Maximality of an ideal for showing that an algebra is in fact a field












5












$begingroup$


I have an algebra $A$ over the field $F$, with the finite dimensionality $n$ as a vector space over $F$. I can also assume that $A$ is an integral domain. Assuming that $v_1,...,v_n$ is a spanning list of vectors and that $v_1=1$, I believe I can represent $A$ as
$F[v_1,v_2,...,v_n]/I$ where $$I=left(v_1-1, v_iv_j forall 2leq i,jleq nright).$$
Now, I'd hope to show that $A$ is a field by proving the maximality of $I$, but I can't figure out how to show that.






abstract-algebra field-theory ideals maximal-and-prime-ideals







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  • $begingroup$
    What is $R$? Your question doesn't define it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 3:43






  • 2




    $begingroup$
    The factor ring as it is defined is not integral $v_iv_j=0$ but $v_ineq0$ and $v_jneq0$.
    $endgroup$
    – mouthetics
    Jan 18 at 3:46






  • 1




    $begingroup$
    Thanks @mouthetics, I see that now. I must have misinterpreted what the lecturer was saying. I'll ask a separate question to try and clarify.
    $endgroup$
    – JFox
    Jan 18 at 4:00






  • 1




    $begingroup$
    Probably you won't like this. Let $0neq alphain A$ then $F[alpha]$ is a finite dimensional subspace of $A$ over $F$. So there is $ngeq 1$ such that ${1,alpha,ldots,alpha^n}$ is linearly dependent. $textit{i.e.}$ there is $a_0,a_1,ldots,a_n$ not all zero such that $a_0+a_1alpha+cdots+a_nalpha^n=0$. Let $0leq i<n$ be the smallest $i$ such that $a_ineq 0$. Then $a_ialpha^i+cdots+a_nalpha^n=0Rightarrow 1+{a_{i+1}over a_i}alpha+cdots+{a_nover a_i}alpha^{n-i}=0$. From this you get that $alpha$ is invertible.
    $endgroup$
    – mouthetics
    Jan 18 at 4:09












  • $begingroup$
    @mouthetics that's definitely another cool way to go about the problem thank you. I do have one question about it. How do we know there will always the powers ${1,alpha,...,alpha^n}$ will be distinct for enough n? For example, there will often be elements with multiplicative order 2, or for example in $mathbb{F_(5^4)}$ there are two elements with order 3 (which I think would still not be enough to necessarily be linearly dependent). How does the argument hold for when $alpha$ has low multiplicative order like in these cases?
    $endgroup$
    – JFox
    Jan 18 at 7:26
















5












$begingroup$


I have an algebra $A$ over the field $F$, with the finite dimensionality $n$ as a vector space over $F$. I can also assume that $A$ is an integral domain. Assuming that $v_1,...,v_n$ is a spanning list of vectors and that $v_1=1$, I believe I can represent $A$ as
$F[v_1,v_2,...,v_n]/I$ where $$I=left(v_1-1, v_iv_j forall 2leq i,jleq nright).$$
Now, I'd hope to show that $A$ is a field by proving the maximality of $I$, but I can't figure out how to show that.






abstract-algebra field-theory ideals maximal-and-prime-ideals







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is $R$? Your question doesn't define it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 3:43






  • 2




    $begingroup$
    The factor ring as it is defined is not integral $v_iv_j=0$ but $v_ineq0$ and $v_jneq0$.
    $endgroup$
    – mouthetics
    Jan 18 at 3:46






  • 1




    $begingroup$
    Thanks @mouthetics, I see that now. I must have misinterpreted what the lecturer was saying. I'll ask a separate question to try and clarify.
    $endgroup$
    – JFox
    Jan 18 at 4:00






