Mixing
Let $R$ be a ring and $X$ a non-empty set. Considering $A=R^X$, describe the ring $(A,+,cdot)$
$begingroup$
An exercise asks me to describe this ring:
Let $R$ be a ring and $X$ a non-empty set. Consider $A=R^X$ with sum and multiplication defined by:
$forall ,,f,gin A, forall xin X$: $,,,,(f+g)(x)=f(x)+g(x)$; $,,,,,(fcdot g)(x)=f(x)cdot g(x)$.
Now $(A,+,cdot)$ is a commutative ring. Identities are $c_0(x)=0_X$ and $c_1(x)=1_X, forall xin X$ and, supposing to know the invertible elements of $X$, those of $A$ are the set of applications which don't assume zero as a value.
About of zero divisors of $A$:
considering an application $f in A, ,,fneq c_0$ such that $f(x)=0$ for some $xin X$ and $f(bar x)neq 0, ,,, forall bar x in X$ with $bar x neq x$. I can assume that exists an application $g in A$ such that:
$$ g(y) = begin{cases}
0 &text{if};;; y neq x \
0 neq k in X &text{if};;; y = x end{cases}$$
so $f,g neq c_0$ and $f cdot g = c_0$. Hence can I say that the set of zero divisors of $A$ is composed by all the applications which assume $0$ as a value?
Let $D$ be that set, is $D$ an ideal of $A$? I am not sure that I can suppose $forall f,g in D$ that $(f-g)in D$ where for $f cdot g$ is trivial.
Lastly, agreed upon $X={0,1}$, show an isomorphism between $R^X$ and $R times R$:
Can I consider $lambda_a :X to R$ such that $lambda_a (x) = a cdot x$, $,, forall x in X$, $forall ain R,,,,$ and $,,,, phi : R^X to R times R$ such that $phi(lambda_a) = (a cdot x,a cdot x)$? Or do I have a loss of informations?
abstract-algebra ideals
edited Jan 2 at 21:53
user26857
39.3k124183
asked Jan 2 at 9:04
F.incF.inc
394110
$endgroup$
add a comment |
$begingroup$
An exercise asks me to describe this ring:
Let $R$ be a ring and $X$ a non-empty set. Consider $A=R^X$ with sum and multiplication defined by:
$forall ,,f,gin A, forall xin X$: $,,,,(f+g)(x)=f(x)+g(x)$; $,,,,,(fcdot g)(x)=f(x)cdot g(x)$.
Now $(A,+,cdot)$ is a commutative ring. Identities are $c_0(x)=0_X$ and $c_1(x)=1_X, forall xin X$ and, supposing to know the invertible elements of $X$, those of $A$ are the set of applications which don't assume zero as a value.
About of zero divisors of $A$:
considering an application $f in A, ,,fneq c_0$ such that $f(x)=0$ for some $xin X$ and $f(bar x)neq 0, ,,, forall bar x in X$ with $bar x neq x$. I can assume that exists an application $g in A$ such that:
$$ g(y) = begin{cases}
0 &text{if};;; y neq x \
0 neq k in X &text{if};;; y = x end{cases}$$
so $f,g neq c_0$ and $f cdot g = c_0$. Hence can I say that the set of zero divisors of $A$ is composed by all the applications which assume $0$ as a value?
Let $D$ be that set, is $D$ an ideal of $A$? I am not sure that I can suppose $forall f,g in D$ that $(f-g)in D$ where for $f cdot g$ is trivial.
Lastly, agreed upon $X={0,1}$, show an isomorphism between $R^X$ and $R times R$:
Can I consider $lambda_a :X to R$ such that $lambda_a (x) = a cdot x$, $,, forall x in X$, $forall ain R,,,,$ and $,,,, phi : R^X to R times R$ such that $phi(lambda_a) = (a cdot x,a cdot x)$? Or do I have a loss of informations?
abstract-algebra ideals
edited Jan 2 at 21:53
user26857
39.3k124183
asked Jan 2 at 9:04
F.incF.inc
394110
$endgroup$
$begingroup$
The elements in ${0,1}$ have no meaning other than just two formal elements, they are not $0_R$ and $1_R$ and you can not multiply them with elements of $R$.
