How to find the extension of an ideal











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I have a very elementary question and I would appreciate understanding.



Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.



Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.



I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.



Thanks! Any help or explanation would be appreciated.










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  • What do you mean by $I^e$?
    – Alex Vong
    Nov 25 at 20:24








  • 2




    The extension of the ideal I
    – S. Prevč
    Nov 25 at 20:25










  • I forget to read the title...
    – Alex Vong
    Nov 25 at 20:25










  • By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
    – Alex Vong
    Nov 25 at 20:58












  • I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
    – S. Prevč
    Nov 25 at 21:13















up vote
0
down vote

favorite
1












I have a very elementary question and I would appreciate understanding.



Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.



Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.



I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.



Thanks! Any help or explanation would be appreciated.










share|cite|improve this question
























  • What do you mean by $I^e$?
    – Alex Vong
    Nov 25 at 20:24








  • 2




    The extension of the ideal I
    – S. Prevč
    Nov 25 at 20:25










  • I forget to read the title...
    – Alex Vong
    Nov 25 at 20:25










  • By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
    – Alex Vong
    Nov 25 at 20:58












  • I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
    – S. Prevč
    Nov 25 at 21:13













up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





I have a very elementary question and I would appreciate understanding.



Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.



Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.



I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.



Thanks! Any help or explanation would be appreciated.










share|cite|improve this question















I have a very elementary question and I would appreciate understanding.



Let $A$ be a commutative and unitary ring, and $I$ an ideal of this ring.



Considering the inclusion $i: Alongrightarrow A[X]$, I need to prove that $I^e=I[X]$, where $I^e$ is the extension of the ideal $I$.



I know this is true because I have sometimes used it in other problems, but I actually don't know the reasoning behind it.



Thanks! Any help or explanation would be appreciated.







abstract-algebra ring-theory ideals






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share|cite|improve this question













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share|cite|improve this question








edited Nov 25 at 20:25

























asked Nov 25 at 20:15









S. Prevč

12




12












  • What do you mean by $I^e$?
    – Alex Vong
    Nov 25 at 20:24








  • 2




    The extension of the ideal I
    – S. Prevč
    Nov 25 at 20:25










  • I forget to read the title...
    – Alex Vong
    Nov 25 at 20:25










  • By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
    – Alex Vong
    Nov 25 at 20:58












  • I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
    – S. Prevč
    Nov 25 at 21:13


















  • What do you mean by $I^e$?
    – Alex Vong
    Nov 25 at 20:24








  • 2




    The extension of the ideal I
    – S. Prevč
    Nov 25 at 20:25










  • I forget to read the title...
    – Alex Vong
    Nov 25 at 20:25










  • By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
    – Alex Vong
    Nov 25 at 20:58












  • I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
    – S. Prevč
    Nov 25 at 21:13
















What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24






What do you mean by $I^e$?
– Alex Vong
Nov 25 at 20:24






2




2




The extension of the ideal I
– S. Prevč
Nov 25 at 20:25




The extension of the ideal I
– S. Prevč
Nov 25 at 20:25












I forget to read the title...
– Alex Vong
Nov 25 at 20:25




I forget to read the title...
– Alex Vong
Nov 25 at 20:25












By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58






By definition, $I^e$ is the ideal generated by $i(I) = I$, so $I^e = {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$. Now we only need to show ${a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]} = I[x]$, can you show it?
– Alex Vong
Nov 25 at 20:58














I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13




I can see one inclusion ($<i(I)>$ included in $I[X]$), but not the other one.
– S. Prevč
Nov 25 at 21:13










2 Answers
2






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1
down vote













To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.






share|cite|improve this answer




























    up vote
    0
    down vote













    Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.



    Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.



    The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.



    Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.



    Combining these pieces of information, we get the following.




    If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.




    Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.






    share|cite|improve this answer





















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      2 Answers
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      To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.






      share|cite|improve this answer

























        up vote
        1
        down vote













        To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.






        share|cite|improve this answer























          up vote
          1
          down vote










          up vote
          1
          down vote









          To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.






          share|cite|improve this answer












          To see the other inclusion, take any $f in I[x]$, then $f = a_0 + a_1 x + dots a_n x^n$ for some $a_j in I$. Now notice that $a_j in I$ and $x^j in A[x]$, so we have $f in {a_1 f_1 + dots + a_n f_n mid a_j in I, f_j in A[x]}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 25 at 21:40









          Alex Vong

          1,157819




          1,157819






















              up vote
              0
              down vote













              Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.



              Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.



              The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.



              Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.



              Combining these pieces of information, we get the following.




              If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.




              Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.






              share|cite|improve this answer

























                up vote
                0
                down vote













                Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.



                Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.



                The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.



                Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.



                Combining these pieces of information, we get the following.




                If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.




                Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.



                  Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.



                  The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.



                  Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.



                  Combining these pieces of information, we get the following.




                  If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.




                  Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.






                  share|cite|improve this answer












                  Alex Vong gave a good suggestion in the comments, but there's another way to see this. Hopefully this answer is helpful, but to be honest, your answer is missing context (by which I mean, you haven't told us what you've tried or what your thoughts are), so when I answer such questions, it's mostly for fun, since without context, I can't know what will be most useful to you.



                  Much of this answer is exposition. If you're familiar with all the material contained here, this argument is much shorter.



                  The extension of an ideal satisfies a universal property. Let $f:Ato B$ be a ring homomorphism, $I$ an ideal of $A$, then $I^e:=f(I)B$ is the smallest ideal of $B$ containing the image of $I$, which means that if $g:Bto C$ is any ring homomorphism such that $f(I)subseteq ker g$, we must have $I^esubseteq ker g$. $I^e$ is the unique ideal with this property, since we can take $g$ to be the quotient $Bto B/I^e$, and then if another ideal $J$ satisfied the property, we would have $f(I)subseteq Jsubseteq ker g= I^e$, and $I^e$ is the smallest ideal of $B$ containing $I^e$, so $J=I^e$.



                  Now the first isomorphism theorem says that if $g:Bto C$ then $I^esubseteq ker g$ is equivalent to the existence of a map $tilde{g} : B/I^e to C$ such that $tilde{g}circ q = g$, where $q:Bto B/I^e$ is the quotient.



                  Combining these pieces of information, we get the following.




                  If $A$ is a commutative, unital ring, and $Isubseteq A$ is an ideal. Then the extension of $I$ along the structure map $iota:Ato A[x]$ is the unique ideal $I^e$ such that for all maps $g$ from $A[x]to C$ such that $Isubseteq ker g$, there exists a map $tilde{g}:A[x]/I^e to C$ with $tilde{g}circ q = g$.




                  Now we recall the universal properties of polynomial rings. Maps $g:A[x]to C$ correspond precisely to pairs $(g',c)$ with $g'$ a map $g' : Ato C$ and $c$ an element of $C$. Maps $g:A[x]to C$ with $Isubseteq ker g$ therefore correspond to pairs $(g',c)$ with $Isubseteq ker g'$. But by the universal property of the quotient, these correspond precisely to pairs $(bar{g},c)$ with $bar{g}$ a map from $A/Ito C$ and $cin C$. Thus maps from $A[x]to C$ which annihilate $I$ correspond precisely to maps from $A/I[x]$ to $C$. Conveniently, we have a surjective map $A[x]to A/I[x]$ whose kernel is precisely $I[x]$ (this part is obvious from the definition of the map from $A[x]to A/I[x]$, which is reduce every coefficient mod $I$). Thus by the universal property (of sorts) of $I^e$, since $A[x]/I[x]$ has the property characterizing $I^e$, $I[x]=I^e$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 25 at 22:32









                  jgon

                  10.6k11739




                  10.6k11739






























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