Proving that $mathbbQ[X] = (X-1) + (X+1)$

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Prove that $I+J = mathbbQ[X]$ with $I$ and $J$ ideals in $mathbbQ[X]$ where $I=(X-1)$, $J=(X+1)$.




I was thinking something along the lines of
$I+J=(1)$ because they are coprime. That means
$$
I+J
= 1 cdot a mid a in mathbbQ[X]
= a mid a in mathbbQ[X]
= mathbbQ[X].
$$



I'm not sure if this is a proper proof for this problem.



This is the proof suggested by my professo, but I can't really figure where the $1/2$ comes from.
A proof suggested by my professor
I should've formated it on the website but it would take me a long time to do so.






abstract-algebra proof-verification ring-theory proof-explanation ideals







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  • Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Sep 3 at 11:11














up vote
-2
down vote

favorite













Prove that $I+J = mathbbQ[X]$ with $I$ and $J$ ideals in $mathbbQ[X]$ where $I=(X-1)$, $J=(X+1)$.




I was thinking something along the lines of
$I+J=(1)$ because they are coprime. That means
$$
I+J
= 1 cdot a mid a in mathbbQ[X]
= a mid a in mathbbQ[X]
= mathbbQ[X].
$$



I'm not sure if this is a proper proof for this problem.



This is the proof suggested by my professo, but I can't really figure where the $1/2$ comes from.
A proof suggested by my professor
I should've formated it on the website but it would take me a long time to do so.






abstract-algebra proof-verification ring-theory proof-explanation ideals







share|cite|improve this question























  • Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Sep 3 at 11:11












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite












Prove that $I+J = mathbbQ[X]$ with $I$ and $J$ ideals in $mathbbQ[X]$ where $I=(X-1)$, $J=(X+1)$.




I was thinking something along the lines of
$I+J=(1)$ because they are coprime. That means
$$
I+J
= 1 cdot a mid a in mathbbQ[X]
= a mid a in mathbbQ[X]
= mathbbQ[X].
$$



I'm not sure if this is a proper proof for this problem.



This is the proof suggested by my professo, but I can't really figure where the $1/2$ comes from.
A proof suggested by my professor
I should've formated it on the website but it would take me a long time to do so.






abstract-algebra proof-verification ring-theory proof-explanation ideals







share|cite|improve this question
















Prove that $I+J = mathbbQ[X]$ with $I$ and $J$ ideals in $mathbbQ[X]$ where $I=(X-1)$, $J=(X+1)$.




I was thinking something along the lines of
$I+J=(1)$ because they are coprime. That means
$$
I+J
= 1 cdot a mid a in mathbbQ[X]
= a mid a in mathbbQ[X]
= mathbbQ[X].
$$



I'm not sure if this is a proper proof for this problem.



This is the proof suggested by my professo, but I can't really figure where the $1/2$ comes from.
A proof suggested by my professor
I should've formated it on the website but it would take me a long time to do so.






abstract-algebra proof-verification ring-theory proof-explanation ideals




abstract-algebra proof-verification ring-theory proof-explanation ideals






share|cite|improve this question















share|cite|improve this question













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edited Sep 3 at 22:23









Scientifica

4,90121331




4,90121331










asked Sep 3 at 11:05









Alex Ionut Gavrila

22




22











  • Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Sep 3 at 11:11
















  • Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
    – José Carlos Santos
    Sep 3 at 11:11















Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 3 at 11:11




Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures.
– José Carlos Santos
Sep 3 at 11:11










2 Answers
2






active

oldest

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up vote
0
down vote



accepted










The number $frac12$ is an element of $mathbbQ[x]$. Hence, you can multiply any element of an ideal contained in $mathbbQ[x]$ by $frac12$ and the result will remain in the ideal by definition of ideal.



What your professor has proved is that $1=-frac12(x-1)+frac12(x+1)$, which is an element of $I+J$ by definition of the ideal. This is sufficient to show $I+J=mathbbQ[x]$ since the inclusion $I+Jsubseteq mathbbQ[x]$ is rather trivial.






share|cite|improve this answer
















  • 1




    Ohhh I wasn't paying enough attention. Thanks for explaining it.
    – Alex Ionut Gavrila
    Sep 3 at 11:20

















up vote
0
down vote













Observe simply that



$$-frac12(x-1)+frac12(x+1)=1implies I+J=Bbb Q[x];$$



I think your professor was doing something like this, but I don't quite follow in romanian...:)






share|cite|improve this answer






















  • Turns out that the OP's professor gave this solution, but he doesn't understand it.
    – Batominovski
    Sep 3 at 11:13











  • @Batominovski I saw the proof: it isn't quite the same...
    – DonAntonio
    Sep 3 at 11:14











  • @Alex Remember: if an ideal of a unitary ring contains the unit $;1;$ , then it is the whole ring.
    – DonAntonio
    Sep 3 at 11:15










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote



accepted










The number $frac12$ is an element of $mathbbQ[x]$. Hence, you can multiply any element of an ideal contained in $mathbbQ[x]$ by $frac12$ and the result will remain in the ideal by definition of ideal.



What your professor has proved is that $1=-frac12(x-1)+frac12(x+1)$, which is an element of $I+J$ by definition of the ideal. This is sufficient to show $I+J=mathbbQ[x]$ since the inclusion $I+Jsubseteq mathbbQ[x]$ is rather trivial.






share|cite|improve this answer
















  • 1




    Ohhh I wasn't paying enough attention. Thanks for explaining it.
    – Alex Ionut Gavrila
    Sep 3 at 11:20














up vote
0
down vote



accepted










The number $frac12$ is an element of $mathbbQ[x]$. Hence, you can multiply any element of an ideal contained in $mathbbQ[x]$ by $frac12$ and the result will remain in the ideal by definition of ideal.



