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Let $\mathit I $$\,$ be the ideal of $\mathbb Z[x] $ generated by $\mathit x^2+1$ and $ 3x+1$. Find the positive integer $\mathit n$ such that $\mathit I$ $\cap \mathbb Z$ = $\mathit n$$\mathbb Z$

I find $(\mathit x^2+1)(9)+ (3x+1)(-3x+1)= 10 $, So It is expected that 'n =10' But I can't find for n=1, 2, 5

Any hint will be appreciated.

s ss
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1 Answers1

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Hint: $\ (x^2\!+\!1)f + (1\!+\!3x)g = n\!\iff\! \bmod x^2\!+\!1\!:\ 1\!+\!3x\mid n.\,$ But $\,\Bbb Z[x]/(x^2\!+\!1)\cong \Bbb Z[i]\,$ so

$$ 1\!+\!3i\mid n\ \, {\rm in}\, \ \Bbb Z[i]\iff \underbrace{\dfrac{n}{1\!+\!3i}\in \Bbb Z[i]\iff \dfrac{n(1\!-\!3i)}{10}\in\Bbb Z[i]}_{\large \text{rationalize denominator}}\iff \color{#c00}{10\mid n}\ \ {\rm in}\ \ \Bbb Z\qquad$$

Bill Dubuque
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    See here for more, include a method using Hermite normal form basis. – Bill Dubuque Jan 29 '20 at 05:54
  • Sorry, What dose "$ \bmod x^2!+!1!:\ 1!+!3x\mid n.,$" mean?? I know only $\bmod $use when '$a \equiv b (\bmod k)$)' – s ss Jan 29 '20 at 06:14
  • Is " $\bmod x^2!+!1!:\ 1!+!3x\mid n.,$" means that $(1+3x)\mathit g \equiv n\ (\bmod (3x+1))$$,$?? – s ss Jan 29 '20 at 06:47
  • @sss It means exactly the same as the first equation to its left, i.e. mod that out by $,x^2+1.,$ You can skip straight to the quotient ring if your prefer. I was simply trying to motivate it additionally with equivalent congruence notation. Generally $,a\mid b,$ in $R,$ means $,ar = b,$ for some $,r\in R,,$ just like in $,R = \Bbb Z\ \ \ $ – Bill Dubuque Jan 29 '20 at 06:59