Mixing
Why do ideals arise naturally in ring theory in the same way that normal subgroups arise naturally in group...
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I guess my question is essentially that why being closed under multiplication by elements in $R$ for $I$ is required for $R/I$ to be a quotient ring.
abstract-algebra
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 2:46
davidhdavidh
3028
$endgroup$
add a comment |
$begingroup$
I guess my question is essentially that why being closed under multiplication by elements in $R$ for $I$ is required for $R/I$ to be a quotient ring.
abstract-algebra
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 2:46
davidhdavidh
3028
$endgroup$
2
$begingroup$
To answer the title: just as normal subgroups arise as kernels of group homomorphisms, ideals arise as kernels of ring homomorphisms. As for the body: if $x = 0$ in $R/I$, then we certainly want $rx = 0$ in $R/I$ for any $r in R$, and this is exactly what the property you mention implies.
$endgroup$
– André 3000
Jan 24 at 4:35
add a comment |
$begingroup$
I guess my question is essentially that why being closed under multiplication by elements in $R$ for $I$ is required for $R/I$ to be a quotient ring.
abstract-algebra
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 2:46
davidhdavidh
3028
$endgroup$
I guess my question is essentially that why being closed under multiplication by elements in $R$ for $I$ is required for $R/I$ to be a quotient ring.
abstract-algebra
abstract-algebra
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 2:46
davidhdavidh
3028
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
asked Jan 24 at 2:46
davidhdavidh
3028
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
edited Jan 24 at 5:47
Antonios-Alexandros Robotis
10.5k41741
10.5k41741
asked Jan 24 at 2:46
davidhdavidh
3028
asked Jan 24 at 2:46
davidhdavidh
3028
asked Jan 24 at 2:46
davidhdavidh
3028
3028
2
$begingroup$
To answer the title: just as normal subgroups arise as kernels of group homomorphisms, ideals arise as kernels of ring homomorphisms. As for the body: if $x = 0$ in $R/I$, then we certainly want $rx = 0$ in $R/I$ for any $r in R$, and this is exactly what the property you mention implies.
$endgroup$
– André 3000
Jan 24 at 4:35
add a comment |
2
$begingroup$
To answer the title: just as normal subgroups arise as kernels of group homomorphisms, ideals arise as kernels of ring homomorphisms. As for the body: if $x = 0$ in $R/I$, then we certainly want $rx = 0$ in $R/I$ for any $r in R$, and this is exactly what the property you mention implies.
$endgroup$
– André 3000
Jan 24 at 4:35
2
2
$begingroup$
To answer the title: just as normal subgroups arise as kernels of group homomorphisms, ideals arise as kernels of ring homomorphisms. As for the body: if $x = 0$ in $R/I$, then we certainly want $rx = 0$ in $R/I$ for any $r in R$, and this is exactly what the property you mention implies.
$endgroup$
– André 3000
Jan 24 at 4:35
$begingroup$
To answer the title: just as normal subgroups arise as kernels of group homomorphisms, ideals arise as kernels of ring homomorphisms. As for the body: if $x = 0$ in $R/I$, then we certainly want $rx = 0$ in $R/I$ for any $r in R$, and this is exactly what the property you mention implies.
$endgroup$
– André 3000
Jan 24 at 4:35
add a comment |
1 Answer
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$begingroup$
Normal subgroups $Nsubseteq G$ of a group $G$ are important precisely because for $H$ a subgroup of $G$, $G/H$ has a group structure compatible with the one on $G$ if and only if $H$ is normal. More precisely, $aHcdot bH=abH$ for any pair $a,bin G$ if and only if $xH=Hx$ for all $xin G$, if and only if $xHx^{-1}=H$ if and only if $H$ is normal.
The same idea explains the importance of ideals. In ring theory, the ring $A$ in question is an abelian group under addition and so any additive subgroup is normal. So, if we fix such a subgroup $Bsubseteq A$ and we try to form a quotient space $A/B$, we notice that $A/B$ has a group structure compatible with the one on $A$. Multiplication becomes more complicated:
$$ (x+B)cdot (y+B)=xy+xB+yB+B.$$
It turns out that the latter expression is equal to $xy+B$ if and only if $xB=B$ for all $xin A$, i.e. $B$ is closed under multiplication by elements of $A$.
