Consider $R=k[x_1,x_2,…x_n]$. Difference between $(f_1,..,f_k)$ and $k[f_1,…f_k]$Why can't the Polynomial...
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Consider $R=k[x_1,x_2,…x_n]$. Difference between $(f_1,..,f_k)$ and $k[f_1,…f_k]$
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I'm confused on the difference between these two objects. My book says that we can see the difference when considering the simple case where $f_1=x^2in k[x]$, and see that $1in k[x^2]$, but $1notin (x^2)$.
However I couldn't figure out why $1in k[x^2]$. Could anyone help illustrate this?
abstract-algebra ring-theory
asked Mar 12 at 22:30
davidhdavidh
3178
$endgroup$
add a comment |
$begingroup$
I'm confused on the difference between these two objects. My book says that we can see the difference when considering the simple case where $f_1=x^2in k[x]$, and see that $1in k[x^2]$, but $1notin (x^2)$.
However I couldn't figure out why $1in k[x^2]$. Could anyone help illustrate this?
abstract-algebra ring-theory
asked Mar 12 at 22:30
davidhdavidh
3178
$endgroup$
add a comment |
$begingroup$
I'm confused on the difference between these two objects. My book says that we can see the difference when considering the simple case where $f_1=x^2in k[x]$, and see that $1in k[x^2]$, but $1notin (x^2)$.
However I couldn't figure out why $1in k[x^2]$. Could anyone help illustrate this?
abstract-algebra ring-theory
asked Mar 12 at 22:30
davidhdavidh
3178
$endgroup$
I'm confused on the difference between these two objects. My book says that we can see the difference when considering the simple case where $f_1=x^2in k[x]$, and see that $1in k[x^2]$, but $1notin (x^2)$.
However I couldn't figure out why $1in k[x^2]$. Could anyone help illustrate this?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Mar 12 at 22:30
davidhdavidh
3178
asked Mar 12 at 22:30
davidhdavidh
3178
asked Mar 12 at 22:30
davidhdavidh
3178
asked Mar 12 at 22:30
davidhdavidh
3178
asked Mar 12 at 22:30
davidhdavidh
3178
3178
add a comment |
add a comment |
1 Answer
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$k[x^2]$ denotes the set of polynomials in $x^2$ with coefficients in the field $k$, i.e. the set of (ordinary) polynomials with only monomials of even degree (+0). $1$ has degree $0$, as all non-zero constants, and if I remember well, $0$ is even. So $1$ belongs to $k[x^2]$.
answered Mar 12 at 22:42
BernardBernard
123k741117
$endgroup$
$begingroup$
lol, yeah that makes sense. Thanks.
$endgroup$
– davidh
Mar 12 at 22:48
add a comment |
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1 Answer
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$begingroup$
$k[x^2]$ denotes the set of polynomials in $x^2$ with coefficients in the field $k$, i.e. the set of (ordinary) polynomials with only monomials of even degree (+0). $1$ has degree $0$, as all non-zero constants, and if I remember well, $0$ is even. So $1$ belongs to $k[x^2]$.
answered Mar 12 at 22:42
BernardBernard
123k741117
$endgroup$
$begingroup$
lol, yeah that makes sense. Thanks.
$endgroup$
– davidh
Mar 12 at 22:48
add a comment |
$begingroup$
$k[x^2]$ denotes the set of polynomials in $x^2$ with coefficients in the field $k$, i.e. the set of (ordinary) polynomials with only monomials of even degree (+0). $1$ has degree $0$, as all non-zero constants, and if I remember well, $0$ is even. So $1$ belongs to $k[x^2]$.
answered Mar 12 at 22:42
BernardBernard
123k741117
$endgroup$
$begingroup$
lol, yeah that makes sense. Thanks.
$endgroup$
– davidh
Mar 12 at 22:48
add a comment |
$begingroup$
$k[x^2]$ denotes the set of polynomials in $x^2$ with coefficients in the field $k$, i.e. the set of (ordinary) polynomials with only monomials of even degree (+0). $1$ has degree $0$, as all non-zero constants, and if I remember well, $0$ is even. So $1$ belongs to $k[x^2]$.
answered Mar 12 at 22:42
BernardBernard
123k741117
$endgroup$
$k[x^2]$ denotes the set of polynomials in $x^2$ with coefficients in the field $k$, i.e. the set of (ordinary) polynomials with only monomials of even degree (+0). $1$ has degree $0$, as all non-zero constants, and if I remember well, $0$ is even. So $1$ belongs to $k[x^2]$.
answered Mar 12 at 22:42
BernardBernard
123k741117
answered Mar 12 at 22:42
BernardBernard
123k741117
answered Mar 12 at 22:42
BernardBernard
123k741117
answered Mar 12 at 22:42
BernardBernard
123k741117
123k741117
$begingroup$
lol, yeah that makes sense. Thanks.
$endgroup$
– davidh
Mar 12 at 22:48
add a comment |
$begingroup$
lol, yeah that makes sense. Thanks.
$endgroup$
– davidh
Mar 12 at 22:48
$begingroup$
lol, yeah that makes sense. Thanks.
$endgroup$
– davidh
Mar 12 at 22:48
$begingroup$
lol, yeah that makes sense. Thanks.
$endgroup$
– davidh
Mar 12 at 22:48
add a comment |
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