$endgroup$
– mouthetics
Jan 2 at 15:41
add a comment |
$begingroup$
An exercise asks me to describe this ring:
Let $R$ be a ring and $X$ a non-empty set. Consider $A=R^X$ with sum and multiplication defined by:
$forall ,,f,gin A, forall xin X$: $,,,,(f+g)(x)=f(x)+g(x)$; $,,,,,(fcdot g)(x)=f(x)cdot g(x)$.
Now $(A,+,cdot)$ is a commutative ring. Identities are $c_0(x)=0_X$ and $c_1(x)=1_X, forall xin X$ and, supposing to know the invertible elements of $X$, those of $A$ are the set of applications which don't assume zero as a value.
About of zero divisors of $A$:
considering an application $f in A, ,,fneq c_0$ such that $f(x)=0$ for some $xin X$ and $f(bar x)neq 0, ,,, forall bar x in X$ with $bar x neq x$. I can assume that exists an application $g in A$ such that:
$$ g(y) = begin{cases}
0 &text{if};;; y neq x \
0 neq k in X &text{if};;; y = x end{cases}$$
so $f,g neq c_0$ and $f cdot g = c_0$. Hence can I say that the set of zero divisors of $A$ is composed by all the applications which assume $0$ as a value?
Let $D$ be that set, is $D$ an ideal of $A$? I am not sure that I can suppose $forall f,g in D$ that $(f-g)in D$ where for $f cdot g$ is trivial.
Lastly, agreed upon $X={0,1}$, show an isomorphism between $R^X$ and $R times R$:
Can I consider $lambda_a :X to R$ such that $lambda_a (x) = a cdot x$, $,, forall x in X$, $forall ain R,,,,$ and $,,,, phi : R^X to R times R$ such that $phi(lambda_a) = (a cdot x,a cdot x)$? Or do I have a loss of informations?
abstract-algebra ideals
edited Jan 2 at 21:53
user26857
39.3k124183
asked Jan 2 at 9:04
F.incF.inc
394110
$endgroup$
An exercise asks me to describe this ring:
Let $R$ be a ring and $X$ a non-empty set. Consider $A=R^X$ with sum and multiplication defined by:
$forall ,,f,gin A, forall xin X$: $,,,,(f+g)(x)=f(x)+g(x)$; $,,,,,(fcdot g)(x)=f(x)cdot g(x)$.
Now $(A,+,cdot)$ is a commutative ring. Identities are $c_0(x)=0_X$ and $c_1(x)=1_X, forall xin X$ and, supposing to know the invertible elements of $X$, those of $A$ are the set of applications which don't assume zero as a value.
About of zero divisors of $A$:
considering an application $f in A, ,,fneq c_0$ such that $f(x)=0$ for some $xin X$ and $f(bar x)neq 0, ,,, forall bar x in X$ with $bar x neq x$. I can assume that exists an application $g in A$ such that:
$$ g(y) = begin{cases}
0 &text{if};;; y neq x \
0 neq k in X &text{if};;; y = x end{cases}$$
so $f,g neq c_0$ and $f cdot g = c_0$. Hence can I say that the set of zero divisors of $A$ is composed by all the applications which assume $0$ as a value?
Let $D$ be that set, is $D$ an ideal of $A$? I am not sure that I can suppose $forall f,g in D$ that $(f-g)in D$ where for $f cdot g$ is trivial.