What your professor has proved is that $1=-frac12(x-1)+frac12(x+1)$, which is an element of $I+J$ by definition of the ideal. This is sufficient to show $I+J=mathbbQ[x]$ since the inclusion $I+Jsubseteq mathbbQ[x]$ is rather trivial.






share|cite|improve this answer
















  • 1




    Ohhh I wasn't paying enough attention. Thanks for explaining it.
    – Alex Ionut Gavrila
    Sep 3 at 11:20












up vote
0
down vote



accepted







up vote
0
down vote



accepted






The number $frac12$ is an element of $mathbbQ[x]$. Hence, you can multiply any element of an ideal contained in $mathbbQ[x]$ by $frac12$ and the result will remain in the ideal by definition of ideal.



What your professor has proved is that $1=-frac12(x-1)+frac12(x+1)$, which is an element of $I+J$ by definition of the ideal. This is sufficient to show $I+J=mathbbQ[x]$ since the inclusion $I+Jsubseteq mathbbQ[x]$ is rather trivial.






share|cite|improve this answer












The number $frac12$ is an element of $mathbbQ[x]$. Hence, you can multiply any element of an ideal contained in $mathbbQ[x]$ by $frac12$ and the result will remain in the ideal by definition of ideal.



What your professor has proved is that $1=-frac12(x-1)+frac12(x+1)$, which is an element of $I+J$ by definition of the ideal. This is sufficient to show $I+J=mathbbQ[x]$ since the inclusion $I+Jsubseteq mathbbQ[x]$ is rather trivial.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 3 at 11:13









Javi

2,2081725




2,2081725







  • 1




    Ohhh I wasn't paying enough attention. Thanks for explaining it.
    – Alex Ionut Gavrila
    Sep 3 at 11:20












  • 1




    Ohhh I wasn't paying enough attention. Thanks for explaining it.
    – Alex Ionut Gavrila
    Sep 3 at 11:20







1




1




Ohhh I wasn't paying enough attention. Thanks for explaining it.
– Alex Ionut Gavrila
Sep 3 at 11:20




Ohhh I wasn't paying enough attention. Thanks for explaining it.
– Alex Ionut Gavrila
Sep 3 at 11:20










up vote
0
down vote













Observe simply that



$$-frac12(x-1)+frac12(x+1)=1implies I+J=Bbb Q[x];$$



I think your professor was doing something like this, but I don't quite follow in romanian...:)






share|cite|improve this answer






















  • Turns out that the OP's professor gave this solution, but he doesn't understand it.
    – Batominovski
    Sep 3 at 11:13











  • @Batominovski I saw the proof: it isn't quite the same...
    – DonAntonio
    Sep 3 at 11:14











  • @Alex Remember: if an ideal of a unitary ring contains the unit $;1;$ , then it is the whole ring.
    – DonAntonio
    Sep 3 at 11:15














up vote
0
down vote













Observe simply that



$$-frac12(x-1)+frac12(x+1)=1implies I+J=Bbb Q[x];$$



I think your professor was doing something like this, but I don't quite follow in romanian...:)






share|cite|improve this answer






















  • Turns out that the OP's professor gave this solution, but he doesn't understand it.
    – Batominovski
    Sep 3 at 11:13











  • @Batominovski I saw the proof: it isn't quite the same...
    – DonAntonio
    Sep 3 at 11:14











  • @Alex Remember: if an ideal of a unitary ring contains the unit $;1;$ , then it is the whole ring.
    – DonAntonio
    Sep 3 at 11:15












up vote
0
down vote










up vote
0
down vote









Observe simply that



$$-frac12(x-1)+frac12(x+1)=1implies I+J=Bbb Q[x];$$



I think your professor was doing something like this, but I don't quite follow in romanian...:)






share|cite|improve this answer














Observe simply that



$$-frac12(x-1)+frac12(x+1)=1implies I+J=Bbb Q[x];$$



I think your professor was doing something like this, but I don't quite follow in romanian...:)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 3 at 11:13

























answered Sep 3 at 11:12









DonAntonio

173k1486220




173k1486220











  • Turns out that the OP's professor gave this solution, but he doesn't understand it.
    – Batominovski
    Sep 3 at 11:13











  • @Batominovski I saw the proof: it isn't quite the same...
    – DonAntonio
    Sep 3 at 11:14











  • @Alex Remember: if an ideal of a unitary ring contains the unit $;1;$ , then it is the whole ring.
    – DonAntonio
    Sep 3 at 11:15
















  • Turns out that the OP's professor gave this solution, but he doesn't understand it.
    – Batominovski
    Sep 3 at 11:13











  • @Batominovski I saw the proof: it isn't quite the same...
    – DonAntonio
    Sep 3 at 11:14











  • @Alex Remember: if an ideal of a unitary ring contains the unit $;1;$ , then it is the whole ring.
    – DonAntonio
    Sep 3 at 11:15















Turns out that the OP's professor gave this solution, but he doesn't understand it.
– Batominovski
Sep 3 at 11:13





Turns out that the OP's professor gave this solution, but he doesn't understand it.
– Batominovski
Sep 3 at 11:13













@Batominovski I saw the proof: it isn't quite the same...
– DonAntonio
Sep 3 at 11:14





@Batominovski I saw the proof: it isn't quite the same...
– DonAntonio
Sep 3 at 11:14













@Alex Remember: if an ideal of a unitary ring contains the unit $;1;$ , then it is the whole ring.
– DonAntonio
Sep 3 at 11:15




@Alex Remember: if an ideal of a unitary ring contains the unit $;1;$ , then it is the whole ring.
– DonAntonio
Sep 3 at 11:15

















 

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