Mixing
Why are units called units?
$begingroup$
Why are units in abstract algebra called units? Is it just because they generalize the notion of $-1$ and $1$?, and the like? There's often a sense that units don't matter, when talking about things like irreducibility- is the name unit supposed to trivialize them somehow?
abstract-algebra terminology
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
asked Oct 5 '15 at 5:47
BobBob
9719
$endgroup$
add a comment |
$begingroup$
Why are units in abstract algebra called units? Is it just because they generalize the notion of $-1$ and $1$?, and the like? There's often a sense that units don't matter, when talking about things like irreducibility- is the name unit supposed to trivialize them somehow?
abstract-algebra terminology
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
asked Oct 5 '15 at 5:47
BobBob
9719
$endgroup$
1
$begingroup$
I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide.
$endgroup$
– Omnomnomnom
Oct 5 '15 at 5:54
$begingroup$
I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers)
$endgroup$
– Eric Wofsey
Oct 5 '15 at 5:58
add a comment |
$begingroup$
Why are units in abstract algebra called units? Is it just because they generalize the notion of $-1$ and $1$?, and the like? There's often a sense that units don't matter, when talking about things like irreducibility- is the name unit supposed to trivialize them somehow?
abstract-algebra terminology
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
asked Oct 5 '15 at 5:47
BobBob
9719
$endgroup$
Why are units in abstract algebra called units? Is it just because they generalize the notion of $-1$ and $1$?, and the like? There's often a sense that units don't matter, when talking about things like irreducibility- is the name unit supposed to trivialize them somehow?
abstract-algebra terminology
abstract-algebra terminology
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
asked Oct 5 '15 at 5:47
BobBob
9719
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
asked Oct 5 '15 at 5:47
BobBob
9719
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
edited Oct 5 '15 at 5:51
Omnomnomnom
129k794188
129k794188
asked Oct 5 '15 at 5:47
BobBob
9719
asked Oct 5 '15 at 5:47
BobBob
9719
asked Oct 5 '15 at 5:47
BobBob
9719
9719
1
$begingroup$
I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide.
$endgroup$
– Omnomnomnom
Oct 5 '15 at 5:54
$begingroup$
I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers)
$endgroup$
– Eric Wofsey
Oct 5 '15 at 5:58
add a comment |
1
$begingroup$
I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide.
$endgroup$
– Omnomnomnom
Oct 5 '15 at 5:54
$begingroup$
I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers)
$endgroup$
– Eric Wofsey
Oct 5 '15 at 5:58
1
1
$begingroup$
I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide.
$endgroup$
– Omnomnomnom
Oct 5 '15 at 5:54
$begingroup$
I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide.
$endgroup$
– Omnomnomnom
Oct 5 '15 at 5:54
$begingroup$
I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers)
$endgroup$
– Eric Wofsey
Oct 5 '15 at 5:58
$begingroup$
I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers)
$endgroup$
– Eric Wofsey
Oct 5 '15 at 5:58
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
$endgroup$
add a comment |
$begingroup$
In the Sciences a unit is a fixed amount of a quantity in question.
Every other amount is a multiple of the fixed amount.
In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.
answered Jul 26 '17 at 19:02
user467419user467419
211
$endgroup$
$begingroup$
Is this true about abstract algebra too?
$endgroup$
– Arman Malekzadeh
Jul 26 '17 at 19:30
add a comment |
$begingroup$
To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.
In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $u$ is a unit of a ring $R$, then there is a $vin R$ such that $uv=vu=1$. Thus, for any $xin R$, one can rewrite $x$ "in terms of the unit $u$" like so: $x=x(vu)=(xv)u$. Thus, $x$ is a multiple of $u$.
This can be understood well in seeing why a ring $F$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are ${0}$ and $F$.
Note every element in a field is a unit $u$. It follows that if we rewrite any element $xin F$ in terms of that unit, we will get $x=ku$ for some $kin F$. Now consider that any ideal is a kernel of a ring homomorphism $f:Fto R$. $f$'s kernel could be ${0}$, in which case $f$ is injective, and the ideal it corresponds to is ${0}$.
Otherwise, some element (and thus unit) $u$ maps to zero. Since we can write any $x$ as $x=ku$, that means $f(x)=f(k)f(u)=0.$ So the kernel is all of $F$.
Therefore the only ideals (homomorphism kernels) of a field are ${0}$ and $F$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.
More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $x$ in a ring $R$ to be $x=ku$, and the behavior of $x$ under any group homomorphism $g$ is determined by the behavior of $k$ and of $u$ under $g$, since $g(x)=g(k)g(u)$. This explains why $u$ is called a "unit" since we can measure behavior of element $x$ in terms of what coefficient $k$ is in front of $u$ when we express $x$ in terms of $u$.
Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:
Theorem (alternate definition of a unit): $uin R$ is a unit iff for all $xin R$, there exists a $vin R$ such that $(xv)u=x$.