Lastly, agreed upon $X={0,1}$, show an isomorphism between $R^X$ and $R times R$:
Can I consider $lambda_a :X to R$ such that $lambda_a (x) = a cdot x$, $,, forall x in X$, $forall ain R,,,,$ and $,,,, phi : R^X to R times R$ such that $phi(lambda_a) = (a cdot x,a cdot x)$? Or do I have a loss of informations?
abstract-algebra ideals
abstract-algebra ideals
edited Jan 2 at 21:53
user26857
39.3k124183
asked Jan 2 at 9:04
F.incF.inc
394110
edited Jan 2 at 21:53
user26857
39.3k124183
asked Jan 2 at 9:04
F.incF.inc
394110
edited Jan 2 at 21:53
user26857
39.3k124183
edited Jan 2 at 21:53
user26857
39.3k124183
edited Jan 2 at 21:53
user26857
39.3k124183
39.3k124183
asked Jan 2 at 9:04
F.incF.inc
394110
asked Jan 2 at 9:04
F.incF.inc
394110
asked Jan 2 at 9:04
F.incF.inc
394110
394110
$begingroup$
The elements in ${0,1}$ have no meaning other than just two formal elements, they are not $0_R$ and $1_R$ and you can not multiply them with elements of $R$.
$endgroup$
– mouthetics
Jan 2 at 15:41
add a comment |
$begingroup$
The elements in ${0,1}$ have no meaning other than just two formal elements, they are not $0_R$ and $1_R$ and you can not multiply them with elements of $R$.
$endgroup$
– mouthetics
Jan 2 at 15:41
$begingroup$
The elements in ${0,1}$ have no meaning other than just two formal elements, they are not $0_R$ and $1_R$ and you can not multiply them with elements of $R$.
$endgroup$
– mouthetics
Jan 2 at 15:41
$begingroup$
The elements in ${0,1}$ have no meaning other than just two formal elements, they are not $0_R$ and $1_R$ and you can not multiply them with elements of $R$.
$endgroup$
– mouthetics
Jan 2 at 15:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
1) The zero divisors of $R^X$ are those functions which either vanishes somewhere or admits a zero divisor as value.
2) $D$ is not stable under addition. Take for example
$$f(x)=begin{cases} 1 ; text{if } xin A \
0 ; text{otherwise} end{cases} qquad text{and} qquad g(x)=begin{cases} 1 ; text{if } xin bar{A} \
0 ; text{otherwise} end{cases},$$
for some $emptyset subsetneq A subsetneq X.$
Then $f,gin D$ while $f+gnotin D.$
3) A function in $fin R^X$ is uniquely determined by its values on ${0,1}$. So if $f(0)=x$ and $f(1)=y$ you can define $phi : R^X longrightarrow R times R, , phi(f)=(x,y) .$ This is a ring homomorphism from the definition of $R^X$, it has inverse
$psi : R times R longrightarrow R^X, (x,y)mapsto (f: f(0)=x, f(1)=y)$.
edited Jan 2 at 21:54
user26857
39.3k124183
answered Jan 2 at 15:32
moutheticsmouthetics
50127
$endgroup$
add a comment |
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1 Answer
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$begingroup$
1) The zero divisors of $R^X$ are those functions which either vanishes somewhere or admits a zero divisor as value.
2) $D$ is not stable under addition. Take for example
$$f(x)=begin{cases} 1 ; text{if } xin A \
0 ; text{otherwise} end{cases} qquad text{and} qquad g(x)=begin{cases} 1 ; text{if } xin bar{A} \
0 ; text{otherwise} end{cases},$$
for some $emptyset subsetneq A subsetneq X.$
Then $f,gin D$ while $f+gnotin D.$
3) A function in $fin R^X$ is uniquely determined by its values on ${0,1}$. So if $f(0)=x$ and $f(1)=y$ you can define $phi : R^X longrightarrow R times R, , phi(f)=(x,y) .$ This is a ring homomorphism from the definition of $R^X$, it has inverse
$psi : R times R longrightarrow R^X, (x,y)mapsto (f: f(0)=x, f(1)=y)$.
edited Jan 2 at 21:54
user26857
39.3k124183
answered Jan 2 at 15:32
moutheticsmouthetics
50127
$endgroup$
add a comment |
$begingroup$
1) The zero divisors of $R^X$ are those functions which either vanishes somewhere or admits a zero divisor as value.
