Mixing
Integral domain.
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Let k a field.
Let $ A = k+ x^2 k[x] $,
Show that A is integral domain and finite type.
An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $
I know that A is a subalgebra of k[x],
Any hint for integral domain
abstract-algebra ring-theory
asked Aug 6 at 19:56
Ali NoumSali Traore
486
add a comment |Â
up vote
1
down vote
favorite
Let k a field.
Let $ A = k+ x^2 k[x] $,
Show that A is integral domain and finite type.
An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $
I know that A is a subalgebra of k[x],
Any hint for integral domain
abstract-algebra ring-theory
asked Aug 6 at 19:56
Ali NoumSali Traore
486
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let k a field.
Let $ A = k+ x^2 k[x] $,
Show that A is integral domain and finite type.
An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $
I know that A is a subalgebra of k[x],
Any hint for integral domain
abstract-algebra ring-theory
asked Aug 6 at 19:56
Ali NoumSali Traore
486
Let k a field.
Let $ A = k+ x^2 k[x] $,
Show that A is integral domain and finite type.
An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $
I know that A is a subalgebra of k[x],
Any hint for integral domain
abstract-algebra ring-theory
asked Aug 6 at 19:56
Ali NoumSali Traore
486
edited Aug 6 at 20:15
asked Aug 6 at 19:56
Ali NoumSali Traore
486
asked Aug 6 at 19:56
Ali NoumSali Traore
486
asked Aug 6 at 19:56
Ali NoumSali Traore
486
486
add a comment |Â
add a comment |Â
1 Answer
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2
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Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.
answered Aug 6 at 20:02
paf
3,9391823
An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
– Ali NoumSali Traore
Aug 6 at 20:06
Yes and such an element is a polynomial in $k[x]$.
– paf
Aug 6 at 20:07
Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
– misogrumpy
Aug 6 at 20:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.
answered Aug 6 at 20:02
paf
3,9391823
An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
– Ali NoumSali Traore
Aug 6 at 20:06
Yes and such an element is a polynomial in $k[x]$.
– paf
Aug 6 at 20:07
Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
– misogrumpy
Aug 6 at 20:36
add a comment |Â
up vote
2
down vote
Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.
answered Aug 6 at 20:02
paf
3,9391823
An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
– Ali NoumSali Traore
Aug 6 at 20:06
Yes and such an element is a polynomial in $k[x]$.
– paf
Aug 6 at 20:07
Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
– misogrumpy
Aug 6 at 20:36
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.
answered Aug 6 at 20:02
paf
3,9391823
Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.
answered Aug 6 at 20:02
paf
3,9391823
answered Aug 6 at 20:02
paf
3,9391823
answered Aug 6 at 20:02
paf
3,9391823
answered Aug 6 at 20:02
paf
3,9391823
3,9391823
An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
– Ali NoumSali Traore
Aug 6 at 20:06
Yes and such an element is a polynomial in $k[x]$.
– paf
Aug 6 at 20:07
Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
– misogrumpy
Aug 6 at 20:36
add a comment |Â
An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
– Ali NoumSali Traore
Aug 6 at 20:06
Yes and such an element is a polynomial in $k[x]$.
– paf
Aug 6 at 20:07
Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
– misogrumpy
Aug 6 at 20:36
An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
– Ali NoumSali Traore
Aug 6 at 20:06
An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
– Ali NoumSali Traore
Aug 6 at 20:06
Yes and such an element is a polynomial in $k[x]$.
– paf
Aug 6 at 20:07
Yes and such an element is a polynomial in $k[x]$.
– paf
Aug 6 at 20:07
Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
– misogrumpy
Aug 6 at 20:36
Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
– misogrumpy
Aug 6 at 20:36
add a comment |Â
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