  • 1




    $begingroup$
    Probably you won't like this. Let $0neq alphain A$ then $F[alpha]$ is a finite dimensional subspace of $A$ over $F$. So there is $ngeq 1$ such that ${1,alpha,ldots,alpha^n}$ is linearly dependent. $textit{i.e.}$ there is $a_0,a_1,ldots,a_n$ not all zero such that $a_0+a_1alpha+cdots+a_nalpha^n=0$. Let $0leq i<n$ be the smallest $i$ such that $a_ineq 0$. Then $a_ialpha^i+cdots+a_nalpha^n=0Rightarrow 1+{a_{i+1}over a_i}alpha+cdots+{a_nover a_i}alpha^{n-i}=0$. From this you get that $alpha$ is invertible.
    $endgroup$
    – mouthetics
    Jan 18 at 4:09












  • $begingroup$
    @mouthetics that's definitely another cool way to go about the problem thank you. I do have one question about it. How do we know there will always the powers ${1,alpha,...,alpha^n}$ will be distinct for enough n? For example, there will often be elements with multiplicative order 2, or for example in $mathbb{F_(5^4)}$ there are two elements with order 3 (which I think would still not be enough to necessarily be linearly dependent). How does the argument hold for when $alpha$ has low multiplicative order like in these cases?
    $endgroup$
    – JFox
    Jan 18 at 7:26














5












5








5





$begingroup$


I have an algebra $A$ over the field $F$, with the finite dimensionality $n$ as a vector space over $F$. I can also assume that $A$ is an integral domain. Assuming that $v_1,...,v_n$ is a spanning list of vectors and that $v_1=1$, I believe I can represent $A$ as
$F[v_1,v_2,...,v_n]/I$ where $$I=left(v_1-1, v_iv_j forall 2leq i,jleq nright).$$
Now, I'd hope to show that $A$ is a field by proving the maximality of $I$, but I can't figure out how to show that.






abstract-algebra field-theory ideals maximal-and-prime-ideals







share|cite|improve this question











$endgroup$




I have an algebra $A$ over the field $F$, with the finite dimensionality $n$ as a vector space over $F$. I can also assume that $A$ is an integral domain. Assuming that $v_1,...,v_n$ is a spanning list of vectors and that $v_1=1$, I believe I can represent $A$ as
$F[v_1,v_2,...,v_n]/I$ where $$I=left(v_1-1, v_iv_j forall 2leq i,jleq nright).$$
Now, I'd hope to show that $A$ is a field by proving the maximality of $I$, but I can't figure out how to show that.






abstract-algebra field-theory ideals maximal-and-prime-ideals




abstract-algebra field-theory ideals maximal-and-prime-ideals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 3:51







JFox

















asked Jan 18 at 3:11









JFoxJFox

1747




1747












  • $begingroup$
    What is $R$? Your question doesn't define it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 3:43






  • 2




    $begingroup$
    The factor ring as it is defined is not integral $v_iv_j=0$ but $v_ineq0$ and $v_jneq0$.
    $endgroup$
    – mouthetics
    Jan 18 at 3:46






  • 1




    $begingroup$
    Thanks @mouthetics, I see that now. I must have misinterpreted what the lecturer was saying. I'll ask a separate question to try and clarify.
    $endgroup$
    – JFox
    Jan 18 at 4:00






  • 1




    $begingroup$
    Probably you won't like this. Let $0neq alphain A$ then $F[alpha]$ is a finite dimensional subspace of $A$ over $F$. So there is $ngeq 1$ such that ${1,alpha,ldots,alpha^n}$ is linearly dependent. $textit{i.e.}$ there is $a_0,a_1,ldots,a_n$ not all zero such that $a_0+a_1alpha+cdots+a_nalpha^n=0$. Let $0leq i<n$ be the smallest $i$ such that $a_ineq 0$. Then $a_ialpha^i+cdots+a_nalpha^n=0Rightarrow 1+{a_{i+1}over a_i}alpha+cdots+{a_nover a_i}alpha^{n-i}=0$. From this you get that $alpha$ is invertible.
    $endgroup$
    – mouthetics
    Jan 18 at 4:09












  • $begingroup$
    @mouthetics that's definitely another cool way to go about the problem thank you. I do have one question about it. How do we know there will always the powers ${1,alpha,...,alpha^n}$ will be distinct for enough n? For example, there will often be elements with multiplicative order 2, or for example in $mathbb{F_(5^4)}$ there are two elements with order 3 (which I think would still not be enough to necessarily be linearly dependent). How does the argument hold for when $alpha$ has low multiplicative order like in these cases?
    $endgroup$
    – JFox
    Jan 18 at 7:26


