2) $D$ is not stable under addition. Take for example
$$f(x)=begin{cases} 1 ; text{if } xin A \
0 ; text{otherwise} end{cases} qquad text{and} qquad g(x)=begin{cases} 1 ; text{if } xin bar{A} \
0 ; text{otherwise} end{cases},$$
for some $emptyset subsetneq A subsetneq X.$
Then $f,gin D$ while $f+gnotin D.$
3) A function in $fin R^X$ is uniquely determined by its values on ${0,1}$. So if $f(0)=x$ and $f(1)=y$ you can define $phi : R^X longrightarrow R times R, , phi(f)=(x,y) .$ This is a ring homomorphism from the definition of $R^X$, it has inverse
$psi : R times R longrightarrow R^X, (x,y)mapsto (f: f(0)=x, f(1)=y)$.
edited Jan 2 at 21:54
user26857
39.3k124183
answered Jan 2 at 15:32
moutheticsmouthetics
50127
$endgroup$
add a comment |
$begingroup$
1) The zero divisors of $R^X$ are those functions which either vanishes somewhere or admits a zero divisor as value.
2) $D$ is not stable under addition. Take for example
$$f(x)=begin{cases} 1 ; text{if } xin A \
0 ; text{otherwise} end{cases} qquad text{and} qquad g(x)=begin{cases} 1 ; text{if } xin bar{A} \
0 ; text{otherwise} end{cases},$$
for some $emptyset subsetneq A subsetneq X.$
Then $f,gin D$ while $f+gnotin D.$
3) A function in $fin R^X$ is uniquely determined by its values on ${0,1}$. So if $f(0)=x$ and $f(1)=y$ you can define $phi : R^X longrightarrow R times R, , phi(f)=(x,y) .$ This is a ring homomorphism from the definition of $R^X$, it has inverse
$psi : R times R longrightarrow R^X, (x,y)mapsto (f: f(0)=x, f(1)=y)$.
edited Jan 2 at 21:54
user26857
39.3k124183
answered Jan 2 at 15:32
moutheticsmouthetics
50127
$endgroup$
1) The zero divisors of $R^X$ are those functions which either vanishes somewhere or admits a zero divisor as value.
2) $D$ is not stable under addition. Take for example
$$f(x)=begin{cases} 1 ; text{if } xin A \
0 ; text{otherwise} end{cases} qquad text{and} qquad g(x)=begin{cases} 1 ; text{if } xin bar{A} \
0 ; text{otherwise} end{cases},$$
for some $emptyset subsetneq A subsetneq X.$
Then $f,gin D$ while $f+gnotin D.$
3) A function in $fin R^X$ is uniquely determined by its values on ${0,1}$. So if $f(0)=x$ and $f(1)=y$ you can define $phi : R^X longrightarrow R times R, , phi(f)=(x,y) .$ This is a ring homomorphism from the definition of $R^X$, it has inverse
$psi : R times R longrightarrow R^X, (x,y)mapsto (f: f(0)=x, f(1)=y)$.
edited Jan 2 at 21:54
user26857
39.3k124183
answered Jan 2 at 15:32
moutheticsmouthetics
50127
edited Jan 2 at 21:54
user26857
39.3k124183
edited Jan 2 at 21:54
user26857
39.3k124183
edited Jan 2 at 21:54
user26857
39.3k124183
39.3k124183
answered Jan 2 at 15:32
moutheticsmouthetics
50127
answered Jan 2 at 15:32
moutheticsmouthetics
50127
answered Jan 2 at 15:32
moutheticsmouthetics
50127
50127
add a comment |
add a comment |
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$begingroup$
The elements in ${0,1}$ have no meaning other than just two formal elements, they are not $0_R$ and $1_R$ and you can not multiply them with elements of $R$.
$endgroup$
– mouthetics
Jan 2 at 15:41