Basis of a polynomial ring
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Considering $mathbb{C}[x,y]$ as a direct sum $bigoplus_{i,j=1}^{infty} mathbb{C} cdot x^iy^j$, it can be viewed as an infinite dimensional vector space. In the same spirit, I am trying to calculate a C-linear basis for a polynomial quotient ring like $mathbb{C}[x,y]/(x^n-y+y^n)$. I know that it should be doable with the division algorithm but in this context it's not clear to me how it should be applied.
linear-algebra abstract-algebra
edited 6 hours ago
Christopher
6,24811628
asked 6 hours ago
Karl
296
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show 1 more comment
up vote
0
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Considering $mathbb{C}[x,y]$ as a direct sum $bigoplus_{i,j=1}^{infty} mathbb{C} cdot x^iy^j$, it can be viewed as an infinite dimensional vector space. In the same spirit, I am trying to calculate a C-linear basis for a polynomial quotient ring like $mathbb{C}[x,y]/(x^n-y+y^n)$. I know that it should be doable with the division algorithm but in this context it's not clear to me how it should be applied.
linear-algebra abstract-algebra
edited 6 hours ago
Christopher
6,24811628
asked 6 hours ago
Karl
296
Hint: it is a bit easier when you think about $mathbb{C}[x,y]$ as $(mathbb{C}[y])[x]$ (the round brackets are superfluous from a mathematical perspective, but try to clarify the 'think about as' part)
– Vincent
5 hours ago
that could help me write $mathbb{C}[x,y]/(x^n-y+y^n)$ as $bigoplus_{i=0}^{n -1}x^i cdot mathbb{C}[y]$ so that $(1,ldots,x^{n-1})$ would be a basis but is this all that can be done?
– Karl
5 hours ago
Sort-of... This way of writing shows that you only need to consider powers of $x$ up to $n-1$, but in order to get a basis over $mathbb{C}$ you still have to 'break up' the $mathbb{C}[y]$ part. But this works just as in your Original Post
– Vincent
5 hours ago
Yes, $oplus_{0 leq i leq n-1, 0 leq j} mathbb{C} cdot x^i y^j $, but I was expecting to find a finite basis. Was I wrong?
– Karl
5 hours ago
Of course I was.
– Karl
4 hours ago
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Considering $mathbb{C}[x,y]$ as a direct sum $bigoplus_{i,j=1}^{infty} mathbb{C} cdot x^iy^j$, it can be viewed as an infinite dimensional vector space. In the same spirit, I am trying to calculate a C-linear basis for a polynomial quotient ring like $mathbb{C}[x,y]/(x^n-y+y^n)$. I know that it should be doable with the division algorithm but in this context it's not clear to me how it should be applied.
linear-algebra abstract-algebra
edited 6 hours ago
Christopher
6,24811628
asked 6 hours ago
Karl
296
Considering $mathbb{C}[x,y]$ as a direct sum $bigoplus_{i,j=1}^{infty} mathbb{C} cdot x^iy^j$, it can be viewed as an infinite dimensional vector space. In the same spirit, I am trying to calculate a C-linear basis for a polynomial quotient ring like $mathbb{C}[x,y]/(x^n-y+y^n)$. I know that it should be doable with the division algorithm but in this context it's not clear to me how it should be applied.
linear-algebra abstract-algebra
linear-algebra abstract-algebra
edited 6 hours ago
Christopher
6,24811628
asked 6 hours ago
Karl
296
edited 6 hours ago
Christopher
6,24811628
asked 6 hours ago
Karl
296
edited 6 hours ago
Christopher
6,24811628
edited 6 hours ago
Christopher
6,24811628
edited 6 hours ago
Christopher
6,24811628
6,24811628
asked 6 hours ago
Karl
296
asked 6 hours ago
Karl
296
asked 6 hours ago
Karl
296
296
Hint: it is a bit easier when you think about $mathbb{C}[x,y]$ as $(mathbb{C}[y])[x]$ (the round brackets are superfluous from a mathematical perspective, but try to clarify the 'think about as' part)
– Vincent
5 hours ago
that could help me write $mathbb{C}[x,y]/(x^n-y+y^n)$ as $bigoplus_{i=0}^{n -1}x^i cdot mathbb{C}[y]$ so that $(1,ldots,x^{n-1})$ would be a basis but is this all that can be done?
