Help proving an alternative version of Chinese Residue Theorem in the ring of polynomials.

Question

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.

Answer 1

$(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.

Answer 2

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 $$\bar\varphi:k[x]/\ker \varphi \xrightarrow{\quad\sim\quad} k[x]/\bigl(f(x)\bigr)\times k[x]/\bigl(f'(x)\bigr)$$

Answer 3

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.