The fowards case of this theorem is proven above, while the backward case is provable letting $x=1$.
answered Feb 3 at 4:31
MathTrainMathTrain
690217
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
$endgroup$
add a comment |
$begingroup$
"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
$endgroup$
add a comment |
$begingroup$
"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
$endgroup$
"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
edited Oct 5 '15 at 6:10
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
answered Oct 5 '15 at 5:59
BrianOBrianO
14.4k1822
14.4k1822
add a comment |
add a comment |
$begingroup$
In the Sciences a unit is a fixed amount of a quantity in question.
Every other amount is a multiple of the fixed amount.
In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.
answered Jul 26 '17 at 19:02
user467419user467419
211
$endgroup$
$begingroup$
Is this true about abstract algebra too?
$endgroup$
– Arman Malekzadeh
Jul 26 '17 at 19:30
add a comment |
$begingroup$
In the Sciences a unit is a fixed amount of a quantity in question.
Every other amount is a multiple of the fixed amount.
In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.
answered Jul 26 '17 at 19:02
user467419user467419
211
$endgroup$
$begingroup$
Is this true about abstract algebra too?
$endgroup$
– Arman Malekzadeh
Jul 26 '17 at 19:30
add a comment |
$begingroup$
In the Sciences a unit is a fixed amount of a quantity in question.
Every other amount is a multiple of the fixed amount.
In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.
answered Jul 26 '17 at 19:02
user467419user467419
211
$endgroup$
In the Sciences a unit is a fixed amount of a quantity in question.
Every other amount is a multiple of the fixed amount.
In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.
answered Jul 26 '17 at 19:02
user467419user467419
211
answered Jul 26 '17 at 19:02
user467419user467419
211
answered Jul 26 '17 at 19:02
user467419user467419
211
answered Jul 26 '17 at 19:02
user467419user467419
211
211
$begingroup$
Is this true about abstract algebra too?
$endgroup$
– Arman Malekzadeh
Jul 26 '17 at 19:30
add a comment |
$begingroup$
Is this true about abstract algebra too?
$endgroup$
– Arman Malekzadeh
Jul 26 '17 at 19:30
$begingroup$
Is this true about abstract algebra too?
$endgroup$
– Arman Malekzadeh
Jul 26 '17 at 19:30
$begingroup$
Is this true about abstract algebra too?
$endgroup$
– Arman Malekzadeh
Jul 26 '17 at 19:30
add a comment |
$begingroup$
To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.
In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $u$ is a unit of a ring $R$, then there is a $vin R$ such that $uv=vu=1$. Thus, for any $xin R$, one can rewrite $x$ "in terms of the unit $u$" like so: $x=x(vu)=(xv)u$. Thus, $x$ is a multiple of $u$.
This can be understood well in seeing why a ring $F$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are ${0}$ and $F$.
Note every element in a field is a unit $u$. It follows that if we rewrite any element $xin F$ in terms of that unit, we will get $x=ku$ for some $kin F$. Now consider that any ideal is a kernel of a ring homomorphism $f:Fto R$. $f$'s kernel could be ${0}$, in which case $f$ is injective, and the ideal it corresponds to is ${0}$.
Otherwise, some element (and thus unit) $u$ maps to zero. Since we can write any $x$ as $x=ku$, that means $f(x)=f(k)f(u)=0.$ So the kernel is all of $F$.
Therefore the only ideals (homomorphism kernels) of a field are ${0}$ and $F$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.
More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $x$ in a ring $R$ to be $x=ku$, and the behavior of $x$ under any group homomorphism $g$ is determined by the behavior of $k$ and of $u$ under $g$, since $g(x)=g(k)g(u)$. This explains why $u$ is called a "unit" since we can measure behavior of element $x$ in terms of what coefficient $k$ is in front of $u$ when we express $x$ in terms of $u$.
Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:
Theorem (alternate definition of a unit): $uin R$ is a unit iff for all $xin R$, there exists a $vin R$ such that $(xv)u=x$.
The fowards case of this theorem is proven above, while the backward case is provable letting $x=1$.
answered Feb 3 at 4:31
MathTrainMathTrain
690217
$endgroup$
add a comment |
$begingroup$
To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.
In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $u$ is a unit of a ring $R$, then there is a $vin R$ such that $uv=vu=1$. Thus, for any $xin R$, one can rewrite $x$ "in terms of the unit $u$" like so: $x=x(vu)=(xv)u$. Thus, $x$ is a multiple of $u$.
This can be understood well in seeing why a ring $F$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are ${0}$ and $F$.
Note every element in a field is a unit $u$. It follows that if we rewrite any element $xin F$ in terms of that unit, we will get $x=ku$ for some $kin F$. Now consider that any ideal is a kernel of a ring homomorphism $f:Fto R$. $f$'s kernel could be ${0}$, in which case $f$ is injective, and the ideal it corresponds to is ${0}$.
