2
$\begingroup$

Im trying to prove the following version of Chinese Residue Theorem given as an exercise by Rotman in his Advanced Algebra Book which is somehow different from the common versions already proved in almost every book on the theme.

1) Prove that for $k$ field and $f(x),f'(x) \in k[x]$ are relatively prime, then given $b(x),b'(x) \in k[x]$, there exists $c(x) \in k[x]$ such $c-b \in \langle f \rangle$ and $c-b´\in \langle f´ \rangle$. Also, if $d(x)$ is a common solution, then $c-d \in \langle ff' \rangle$.

2)Prove that for $k$ field and $f(x),g(x) \in k[x]$ relatively prime, then

$$\frac{k[x]}{\langle f(x)g(x) \rangle} \cong \frac{k[x]}{\langle f(x) \rangle} \times \frac{k[x]}{ \langle g(x) \rangle}$$.

For the first part I got that there exist $r(x),s(x) \in k[x]$ such $f(x)r(x)+f'(x)s(x)=1$ then I should prove the existence of one $c(x) \in k[x]$ such $c(x)-b(x)=f(x)p(x)$ and $c(x)-b'(x)=f'(x)q(x)$ for some $p(x),q(x) \in k[x]$. Im troubled finding the desired $c(x)$. I understand that im required to prove that for an $d(x) \in k[x]$ such $c(x)-b(x)=f(x)d(x)$ and $c(x)-b'(x)=f'(x)d(x)$ then I should find a $g(x) \in k[x]$ such $c(x)-d(x)=[f(x)f'(x)]g(x)$. Here Im also troubled finding the desired $g(x)$ even if I suppose I already had proved the existence of $c(x)$ mentioned before.

For the second part Im trying to define a ring morphism $$\phi:\frac{k[x]}{\langle f(x) \rangle} \times \frac{k[x]}{ \langle g(x) \rangle} \to k[x]$$ which satisfies the first isomorphism theorem in order to prove the required isomorphism. My intuition says this morphism is given by something like $$\phi(a(x)+ \langle f(x)g(x) \rangle)= (a+ \langle f(x) \rangle, a+ \langle g(x) \rangle)$$. Rotman´s Advanced Algebra book mentions as an hint to think about the fact $\mathbb{Z}_{m} \mathbb{Z}_{n} \cong \mathbb{Z}_{mn}$ for $m$ and $n$ relatively prime.

$\endgroup$
  • $\begingroup$ If none of the answer help, then please let us know where elaboration is needed. $\endgroup$ – Bill Dubuque Apr 6 at 1:50
  • $\begingroup$ I find really good explained answers by you, specially at point (1a) and (1b). @BillDubuque. Also, the isomorphism you described in (2) was really helpful along with DavidHill. No problem finishing the details of the proof with the help given by both. I really aprecciate your help! $\endgroup$ – Cos Apr 6 at 1:56
  • $\begingroup$ Glad to help. Don't forget to upvote and accept (without any such it is possible the question might get auto-deleted by the system).. Worth emphasis: the proofs are nor essentially different than the classical proof for integers, except some use more ideal-theoretic language. You might find it instructive to carefully compare with the integer case. The same proofs work in any PID. More generally the rings where CRT holds are known as Prufer domains, which can be characterized by many interesting properties ($27$ listed there). $\endgroup$ – Bill Dubuque Apr 6 at 2:08
  • $\begingroup$ @BillDubuque Already upvoted your answers! Wow, Interesting . Didnt have heard about Prufer Rings, I will be reading more about this for sure. Thanks for pointing those facts Yep, im gonna work in comparing this proof with the integers case and ideal oriented soon, is just i dont see the connection with ideals generated by one single polynomial version, yet :P $\endgroup$ – Cos Apr 6 at 4:06
1
$\begingroup$

$(1a)\ \ \ b-b'\in(f,f')=(1)\,\Rightarrow\, b-b' = hf+h'f',\ $ so $\ b-hf = b'+hf' =: c\,$ works.

$(1b)\ \ \ c-b,d-b \in (f)\,\Rightarrow\,c-d\in (f).\,$ Similarly $\,c-d\in (f')\ $ thus $$\,c-d\in (f)\cap (f') = ({\rm lcm}(f,f')) = (ff')\ \ {\rm by}\ \ \gcd(f,f')=1$$

$(2)\ \ $ By $(1a)$ we know $\,c\mapsto (c+(f),c+(f'))\,$ maps $\,k[x]\,$ onto $(k[x]/f,\, k[x]/f')$

By $(1b)$ it has kernel $\,(ff')\,$ so $\,k[x]/(ff')\cong (k[x]/f,\, k[x]/f')\,$ by the First Isomorphism Theorem.

$\endgroup$
1
$\begingroup$

For the first part, write $1=fr+f's$. Then $fr+(f')=1+(f')$ and $f's+(f)=1+(f)$. Set $c=frb'+f'rb$. Then $c+(f')=frb'+(f')=b'+(f')$ and $c+(f)=f'sb+(f)=b+(f)$.

You should define $\phi:k[x]\to k[x]/(f)\times k[x]/(g)$, by $\phi(h)=(h+(f),h+(g))$.

First, note that $\phi$ is surjective. Since $\gcd(f,g)=1$ you can write $1=fa+gb$ for some polynomials $a,b\in k[x]$. It follows that $fa+(g)=1+(g)$ and $gb+(f)=1+(f)$. Therefore, given $(p+(f),q+(g))\in k[x]/(f)\times k[x]/(g)$, let $r=faq+gbp$ as in part 1. Then, $r+(f)=gbp+(f)=p+(f)$ and $r+(g)=faq+(g)=q+(g)$. Hence, $$\phi(r)=(p+(f),q+(g)).$$

Now, we just need to check that $\ker\phi=(fg)$. Certainly, $(fg)\subseteq\ker\phi$. For the reverse inclusion, note that if $h\in\ker\phi$, then $f|h$ and $g|h$. Since $\gcd(f,g)=1$, it follows that $fg|h$, so $h\in(fg)$. The result now follows from the first isomorphism theorem.

$\endgroup$
0
$\begingroup$

Hints:

Question 1: start from a Bézout's relation $\;f(x)r(x)+f'(x)s(x)=1$, and calculate the congruence classes of $ b'(x)f(x)r(x)+b(x)f'(x) s(x)$ modulo $f(x)$ and modulo $f'(x)$.

Question 2: Determine the kernel of the natural homomorphism \begin{align} \varphi:k (x )&\longrightarrow k[x]/\bigl(f(x)\bigr)\times k[x]/\bigl(f'(x)\bigr)\\ g(x)&\longmapsto \bigl(g(x)\bmod f(x), g(x)\bmod f'(x)\bigr) \end{align} and use question 1 to determine the inverse isomorphism of the quotient map $$\bar\varphi:k[x]/\ker \varphi \xrightarrow{\quad\sim\quad} k[x]/\bigl(f(x)\bigr)\times k[x]/\bigl(f'(x)\bigr)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.