# whether the map is surjective?

question :

the natural map defined by

$\phi :\mathbb Z[x] \rightarrow$ $\mathbb Z[x]/<2> \times$ $\mathbb Z[x]/<x>$ is not surjective .

ans

$\phi(f(x))=(\pi_1(f(x)),\pi_2(f(x)))$ where $\pi_1: \mathbb Z[x] \rightarrow \mathbb Z[x]/<2>$ and $\pi_2: \mathbb Z[x] \rightarrow \mathbb Z[x]/<x>$

now $(0,1) \in \mathbb Z[x]/<2> \times$ $\mathbb Z[x]/<x>$ but $\nexists f(x) \in \mathbb Z[x]$ such that $\phi(f(x))=(0,1)$ as

$f(x) =0 (\mod 2)\tag{1}$ and $f(x) =1 (\mod x)\tag{2}$

from$(1) \implies$ constant term of $f(x)$ is even where as $(2) \implies$ constant term of $f(x)$ is odd .

is this correct?

and also i know chinese remainder thrm. holds in P.I.D does it hold in U.F.D?

here $\mathbb Z[x]$ is a U.F.D. and not a P.I.D also $<x>$ and $<2>$ are prime ideals in $\mathbb Z[x]$

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Your proof is correct. The Chinese remainder theorem (properly stated) is valid in all rings (but congruences are mod ideals, not mod numbers), but nevermind that, you aren't using the Chinese remainder theorem at all. You are saying there's no polynomial that's 1 mod $x$ and 0 mod $2$ because the constant coefficient of such a polynomial must be both odd and even. That's true and doesn't use the Chinese remainder theorem.
actually this is a exercise which is present in the part of the book which talks about chinese remainder thrm. here the map is not surjective but $\ker(\phi)=<2x>$ – jim Feb 16 '13 at 6:35