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I know that $\mathbb Z[t]$ is factoriel, but is it principal ? For example, let consider the ideal $(5,t)$ it look that is not principal, but after all, $5$ and $t$ are co-prime, so by Bezout, there is $q(x),r(x)$ s.t. $$5q(t)+tk(t)=1$$ and thus $(5,t)=\mathbb Z[t]$ ?

user621345
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2 Answers2

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Bezout's doesn't apply here. We can clearly see that in the polynomial $5q(t) + tk(t)$, the constant term is divisible by $5$. More formally, consider the evaluation homomorphism $v_0:\Bbb Z[t]\to \Bbb Z$ at $t = 0$. We see that $5\mid v_0(5q(t) + tk(t))$, but $5\nmid v_0(1)$, so we cannot have $5q(t) + tk(t) = 1$.

And no, it's not a PID. The only principal ideal which contains both $5$ and $t$ is $(1)$, and as we have seen, $1\notin (5, t)$.

Arthur
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  • Sorry, I don't get why Bezout doesn't apply there. Don't we have that $\gcd(f(t),g(t))=1\iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $\mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring. – user621345 Jan 04 '19 at 12:17
  • @NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $\Bbb Z$ has and $\Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important. – Arthur Jan 04 '19 at 12:23
  • yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $I\cap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJ\subset (a,b)$. Could it be $(a,b)=I+J+IJ$ ? – user621345 Jan 04 '19 at 12:43
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    @NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing. – Arthur Jan 04 '19 at 12:57
  • Thank you very much dear @arthur. It's perfectly clear. – user621345 Jan 04 '19 at 13:11
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In the ring ${\Bbb Z}[t]$, we have $\gcd(2, t) = 1$. But there is no representation of the form $2\cdot f + t\cdot g =1$, since the constant term on the left hand side is even.

Wuestenfux
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