Representing an algebra over a field using a quotient ring
$begingroup$
My professor was talking about different ways to think about an algebra $A$ over a field $k$. One that she mentioned briefly but didn't go into much detail on was that of a quotient ring. Roughly the idea was to take vectors $v_1,...,v_n$ as a basis for $A$ and construct the algebra as $A = k[v_1,...,v_n]/I$, where $I$ was an ideal for which we didn't get a specific definition in lecture.
Is there a way we can pick $I$ such that $A$ will be an algebra?
(P.S. we were specifically talking about algebras that are also integral domains, but AFAIK that's super relevant)
@Enkidu's answer was really helpful in refining the question I wanted to be asking, which was specifically about the finite-dimensional case. I'm going to leave this question be in case it's useful for others. If you're interested in the finite dimensional case see this revised question.
abstract-algebra field-theory ideals algebras
asked Jan 18 at 7:34
JFoxJFox
1747
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add a comment |
$begingroup$
My professor was talking about different ways to think about an algebra $A$ over a field $k$. One that she mentioned briefly but didn't go into much detail on was that of a quotient ring. Roughly the idea was to take vectors $v_1,...,v_n$ as a basis for $A$ and construct the algebra as $A = k[v_1,...,v_n]/I$, where $I$ was an ideal for which we didn't get a specific definition in lecture.
Is there a way we can pick $I$ such that $A$ will be an algebra?
(P.S. we were specifically talking about algebras that are also integral domains, but AFAIK that's super relevant)
@Enkidu's answer was really helpful in refining the question I wanted to be asking, which was specifically about the finite-dimensional case. I'm going to leave this question be in case it's useful for others. If you're interested in the finite dimensional case see this revised question.
abstract-algebra field-theory ideals algebras
asked Jan 18 at 7:34
JFoxJFox
1747
$endgroup$
add a comment |
$begingroup$
My professor was talking about different ways to think about an algebra $A$ over a field $k$. One that she mentioned briefly but didn't go into much detail on was that of a quotient ring. Roughly the idea was to take vectors $v_1,...,v_n$ as a basis for $A$ and construct the algebra as $A = k[v_1,...,v_n]/I$, where $I$ was an ideal for which we didn't get a specific definition in lecture.
Is there a way we can pick $I$ such that $A$ will be an algebra?
(P.S. we were specifically talking about algebras that are also integral domains, but AFAIK that's super relevant)
@Enkidu's answer was really helpful in refining the question I wanted to be asking, which was specifically about the finite-dimensional case. I'm going to leave this question be in case it's useful for others. If you're interested in the finite dimensional case see this revised question.
abstract-algebra field-theory ideals algebras
asked Jan 18 at 7:34
JFoxJFox
1747
$endgroup$
My professor was talking about different ways to think about an algebra $A$ over a field $k$. One that she mentioned briefly but didn't go into much detail on was that of a quotient ring. Roughly the idea was to take vectors $v_1,...,v_n$ as a basis for $A$ and construct the algebra as $A = k[v_1,...,v_n]/I$, where $I$ was an ideal for which we didn't get a specific definition in lecture.
Is there a way we can pick $I$ such that $A$ will be an algebra?
(P.S. we were specifically talking about algebras that are also integral domains, but AFAIK that's super relevant)
@Enkidu's answer was really helpful in refining the question I wanted to be asking, which was specifically about the finite-dimensional case. I'm going to leave this question be in case it's useful for others. If you're interested in the finite dimensional case see this revised question.
abstract-algebra field-theory ideals algebras
abstract-algebra field-theory ideals algebras
asked Jan 18 at 7:34
JFoxJFox
1747
asked Jan 18 at 7:34
JFoxJFox
1747
edited Jan 18 at 23:16
JFox
asked Jan 18 at 7:34
JFoxJFox
1747
asked Jan 18 at 7:34
JFoxJFox
1747
asked Jan 18 at 7:34
JFoxJFox
1747
1747
add a comment |
add a comment |
1 Answer
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yep, any Ideal of an algebra will give you an algebra again, and as $k[x_1,...,x_n]$ is an algebra that works out (however, possibly not an integral domain), since this ideal would also be a vector space apart from just being an Ideal.
To get an integral domain it actually suffices for the ideal to be prime, and if it even is maximal, your algebra even becomes a field. Hwoever, I recommend looking up the definition of an Ideal, because it is a pretty powerful notion and allows fun exercises like (for $R$ comm. Ring, and $Isubset R$ Ideal):
$$R/I textrm{ integral domain } iff I textrm{ prime}\R/I textrm{ field } iff I textrm{ maximal}$$
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
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$begingroup$
yep, any Ideal of an algebra will give you an algebra again, and as $k[x_1,...,x_n]$ is an algebra that works out (however, possibly not an integral domain), since this ideal would also be a vector space apart from just being an Ideal.
To get an integral domain it actually suffices for the ideal to be prime, and if it even is maximal, your algebra even becomes a field. Hwoever, I recommend looking up the definition of an Ideal, because it is a pretty powerful notion and allows fun exercises like (for $R$ comm. Ring, and $Isubset R$ Ideal):
$$R/I textrm{ integral domain } iff I textrm{ prime}\R/I textrm{ field } iff I textrm{ maximal}$$
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
$endgroup$
add a comment |
$begingroup$
yep, any Ideal of an algebra will give you an algebra again, and as $k[x_1,...,x_n]$ is an algebra that works out (however, possibly not an integral domain), since this ideal would also be a vector space apart from just being an Ideal.
To get an integral domain it actually suffices for the ideal to be prime, and if it even is maximal, your algebra even becomes a field. Hwoever, I recommend looking up the definition of an Ideal, because it is a pretty powerful notion and allows fun exercises like (for $R$ comm. Ring, and $Isubset R$ Ideal):
$$R/I textrm{ integral domain } iff I textrm{ prime}\R/I textrm{ field } iff I textrm{ maximal}$$
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
$endgroup$
add a comment |
$begingroup$
yep, any Ideal of an algebra will give you an algebra again, and as $k[x_1,...,x_n]$ is an algebra that works out (however, possibly not an integral domain), since this ideal would also be a vector space apart from just being an Ideal.
To get an integral domain it actually suffices for the ideal to be prime, and if it even is maximal, your algebra even becomes a field. Hwoever, I recommend looking up the definition of an Ideal, because it is a pretty powerful notion and allows fun exercises like (for $R$ comm. Ring, and $Isubset R$ Ideal):
$$R/I textrm{ integral domain } iff I textrm{ prime}\R/I textrm{ field } iff I textrm{ maximal}$$
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
$endgroup$
yep, any Ideal of an algebra will give you an algebra again, and as $k[x_1,...,x_n]$ is an algebra that works out (however, possibly not an integral domain), since this ideal would also be a vector space apart from just being an Ideal.
To get an integral domain it actually suffices for the ideal to be prime, and if it even is maximal, your algebra even becomes a field. Hwoever, I recommend looking up the definition of an Ideal, because it is a pretty powerful notion and allows fun exercises like (for $R$ comm. Ring, and $Isubset R$ Ideal):
$$R/I textrm{ integral domain } iff I textrm{ prime}\R/I textrm{ field } iff I textrm{ maximal}$$
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
answered Jan 18 at 8:34
EnkiduEnkidu
1,44429
1,44429
add a comment |
add a comment |
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