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I have the following congruence $P\equiv x+1\pmod 5,$ $P\equiv3 \pmod {x^2+1}$, $P\equiv 3x\pmod{x-2}$ and I am asked if I can solve it in $\mathbb{Z}[X]$.

I know how to solve it in $\mathbb{Z}$ but in this case I am completely lost. Thanks in advance!

Bernard
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hallomate87
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    Hint: Consider what you can deduce about $P(2).$ – Thomas Andrews Oct 26 '21 at 21:42
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    You can apply the technique from integers to solve this in $\mathbb Q[x],$ but there might or might not be a solution in $\mathbb Z[x].$ This is because the Chinese Remainder Theorem algorithm relies on the ring being a Euclidean Domain. – Thomas Andrews Oct 26 '21 at 21:52
  • @ThomasAndrews That final sentence is false. See here for a correct statement. – Bill Dubuque Oct 27 '21 at 21:09
  • @BillDubuque I didn’t say Chinese Remainder Theorem, which only says that there is a solution, I said the algorithm we use in integers to find solutions. For example, there are PIDs which are not Euclidean Domains, but it is clear the being a PID is enough. But the algorithm we use in $\mathbb Z$ to solve $ax+by=\gcd(a,b)$ is to use the Euclidean algorithm. So to use the same algorithm to find solutions, we need a Euclidean domain. Since this question is about solving a CRT problem, not saying a solution exists, I wanted to say when we can generalize the algorithm we use in $\mathbb Z.$ – Thomas Andrews Oct 27 '21 at 21:27
  • It's not clear what "algorithm" you refer to, but as the linked post shows, CRT holds more generally. – Bill Dubuque Oct 27 '21 at 21:33
  • @BillDubuque You’d be correct if I said anything about CRT alone. How do you solve CRT problems? $$x\equiv m_i\pmod{n_i},i=1,2,$$ with $\gcd(n_1,n_2)=1,$ involves solving $$n_1x+n_2y=1.\tag1$$ Then the solution is $$x\equiv n_1xm_2+n_2ym_1\pmod{n_1n_2}.$$ Solving (1) in integers is done using division algorithm, or some descent argument which is equivalent to the division algorithm. The existence of a solution is not enough to solve the equation. – Thomas Andrews Oct 27 '21 at 21:40
  • My remarks refer to the general Theorem (the T in CRT) but you refer to an algorithm for a specific case. The point of my remark was to clarify this distinction to help readers avoid any misunderstandings about the generality of CRT, – Bill Dubuque Oct 27 '21 at 22:09

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