  • $begingroup$
    What is $R$? Your question doesn't define it. Cheers!
    $endgroup$
    – Robert Lewis
    Jan 18 at 3:43






  • 2




    $begingroup$
    The factor ring as it is defined is not integral $v_iv_j=0$ but $v_ineq0$ and $v_jneq0$.
    $endgroup$
    – mouthetics
    Jan 18 at 3:46






  • 1




    $begingroup$
    Thanks @mouthetics, I see that now. I must have misinterpreted what the lecturer was saying. I'll ask a separate question to try and clarify.
    $endgroup$
    – JFox
    Jan 18 at 4:00






  • 1




    $begingroup$
    Probably you won't like this. Let $0neq alphain A$ then $F[alpha]$ is a finite dimensional subspace of $A$ over $F$. So there is $ngeq 1$ such that ${1,alpha,ldots,alpha^n}$ is linearly dependent. $textit{i.e.}$ there is $a_0,a_1,ldots,a_n$ not all zero such that $a_0+a_1alpha+cdots+a_nalpha^n=0$. Let $0leq i<n$ be the smallest $i$ such that $a_ineq 0$. Then $a_ialpha^i+cdots+a_nalpha^n=0Rightarrow 1+{a_{i+1}over a_i}alpha+cdots+{a_nover a_i}alpha^{n-i}=0$. From this you get that $alpha$ is invertible.
    $endgroup$
    – mouthetics
    Jan 18 at 4:09












  • $begingroup$
    @mouthetics that's definitely another cool way to go about the problem thank you. I do have one question about it. How do we know there will always the powers ${1,alpha,...,alpha^n}$ will be distinct for enough n? For example, there will often be elements with multiplicative order 2, or for example in $mathbb{F_(5^4)}$ there are two elements with order 3 (which I think would still not be enough to necessarily be linearly dependent). How does the argument hold for when $alpha$ has low multiplicative order like in these cases?
    $endgroup$
    – JFox
    Jan 18 at 7:26
















$begingroup$
What is $R$? Your question doesn't define it. Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 3:43




$begingroup$
What is $R$? Your question doesn't define it. Cheers!
$endgroup$
– Robert Lewis
Jan 18 at 3:43




2




2




$begingroup$
The factor ring as it is defined is not integral $v_iv_j=0$ but $v_ineq0$ and $v_jneq0$.
$endgroup$
– mouthetics
Jan 18 at 3:46




$begingroup$
The factor ring as it is defined is not integral $v_iv_j=0$ but $v_ineq0$ and $v_jneq0$.
$endgroup$
– mouthetics
Jan 18 at 3:46




1




1




$begingroup$
Thanks @mouthetics, I see that now. I must have misinterpreted what the lecturer was saying. I'll ask a separate question to try and clarify.
$endgroup$
– JFox
Jan 18 at 4:00




$begingroup$
Thanks @mouthetics, I see that now. I must have misinterpreted what the lecturer was saying. I'll ask a separate question to try and clarify.
$endgroup$
– JFox
Jan 18 at 4:00




1




1




$begingroup$
Probably you won't like this. Let $0neq alphain A$ then $F[alpha]$ is a finite dimensional subspace of $A$ over $F$. So there is $ngeq 1$ such that ${1,alpha,ldots,alpha^n}$ is linearly dependent. $textit{i.e.}$ there is $a_0,a_1,ldots,a_n$ not all zero such that $a_0+a_1alpha+cdots+a_nalpha^n=0$. Let $0leq i<n$ be the smallest $i$ such that $a_ineq 0$. Then $a_ialpha^i+cdots+a_nalpha^n=0Rightarrow 1+{a_{i+1}over a_i}alpha+cdots+{a_nover a_i}alpha^{n-i}=0$. From this you get that $alpha$ is invertible.
$endgroup$
– mouthetics
Jan 18 at 4:09