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
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$begingroup$
Normal subgroups $Nsubseteq G$ of a group $G$ are important precisely because for $H$ a subgroup of $G$, $G/H$ has a group structure compatible with the one on $G$ if and only if $H$ is normal. More precisely, $aHcdot bH=abH$ for any pair $a,bin G$ if and only if $xH=Hx$ for all $xin G$, if and only if $xHx^{-1}=H$ if and only if $H$ is normal.
The same idea explains the importance of ideals. In ring theory, the ring $A$ in question is an abelian group under addition and so any additive subgroup is normal. So, if we fix such a subgroup $Bsubseteq A$ and we try to form a quotient space $A/B$, we notice that $A/B$ has a group structure compatible with the one on $A$. Multiplication becomes more complicated:
$$ (x+B)cdot (y+B)=xy+xB+yB+B.$$
It turns out that the latter expression is equal to $xy+B$ if and only if $xB=B$ for all $xin A$, i.e. $B$ is closed under multiplication by elements of $A$.
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Normal subgroups $Nsubseteq G$ of a group $G$ are important precisely because for $H$ a subgroup of $G$, $G/H$ has a group structure compatible with the one on $G$ if and only if $H$ is normal. More precisely, $aHcdot bH=abH$ for any pair $a,bin G$ if and only if $xH=Hx$ for all $xin G$, if and only if $xHx^{-1}=H$ if and only if $H$ is normal.
The same idea explains the importance of ideals. In ring theory, the ring $A$ in question is an abelian group under addition and so any additive subgroup is normal. So, if we fix such a subgroup $Bsubseteq A$ and we try to form a quotient space $A/B$, we notice that $A/B$ has a group structure compatible with the one on $A$. Multiplication becomes more complicated:
$$ (x+B)cdot (y+B)=xy+xB+yB+B.$$
It turns out that the latter expression is equal to $xy+B$ if and only if $xB=B$ for all $xin A$, i.e. $B$ is closed under multiplication by elements of $A$.
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
add a comment |
$begingroup$
Normal subgroups $Nsubseteq G$ of a group $G$ are important precisely because for $H$ a subgroup of $G$, $G/H$ has a group structure compatible with the one on $G$ if and only if $H$ is normal. More precisely, $aHcdot bH=abH$ for any pair $a,bin G$ if and only if $xH=Hx$ for all $xin G$, if and only if $xHx^{-1}=H$ if and only if $H$ is normal.
The same idea explains the importance of ideals. In ring theory, the ring $A$ in question is an abelian group under addition and so any additive subgroup is normal. So, if we fix such a subgroup $Bsubseteq A$ and we try to form a quotient space $A/B$, we notice that $A/B$ has a group structure compatible with the one on $A$. Multiplication becomes more complicated:
$$ (x+B)cdot (y+B)=xy+xB+yB+B.$$
It turns out that the latter expression is equal to $xy+B$ if and only if $xB=B$ for all $xin A$, i.e. $B$ is closed under multiplication by elements of $A$.
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
$endgroup$
Normal subgroups $Nsubseteq G$ of a group $G$ are important precisely because for $H$ a subgroup of $G$, $G/H$ has a group structure compatible with the one on $G$ if and only if $H$ is normal. More precisely, $aHcdot bH=abH$ for any pair $a,bin G$ if and only if $xH=Hx$ for all $xin G$, if and only if $xHx^{-1}=H$ if and only if $H$ is normal.
The same idea explains the importance of ideals. In ring theory, the ring $A$ in question is an abelian group under addition and so any additive subgroup is normal. So, if we fix such a subgroup $Bsubseteq A$ and we try to form a quotient space $A/B$, we notice that $A/B$ has a group structure compatible with the one on $A$. Multiplication becomes more complicated:
$$ (x+B)cdot (y+B)=xy+xB+yB+B.$$
It turns out that the latter expression is equal to $xy+B$ if and only if $xB=B$ for all $xin A$, i.e. $B$ is closed under multiplication by elements of $A$.
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
answered Jan 24 at 5:55
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.5k41741
10.5k41741
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$begingroup$
To answer the title: just as normal subgroups arise as kernels of group homomorphisms, ideals arise as kernels of ring homomorphisms. As for the body: if $x = 0$ in $R/I$, then we certainly want $rx = 0$ in $R/I$ for any $r in R$, and this is exactly what the property you mention implies.
$endgroup$
– André 3000
Jan 24 at 4:35