Is $mathbb Z[t]$ principal? [duplicate]












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  • 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 $(5,t)=mathbb Z[t]$ ?










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Jan 4 at 11:44


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    • 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 $(5,t)=mathbb Z[t]$ ?










    share|cite|improve this question









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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.


















      0












      0








      0





      $begingroup$



      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 $(5,t)=mathbb Z[t]$ ?










      share|cite|improve this question









      $endgroup$





      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






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      asked Jan 4 at 11:16









      NewMathNewMath

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      Jan 4 at 11:44


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      Jan 4 at 11:44


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          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          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, $1notin (5, t)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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



















          0












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$




















            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            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, $1notin (5, t)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
















            1












            $begingroup$

            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, $1notin (5, t)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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














            1












            1








            1





            $begingroup$

            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, $1notin (5, t)$.






            share|cite|improve this answer









            $endgroup$



            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, $1notin (5, t)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • $begingroup$
              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











            0












            $begingroup$

            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.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              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.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                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.






                share|cite|improve this answer











                $endgroup$



                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.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 4 at 11:24

























                answered Jan 4 at 11:22









                WuestenfuxWuestenfux

                4,2161413




                4,2161413















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