant $Z_{15}$. $T$ is the subring of $Z_{15}$ generated by 10 (basically, that corresponds to the residue classes of 0, 5, 10). – Akhil Mathew Jul 21 '10 at 13:30
Okay, so the annihilation leads to the mapping being a homomorphism as both rings are cyclic. – Casebash Jul 21 '10 at 13:43
Precisely (to give a homomorphism from $Z_n$ to a ring is to give an idempotent element of the ring which is annihilated by $n$). – Akhil Mathew Jul 21 '10 at 13:54
@Akhil: It is not a subring. It is a subrng, or a pseudo-subring. – user126 Jul 21 '10 at 17:09

Continuing Akhil M's answer above (the comments under it are getting pushed down out of sight): it is also not hard to systematically find all the idempotents in ${\mathbb Z}/n$. Namely, for example with $n=pq$ with distinct primes $p,q$, $\mathbb Z/n \approx \mathbb Z/p \oplus \mathbb Z/q$, by Sun-Ze's theorem (altho' one might carp about what kind of "sum" it is). So, solving the idempotent condition $x^2=x$ mod $pq$ is equivalent to solving that equation mod $p$ and mod $q$. The integers mod a prime form a field, so we know that there are only the two solutions, the obvious ones, $0,1$. Thus, the idempotents mod $pq$ are 0-or-1 mod $p$ and $0-or-1$ mod q. Obviously 0 and 1 mod pq work, but/and also 0 mod p but/and 1 mod q, and vice-versa. In the case at hand, both 6 and 10 are non-obvious idempotents.

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I will quote Wikipedia on the definition of a ring homomorphism.

More precisely, if R and S are rings, then a ring homomorphism is a function f : R → S such that1

  • f(a + b) = f(a) + f(b) for all a and b in R
  • f(ab) = f(a) f(b) for all a and b in R
  • f(1) = 1

The last requirement is being relaxed. We can then sub in our f, in this case f(a1)=5a or f(a1)=10a and see if these relations hold. Given that these are small, finite and well understood groups, the problem is easy to solve from here.

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