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show that in $\mathbb{Z}[x]$, $(x^2+1,3x+2)=(3x+2,13)$ directly by ideal manipulation (or any easy method). More generally, can we show $(x^2+1,ax+b)=(ax+b,a^2+b^2)$? Should I impose some extra conditions on $a,b$ like $\gcd(a,b)=1$? And even more, can we replace $x^2+1$ by any irreducible polynomial over $\mathbb Z$?

To supplement the background, I want to show $\mathbb Z_i/(3i+2)$ is a field with 13 elements (I know this question is duplicated, but i'm asking for a specific technical detail which many other answers skip).

Bill Dubuque
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Westlifer
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  • Duplicate e.g. $,(a,b)!=!(1),\Rightarrow, (\color{#0a0}{ax}!+!\color{#c00}b,x^2!+!1) = (ax!+!b,\color{#0a0}{a^2}(\color{#0a0}{x^2}!+!1)) = (ax!+!b,(-\color{#c00}b)^2!+!a^2).,$ But if $(a,b)=c>1$ then they are unequal, else reducing $!\bmod c,$ yields $,(x^2!+!1) = (0).\ $ – Bill Dubuque Nov 17 '24 at 10:40
  • Above we used $,(r,st) = (r,(r,s)t),$ in the coprime case $,(r,s)=(ax+b,a) =(b,a)=1,,$ then followed by ideal mod reduction $(r,s) = (r,s'),$ if $,s\equiv s'\pmod{!r},,$ here $,\color{#0a0}{ax}\equiv -\color{#c00}b\pmod{!ax!+!b}.\ \ $ – Bill Dubuque Nov 17 '24 at 11:06
  • See the nonmonic generalization of the Euclidean algorithm here in the linked dupe, cf. above comments. $\ \ $ – Bill Dubuque Nov 17 '24 at 11:14
  • @Bill Dubuque Thx for your answer, I know $(r,st)=(r,(r,s)t)$ in number theory, but in proving it (or it's special case where $(r,s)=1$) I need bezout's identity which doesnt hold in $\Bbb Z[x]$. you quoted a link to show this without bezout's, but that post is about GCD, not ideal, he used $(f,g)=(\gcd(f,g))$. I don't really get it, since in $\Bbb Z[x], (f,g)\neq (\gcd(f,g))$. I appreciate your patient explaination. – Westlifer Dec 19 '24 at 12:25
  • The linked proof works for both gcds and ideals because the laws used hold for both, so just read $(f,g)$ as an ideal vs. gcd. Above the ideal $(r,s) = (ax+b,a) = (b,a)=(1)$ because $,(b,a) = (\gcd(b,a)),$ in $\Bbb Z$ (being Euclidean $\Rightarrow$ PID). If anything remains unclear please identify the step that you doubt. $\ \ $ – Bill Dubuque Dec 19 '24 at 20:31
  • @Bill Dubuque OK so we want to show $(f,gh)=(f,\gcd(f,g)h)$ as ideals in $\Bbb Z[x]$ right? What i'm not sure is the step where $(\gcd(a,b)c,d)=(ac,bc,d)$ in the linked proof. This of course is true in $\Bbb Z$ when '(a,b)' means gcd. but I doubt this in $\Bbb Z[x]$ as ideals. He used so-called 'associative law': $\gcd(\gcd(a,b),c)=\gcd(a,b,c)$, but for ideals in $\Bbb Z[x]$ we don't have $(\gcd(f,g),h)=(f,g,h)$. Thanks in advance – Westlifer Dec 20 '24 at 07:31
  • For ideals $F,G,H$ we have $(F,GH) = (F,FH,GH) = (F,(F,G)H),,$ i.e. $,F+GH = F+FH+GH = F+(F+G)H.,$ OP is principal case $,F,G,H = (f),(g),(h).,$ See also here. $\ \ $ – Bill Dubuque Dec 20 '24 at 07:50
  • @Bill Dubuque Oh I got it. $(f,g)h$ is meant to be ideal product, not $\gcd(f,g)h$ in this case. I made a silly misunderstanding. Now it's all clear to me. Thanks for your patient help! – Westlifer Dec 20 '24 at 08:38

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