$begingroup$
Probably you won't like this. Let $0neq alphain A$ then $F[alpha]$ is a finite dimensional subspace of $A$ over $F$. So there is $ngeq 1$ such that ${1,alpha,ldots,alpha^n}$ is linearly dependent. $textit{i.e.}$ there is $a_0,a_1,ldots,a_n$ not all zero such that $a_0+a_1alpha+cdots+a_nalpha^n=0$. Let $0leq i<n$ be the smallest $i$ such that $a_ineq 0$. Then $a_ialpha^i+cdots+a_nalpha^n=0Rightarrow 1+{a_{i+1}over a_i}alpha+cdots+{a_nover a_i}alpha^{n-i}=0$. From this you get that $alpha$ is invertible.
$endgroup$
– mouthetics
Jan 18 at 4:09














$begingroup$
@mouthetics that's definitely another cool way to go about the problem thank you. I do have one question about it. How do we know there will always the powers ${1,alpha,...,alpha^n}$ will be distinct for enough n? For example, there will often be elements with multiplicative order 2, or for example in $mathbb{F_(5^4)}$ there are two elements with order 3 (which I think would still not be enough to necessarily be linearly dependent). How does the argument hold for when $alpha$ has low multiplicative order like in these cases?
$endgroup$
– JFox
Jan 18 at 7:26




$begingroup$
@mouthetics that's definitely another cool way to go about the problem thank you. I do have one question about it. How do we know there will always the powers ${1,alpha,...,alpha^n}$ will be distinct for enough n? For example, there will often be elements with multiplicative order 2, or for example in $mathbb{F_(5^4)}$ there are two elements with order 3 (which I think would still not be enough to necessarily be linearly dependent). How does the argument hold for when $alpha$ has low multiplicative order like in these cases?
$endgroup$
– JFox
Jan 18 at 7:26










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$begingroup$

Here's a different way to show $A$ is a field. We will simply show that if $ain A$ is nonzero, then $a$ has a multiplicative inverse in $A$. Consider the $F$-linear map $M_a:Ato A$ given by $M_a(b) = acdot b$. It is an injective $F$-linear map between $F$-vector spaces of the same (finite) dimension, and so it is also surjective. Therefore, $M_a(b) = 1$ for some $bin A$, i.e., $acdot b = 1$.






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    $begingroup$

    Here's a different way to show $A$ is a field. We will simply show that if $ain A$ is nonzero, then $a$ has a multiplicative inverse in $A$. Consider the $F$-linear map $M_a:Ato A$ given by $M_a(b) = acdot b$. It is an injective $F$-linear map between $F$-vector spaces of the same (finite) dimension, and so it is also surjective. Therefore, $M_a(b) = 1$ for some $bin A$, i.e., $acdot b = 1$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Here's a different way to show $A$ is a field. We will simply show that if $ain A$ is nonzero, then $a$ has a multiplicative inverse in $A$. Consider the $F$-linear map $M_a:Ato A$ given by $M_a(b) = acdot b$. It is an injective $F$-linear map between $F$-vector spaces of the same (finite) dimension, and so it is also surjective. Therefore, $M_a(b) = 1$ for some $bin A$, i.e., $acdot b = 1$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Here's a different way to show $A$ is a field. We will simply show that if $ain A$ is nonzero, then $a$ has a multiplicative inverse in $A$. Consider the $F$-linear map $M_a:Ato A$ given by $M_a(b) = acdot b$. It is an injective $F$-linear map between $F$-vector spaces of the same (finite) dimension, and so it is also surjective. Therefore, $M_a(b) = 1$ for some $bin A$, i.e., $acdot b = 1$.






        share|cite|improve this answer









        $endgroup$



        Here's a different way to show $A$ is a field. We will simply show that if $ain A$ is nonzero, then $a$ has a multiplicative inverse in $A$. Consider the $F$-linear map $M_a:Ato A$ given by $M_a(b) = acdot b$. It is an injective $F$-linear map between $F$-vector spaces of the same (finite) dimension, and so it is also surjective. Therefore, $M_a(b) = 1$ for some $bin A$, i.e., $acdot b = 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 3:33









        cspruncsprun

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