Otherwise, some element (and thus unit) $u$ maps to zero. Since we can write any $x$ as $x=ku$, that means $f(x)=f(k)f(u)=0.$ So the kernel is all of $F$.
Therefore the only ideals (homomorphism kernels) of a field are ${0}$ and $F$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.
More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $x$ in a ring $R$ to be $x=ku$, and the behavior of $x$ under any group homomorphism $g$ is determined by the behavior of $k$ and of $u$ under $g$, since $g(x)=g(k)g(u)$. This explains why $u$ is called a "unit" since we can measure behavior of element $x$ in terms of what coefficient $k$ is in front of $u$ when we express $x$ in terms of $u$.
Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:
Theorem (alternate definition of a unit): $uin R$ is a unit iff for all $xin R$, there exists a $vin R$ such that $(xv)u=x$.
The fowards case of this theorem is proven above, while the backward case is provable letting $x=1$.
answered Feb 3 at 4:31
MathTrainMathTrain
690217
$endgroup$
add a comment |
$begingroup$
To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.
In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $u$ is a unit of a ring $R$, then there is a $vin R$ such that $uv=vu=1$. Thus, for any $xin R$, one can rewrite $x$ "in terms of the unit $u$" like so: $x=x(vu)=(xv)u$. Thus, $x$ is a multiple of $u$.
This can be understood well in seeing why a ring $F$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are ${0}$ and $F$.
Note every element in a field is a unit $u$. It follows that if we rewrite any element $xin F$ in terms of that unit, we will get $x=ku$ for some $kin F$. Now consider that any ideal is a kernel of a ring homomorphism $f:Fto R$. $f$'s kernel could be ${0}$, in which case $f$ is injective, and the ideal it corresponds to is ${0}$.
Otherwise, some element (and thus unit) $u$ maps to zero. Since we can write any $x$ as $x=ku$, that means $f(x)=f(k)f(u)=0.$ So the kernel is all of $F$.
Therefore the only ideals (homomorphism kernels) of a field are ${0}$ and $F$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.
More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $x$ in a ring $R$ to be $x=ku$, and the behavior of $x$ under any group homomorphism $g$ is determined by the behavior of $k$ and of $u$ under $g$, since $g(x)=g(k)g(u)$. This explains why $u$ is called a "unit" since we can measure behavior of element $x$ in terms of what coefficient $k$ is in front of $u$ when we express $x$ in terms of $u$.
Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:
Theorem (alternate definition of a unit): $uin R$ is a unit iff for all $xin R$, there exists a $vin R$ such that $(xv)u=x$.
The fowards case of this theorem is proven above, while the backward case is provable letting $x=1$.
answered Feb 3 at 4:31
MathTrainMathTrain
690217
$endgroup$
To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.
In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $u$ is a unit of a ring $R$, then there is a $vin R$ such that $uv=vu=1$. Thus, for any $xin R$, one can rewrite $x$ "in terms of the unit $u$" like so: $x=x(vu)=(xv)u$. Thus, $x$ is a multiple of $u$.
This can be understood well in seeing why a ring $F$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are ${0}$ and $F$.
Note every element in a field is a unit $u$. It follows that if we rewrite any element $xin F$ in terms of that unit, we will get $x=ku$ for some $kin F$. Now consider that any ideal is a kernel of a ring homomorphism $f:Fto R$. $f$'s kernel could be ${0}$, in which case $f$ is injective, and the ideal it corresponds to is ${0}$.
Otherwise, some element (and thus unit) $u$ maps to zero. Since we can write any $x$ as $x=ku$, that means $f(x)=f(k)f(u)=0.$ So the kernel is all of $F$.
Therefore the only ideals (homomorphism kernels) of a field are ${0}$ and $F$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.
More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $x$ in a ring $R$ to be $x=ku$, and the behavior of $x$ under any group homomorphism $g$ is determined by the behavior of $k$ and of $u$ under $g$, since $g(x)=g(k)g(u)$. This explains why $u$ is called a "unit" since we can measure behavior of element $x$ in terms of what coefficient $k$ is in front of $u$ when we express $x$ in terms of $u$.
Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:
Theorem (alternate definition of a unit): $uin R$ is a unit iff for all $xin R$, there exists a $vin R$ such that $(xv)u=x$.
The fowards case of this theorem is proven above, while the backward case is provable letting $x=1$.
answered Feb 3 at 4:31
MathTrainMathTrain
690217
answered Feb 3 at 4:31
MathTrainMathTrain
690217
answered Feb 3 at 4:31
MathTrainMathTrain
690217
answered Feb 3 at 4:31
MathTrainMathTrain
690217
690217
add a comment |
add a comment |
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I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide.
$endgroup$
– Omnomnomnom
Oct 5 '15 at 5:54
$begingroup$
I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers)
$endgroup$
– Eric Wofsey
Oct 5 '15 at 5:58