Why is $f(x) = 5x$ not a homomorphism?
0
$begingroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,03792030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
$endgroup$
add a comment |
0
$begingroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,03792030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
$endgroup$
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
add a comment |
0
0
0
$begingroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,03792030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
$endgroup$
Why is the function $f: mathbb Z to mathbb Z$ given by $f(x) = 5x$ not a homorphism, since $f(a+b) = 5(a+b) = 5a + 5b = f(a) + f(b)$, and same for $f(a*b)$.
abstract-algebra ring-theory ring-homomorphism
abstract-algebra ring-theory ring-homomorphism
edited Dec 13 '18 at 21:52
Henrik
6,03792030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
edited Dec 13 '18 at 21:52
Henrik
6,03792030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
edited Dec 13 '18 at 21:52
Henrik
6,03792030
edited Dec 13 '18 at 21:52
Henrik
6,03792030
edited Dec 13 '18 at 21:52
Henrik
6,03792030
6,03792030
asked Dec 13 '18 at 21:48
imariptideimariptide
72
asked Dec 13 '18 at 21:48
imariptideimariptide
72
asked Dec 13 '18 at 21:48
imariptideimariptide
72
72
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
add a comment |
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
1
1
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17
add a comment |
1 Answer
1
active
oldest
votes
3
$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
903313
$endgroup$
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
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1 Answer
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$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
903313
$endgroup$
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
3
$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
903313
$endgroup$
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
3
3
3
$begingroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
903313
$endgroup$
Since $mathbb{Z}$ is a ring, to be an homomorphism you must have $f(a+b)=f(a)+f(b)$ and $f(a*b)=f(a)*f(b)$ and $f(1)=1$.
But $30=f(6)=f(2*3) neq f(2)*f(3)=10*15=150$ and $f(1)=5$
answered Dec 13 '18 at 21:52
user289143user289143
903313
answered Dec 13 '18 at 21:52
user289143user289143
903313
answered Dec 13 '18 at 21:52
user289143user289143
903313
answered Dec 13 '18 at 21:52
user289143user289143
903313
903313
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
1
1
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
Even without requiring $f(1)=1$ (not all authors require this of ring homomorphisms), we still need to have $f(1)=f(1)^2$, which we clearly don't have.
$endgroup$
– Arthur
Dec 13 '18 at 21:58
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
@Arthur indeed, but the alternative $f(1)=0$ would lead to trivial $f=0$, so it is quite an obvious requirement to force $f(1)=1$.
$endgroup$
– zwim
Dec 13 '18 at 22:15
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
In this case, sure. But as a general requirement for ring homomorphisms, $f(1)=1$ is not always assumed.
$endgroup$
– Arthur
Dec 13 '18 at 22:17
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
It depends if you consider unit rings or just rings. In this case, since $mathbb{Z}$ is a unit ring, $f(1)$ must be equal to $1$. But even if we consider $mathbb{Z}$ as a ring ("forgetting" the unit), this map doesn't preserve multiplication, so it's not an homomorphism.
$endgroup$
– user289143
Dec 13 '18 at 22:19
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
$begingroup$
@user289143 Not quite right. There are rings with units and homomorphisms inbetween not being unit-preserving. Eg: take any ring $R$ with unit $1_R$ and send $R$ to $text{M}_2(R)$ by embedding into the upper-left corner with zeroes elsewhere. This is an injective homomorphism not sending $1_R$ to the unit in $text{M}_2(R)$.
$endgroup$
– Munk
Dec 14 '18 at 18:46
add a comment |
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I would assume that "rings" in your definition must have $1$. And "ring homomorphisms" in your case must also map $1$ to $1$. Clearly, $f(1)=5$ does not meet this requirement.
$endgroup$
– user614671
Dec 13 '18 at 21:49
$begingroup$
You said same for f(ab), you should have written it : $f(ab)=5(ab)neq f(a)f(b)=5a5b=25(ab)$ to notice it is not the case.
$endgroup$
– zwim
Dec 13 '18 at 22:17