Prove the Ring Homomorphism is Surjective

Question

Prove the following homomorphism is surjective. $$f : Z → \frac{Z}{3Z} × \frac{Z}{5Z}$$

I completely get the questions and i can prove it by working out a corresponding pre-image for all of $\frac{Z}{3Z} × \frac{Z}{5Z}$.

But how do i give a formal proof if i change the question and ask

for any $(p,q)=1$ Prove the following homomorphism is surjective. $$f : Z → \frac{Z}{pZ} × \frac{Z}{qZ}$$

I feel we may have to use Bezout's identity but don't know how to. any help appreciated

Thanks

@PedroTamaroff I think the idea is to send z to (z+pZ,z+qZ). — Jacob Wakem, Nov 19 '16 at 18:33
It is sufficient to produce (1,0) as a homomorphic image. (0,1) is produced similarly. — Jacob Wakem, Nov 19 '16 at 18:34

score 1 · Accepted Answer · answered Mar 03 '15 at 00:10

1

Hint: Use the Chinese remainder theorem to show that $f: {Z/pZ} \times {Z/qZ} \to Z/(pqZ)$ is bijective. That's all you should need.

answered Mar 03 '15 at 00:10

user2566092

26,450

i had that hunch but cant get to see it properly can u elaborate please – Mr. Math Mar 03 '15 at 00:11
elaborated a bit. – user2566092 Mar 03 '15 at 00:13
Hmmm i dont see why thats enough to show its surjective – Mr. Math Mar 03 '15 at 00:14
sorry i meant to say BIJECTIVE. edited again lol – user2566092 Mar 03 '15 at 00:14
i get it is enough now but how i can show this now ? – Mr. Math Mar 03 '15 at 00:15
It really is, in a strong sense, equivalent between your original claim and the chinese remainder theorem. You can look up proofs of the chinese remainder theorem. Just don't get the abstract algebra versions in terms of ideals, you want the elementary number thoery versions in terms of integers and modulo. – user2566092 Mar 03 '15 at 00:17
i got it man/woman Thanks – Mr. Math Mar 03 '15 at 00:18
@MRK IF you don't yet know CRT you get get the surjectiveness immediately from the Bezout idenity - see my answer. – Bill Dubuque Mar 03 '15 at 00:43

Bill Dubuque · Answer 2 · 2024-07-04T22:00:11.513

Yes, we can trivially transform a Bezout identity to a CRT solution as below

$\quad \color{c00}{jp} + kq = 1\,\Rightarrow\, \begin{align}{\color{#c00}{kq\equiv 1}\pmod p}\\ \color{c00}{jp\equiv 1}\pmod q\end{align}\,\ $ thus $\,\ x = a\color{#c00}{kq} + bjp \ \Rightarrow \begin{align}&x\equiv a\!\!\!\pmod p\\ &x\equiv b\!\!\!\pmod q\end{align}$

i.e. $\ \ \ {\rm mod}\,\ (p,q)\!:\,\ \begin{align}\color{#c00}{kq \equiv (\color{#c00}1,0)}\\ \color{c00}{jp\equiv (0,1)}\end{align} \,\Longrightarrow\, (a,b) = a\color{#c00}{(1,0)}+b(0,1) \equiv a\color{#c00}{kq} + bjp $

which reveals innate CRT linearity, i.e. how the general solution $\,(a,b)\,$ can be generated as a linear combination of the "basis" solutions $\,(1,0)\,$ and $\,(0,1).\,$ This will become clearer when one studies the ring-theoretic form of CRT (and modules).

score 0 · Answer 3 · answered Mar 03 '15 at 01:25

Since $\ker f = 15 \mathbb Z$ has index $15$, the image of $f$ has size $15$ and so $f$ is surjective.

The same argument works in general when $\gcd(p,q)=1$.

(Invoking the Chinese remainder theorem sounds wrong to me because the argument above gives the simplest proof of the Chinese remainder theorem when the ring is $\mathbb Z$.)

Prove the Ring Homomorphism is Surjective

3 Answers3

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