– Karl
5 hours ago
Sort-of... This way of writing shows that you only need to consider powers of $x$ up to $n-1$, but in order to get a basis over $mathbb{C}$ you still have to 'break up' the $mathbb{C}[y]$ part. But this works just as in your Original Post
– Vincent
5 hours ago
Yes, $oplus_{0 leq i leq n-1, 0 leq j} mathbb{C} cdot x^i y^j $, but I was expecting to find a finite basis. Was I wrong?
– Karl
5 hours ago
Of course I was.
– Karl
4 hours ago
|
show 1 more comment
Hint: it is a bit easier when you think about $mathbb{C}[x,y]$ as $(mathbb{C}[y])[x]$ (the round brackets are superfluous from a mathematical perspective, but try to clarify the 'think about as' part)
– Vincent
5 hours ago
that could help me write $mathbb{C}[x,y]/(x^n-y+y^n)$ as $bigoplus_{i=0}^{n -1}x^i cdot mathbb{C}[y]$ so that $(1,ldots,x^{n-1})$ would be a basis but is this all that can be done?
– Karl
5 hours ago
Sort-of... This way of writing shows that you only need to consider powers of $x$ up to $n-1$, but in order to get a basis over $mathbb{C}$ you still have to 'break up' the $mathbb{C}[y]$ part. But this works just as in your Original Post
– Vincent
5 hours ago
Yes, $oplus_{0 leq i leq n-1, 0 leq j} mathbb{C} cdot x^i y^j $, but I was expecting to find a finite basis. Was I wrong?
– Karl
5 hours ago
Of course I was.
– Karl
4 hours ago
Hint: it is a bit easier when you think about $mathbb{C}[x,y]$ as $(mathbb{C}[y])[x]$ (the round brackets are superfluous from a mathematical perspective, but try to clarify the 'think about as' part)
– Vincent
5 hours ago
Hint: it is a bit easier when you think about $mathbb{C}[x,y]$ as $(mathbb{C}[y])[x]$ (the round brackets are superfluous from a mathematical perspective, but try to clarify the 'think about as' part)
– Vincent
5 hours ago
that could help me write $mathbb{C}[x,y]/(x^n-y+y^n)$ as $bigoplus_{i=0}^{n -1}x^i cdot mathbb{C}[y]$ so that $(1,ldots,x^{n-1})$ would be a basis but is this all that can be done?
– Karl
5 hours ago
that could help me write $mathbb{C}[x,y]/(x^n-y+y^n)$ as $bigoplus_{i=0}^{n -1}x^i cdot mathbb{C}[y]$ so that $(1,ldots,x^{n-1})$ would be a basis but is this all that can be done?
– Karl
5 hours ago
Sort-of... This way of writing shows that you only need to consider powers of $x$ up to $n-1$, but in order to get a basis over $mathbb{C}$ you still have to 'break up' the $mathbb{C}[y]$ part. But this works just as in your Original Post
– Vincent
5 hours ago
Sort-of... This way of writing shows that you only need to consider powers of $x$ up to $n-1$, but in order to get a basis over $mathbb{C}$ you still have to 'break up' the $mathbb{C}[y]$ part. But this works just as in your Original Post
– Vincent
5 hours ago
Yes, $oplus_{0 leq i leq n-1, 0 leq j} mathbb{C} cdot x^i y^j $, but I was expecting to find a finite basis. Was I wrong?
– Karl
5 hours ago
Yes, $oplus_{0 leq i leq n-1, 0 leq j} mathbb{C} cdot x^i y^j $, but I was expecting to find a finite basis. Was I wrong?
– Karl
5 hours ago
Of course I was.
– Karl
4 hours ago
Of course I was.
– Karl
4 hours ago
|
show 1 more comment
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Hint: it is a bit easier when you think about $mathbb{C}[x,y]$ as $(mathbb{C}[y])[x]$ (the round brackets are superfluous from a mathematical perspective, but try to clarify the 'think about as' part)
– Vincent
5 hours ago
that could help me write $mathbb{C}[x,y]/(x^n-y+y^n)$ as $bigoplus_{i=0}^{n -1}x^i cdot mathbb{C}[y]$ so that $(1,ldots,x^{n-1})$ would be a basis but is this all that can be done?
– Karl
5 hours ago
Sort-of... This way of writing shows that you only need to consider powers of $x$ up to $n-1$, but in order to get a basis over $mathbb{C}$ you still have to 'break up' the $mathbb{C}[y]$ part. But this works just as in your Original Post
– Vincent
5 hours ago
Yes, $oplus_{0 leq i leq n-1, 0 leq j} mathbb{C} cdot x^i y^j $, but I was expecting to find a finite basis. Was I wrong?
– Karl
5 hours ago
Of course I was.
– Karl
4 hours ago