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Is this possible? I used this as an example to show that there is a UFD $R$ where $2$ elements $a$ and $b$ are coprime but there may not be $r$, $s$ in $R$ such that $ra + sb = 1$. I know that $2$ and $x$ is one example (but I didn't come up with that during my exam).

Gibbs
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Nam Le
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    If $1$ is a linear combination of $x^2+1$ and $x^2+4$, then $1$ is also a linear combination of $x^2+1$ and $3$. Going mod 3, is $x^2+1$ a unit in $\mathbb{Z}/3\mathbb{Z}[x]$? – math54321 Mar 02 '23 at 03:55
  • Or instead of reducing mod $3$ (as in prior comment) we could instead reduce mod $x^2+1$ to reach the contradiction that $,3\mid 1,$ in $,\Bbb Z[i]\cong \Bbb Z[x]/(x^2+1) = $ Gaussian integers. – Bill Dubuque Mar 02 '23 at 04:03
  • Or combine the above two comments using quotient reciprocity to get

    $$\Bbb Z[i]/3 ,\cong, \Bbb Z[x]/(3,x^2+1) ,\cong, \Bbb Z_3[x]/(x^2+1)\qquad\qquad$$ so $,(x^2+4,x^2+1) = (3,x^2+1)\neq (1),$ by LHS (or RHS) ring above is not the zero ring.

     – Bill Dubuque Mar 02 '23 at 04:13
  • Thank you all! I get it now! I'll probably get points docked for not enough justification for my claim but at least now I know I didn't make any false claims. – Nam Le Mar 02 '23 at 06:09

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The linear combinations of $x^2+1$ and $x^2+4$ are the same as those of $x^2+1$ and 3 (in both cases with coefficients in $\mathbb Z[x]$ - this detail is missing in the OP but it seems clear that this is what is meant) ... in other words the two pairs generate the same ideal of $\mathbb Z[x]$. (It seems also clear that $x$ is an indeterminate here - it seems to me more usual to use a capital $X$ instead.) Now in a relationship like $PA+QB=1$ one can take values at any element in any commutative associative algebra with unit over the base ring (here $\mathbb Z$) - in this case this means an element of any commutative ring. Take the imaginary unit $i$ in the ring $\mathbb Z[i]$ of Gaussian integers mentioned in a comment above. We get $P(i)A(i)+Q(i)B(i)=1$ and in case that $A=x^2+1$ and $B=3$ it follows from $A(i)=0$ that $3 Q(i) = 1$. But values of polynomials are in the considered algebra, i.e. here in the ring of Gaussian integers and we get $3(a+bi)=1$ with $a, b$ ordinary integers, which is clearly impossible.

Ulysse Keller
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