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Is $mathbb Z[t]$ principal? [duplicate]
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This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
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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 $(5,t)=mathbb Z[t]$ ?
abstract-algebra
asked Jan 4 at 11:16
NewMathNewMath
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marked as duplicate by Pierre-Guy Plamondon, Dietrich Burde abstract-algebra
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Jan 4 at 11:44
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This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
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 $(5,t)=mathbb Z[t]$ ?
abstract-algebra
asked Jan 4 at 11:16
NewMathNewMath
4059
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Jan 4 at 11:44
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
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 $(5,t)=mathbb Z[t]$ ?
abstract-algebra
asked Jan 4 at 11:16
NewMathNewMath
4059
$endgroup$
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
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 $(5,t)=mathbb Z[t]$ ?
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
abstract-algebra
abstract-algebra
asked Jan 4 at 11:16
NewMathNewMath
4059
asked Jan 4 at 11:16
NewMathNewMath
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asked Jan 4 at 11:16
NewMathNewMath
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asked Jan 4 at 11:16
NewMathNewMath
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asked Jan 4 at 11:16
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Jan 4 at 11:44
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Jan 4 at 11:44
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2 Answers
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$begingroup$
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 $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, 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, $1notin (5, t)$.
answered Jan 4 at 11:24
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
$endgroup$
– NewMath
Jan 4 at 12:17
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:23
$begingroup$
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 $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
$endgroup$
– NewMath
Jan 4 at 12:43
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:57
$begingroup$
Thank you very much dear @arthur. It's perfectly clear.
$endgroup$
– NewMath
Jan 4 at 13:11
add a comment |
$begingroup$
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, 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, $1notin (5, t)$.
answered Jan 4 at 11:24
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
$endgroup$
– NewMath
Jan 4 at 12:17
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:23
$begingroup$
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 $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
$endgroup$
– NewMath
Jan 4 at 12:43
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:57
$begingroup$
Thank you very much dear @arthur. It's perfectly clear.
$endgroup$
– NewMath
Jan 4 at 13:11
add a comment |
$begingroup$
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 $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, 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, $1notin (5, t)$.
answered Jan 4 at 11:24
ArthurArthur
112k7109191
$endgroup$
$begingroup$
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
$endgroup$
– NewMath
Jan 4 at 12:17
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:23
$begingroup$
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 $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
$endgroup$
– NewMath
Jan 4 at 12:43
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:57
$begingroup$
Thank you very much dear @arthur. It's perfectly clear.
$endgroup$
– NewMath
Jan 4 at 13:11
add a comment |
$begingroup$
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 $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, 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, $1notin (5, t)$.
answered Jan 4 at 11:24
ArthurArthur
112k7109191
$endgroup$
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 $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, 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, $1notin (5, t)$.
answered Jan 4 at 11:24
ArthurArthur
112k7109191
answered Jan 4 at 11:24
ArthurArthur
112k7109191
answered Jan 4 at 11:24
ArthurArthur
112k7109191
answered Jan 4 at 11:24
ArthurArthur
112k7109191
112k7109191
$begingroup$
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
$endgroup$
– NewMath
Jan 4 at 12:17
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:23
$begingroup$
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 $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
$endgroup$
– NewMath
Jan 4 at 12:43
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:57
$begingroup$
Thank you very much dear @arthur. It's perfectly clear.
$endgroup$
– NewMath
Jan 4 at 13:11
add a comment |
$begingroup$
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
$endgroup$
– NewMath
Jan 4 at 12:17
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:23
$begingroup$
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 $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
$endgroup$
– NewMath
Jan 4 at 12:43
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:57
$begingroup$
Thank you very much dear @arthur. It's perfectly clear.
$endgroup$
– NewMath
Jan 4 at 13:11
$begingroup$
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
$endgroup$
– NewMath
Jan 4 at 12:17
$begingroup$
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
$endgroup$
– NewMath
Jan 4 at 12:17
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:23
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:23
$begingroup$
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 $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
$endgroup$
– NewMath
Jan 4 at 12:43
$begingroup$
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 $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
$endgroup$
– NewMath
Jan 4 at 12:43
1
1
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:57
$begingroup$
@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.
$endgroup$
– Arthur
Jan 4 at 12:57
$begingroup$
Thank you very much dear @arthur. It's perfectly clear.
$endgroup$
– NewMath
Jan 4 at 13:11
$begingroup$
Thank you very much dear @arthur. It's perfectly clear.
$endgroup$
– NewMath
Jan 4 at 13:11
add a comment |
$begingroup$
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
$endgroup$
add a comment |
$begingroup$
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
$endgroup$
add a comment |
$begingroup$
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
$endgroup$
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
edited Jan 4 at 11:24
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
answered Jan 4 at 11:22
WuestenfuxWuestenfux
4,2161413
4,2161413
add a comment |
add a comment |
