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Conditions for Which $mathbbF[T]$ Is a Field
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Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $mathbbF,$ consider the ring $mathbbF[T]$ of polynomials in the linear operator $T.$ Under what conditions is $mathbbF[T]$ a field? One sufficient condition with which I am already familiar is that $T$ maps no nonzero subspace of $V$ into itself.
Proof. Given an element $p(T) in mathbbF[T],$ let $v$ be a vector in $ker p(T).$ Considering that $T$ commutes with $p(T),$ we have that $p(T)(T(v)) = T(p(T)(v)) = T(0) = 0,$ from which it follows that $T(ker p(T)) subseteq ker p(T).$ Our assumption that $T$ maps no nonzero subspace of $V$ into itself implies that $ker p(T) = 0,$ i.e., $p(T)$ is injective. Of course, $p(T)$ is a linear operator on $V,$ hence $p(T)$ is invertible. We conclude that every element of $mathbbF[T]$ is invertible. Furthermore, $mathbbF[T]$ is an integral domain since we have that $p(T) q(T) = 0$ if and only if $p(T) q(T)(v) = 0$ for every nonzero vector $v$ in $V$ if and only if $p(T) = 0$ or $q(T) = 0.$ So, $mathbbF[T]$ is a field.
Lately, I have been thinking about whether it is sufficient that $T$ is invertible. Using the evaluation map $varphi_T : mathbbF[x] to mathbbF[T]$ defined by $varphi_T(p(x)) = p(T)$ in tandem with the fact that $mathbbF[x]$ is a PID, it is clear that $mathbbF[T] cong mathbbF[x]/(mu_T(x)),$ where $mu_T(x)$ is the minimal polynomial of $T.$ We have therefore that $mathbbF[T]$ is a finite-dimensional vector space over $mathbbF$ with a basis $I, T, T^2, dots, T^d-1 ,$ where $d = deg mu_T.$ We could conclude that $mathbbF[T]$ is a field if we could show that $mathbbF[T]$ is an integral domain since every integral domain that is a finite-dimensional vector space over a field is itself a field. But I am not entirely sure how to prove that $mathbbF[T]$ is an integral domain in this case.
Generally, are there any other conditions than the one I initially stated for which $mathbbF[T]$ is a field?
vector-spaces field-theory linear-transformations minimal-polynomials
asked Jul 26 at 2:14
Dylan_Carlo_Beck
361211
add a comment |Â
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3
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Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $mathbbF,$ consider the ring $mathbbF[T]$ of polynomials in the linear operator $T.$ Under what conditions is $mathbbF[T]$ a field? One sufficient condition with which I am already familiar is that $T$ maps no nonzero subspace of $V$ into itself.
Proof. Given an element $p(T) in mathbbF[T],$ let $v$ be a vector in $ker p(T).$ Considering that $T$ commutes with $p(T),$ we have that $p(T)(T(v)) = T(p(T)(v)) = T(0) = 0,$ from which it follows that $T(ker p(T)) subseteq ker p(T).$ Our assumption that $T$ maps no nonzero subspace of $V$ into itself implies that $ker p(T) = 0,$ i.e., $p(T)$ is injective. Of course, $p(T)$ is a linear operator on $V,$ hence $p(T)$ is invertible. We conclude that every element of $mathbbF[T]$ is invertible. Furthermore, $mathbbF[T]$ is an integral domain since we have that $p(T) q(T) = 0$ if and only if $p(T) q(T)(v) = 0$ for every nonzero vector $v$ in $V$ if and only if $p(T) = 0$ or $q(T) = 0.$ So, $mathbbF[T]$ is a field.
Lately, I have been thinking about whether it is sufficient that $T$ is invertible. Using the evaluation map $varphi_T : mathbbF[x] to mathbbF[T]$ defined by $varphi_T(p(x)) = p(T)$ in tandem with the fact that $mathbbF[x]$ is a PID, it is clear that $mathbbF[T] cong mathbbF[x]/(mu_T(x)),$ where $mu_T(x)$ is the minimal polynomial of $T.$ We have therefore that $mathbbF[T]$ is a finite-dimensional vector space over $mathbbF$ with a basis $I, T, T^2, dots, T^d-1 ,$ where $d = deg mu_T.$ We could conclude that $mathbbF[T]$ is a field if we could show that $mathbbF[T]$ is an integral domain since every integral domain that is a finite-dimensional vector space over a field is itself a field. But I am not entirely sure how to prove that $mathbbF[T]$ is an integral domain in this case.
Generally, are there any other conditions than the one I initially stated for which $mathbbF[T]$ is a field?
vector-spaces field-theory linear-transformations minimal-polynomials
asked Jul 26 at 2:14
Dylan_Carlo_Beck
361211
3
$mathbbF[T]$ is isomorphic to $mathbbF[x]/(m(x))$, where $m(x)$ is the minimal polynomial of $T$, by sending $T$ to the class of $x$. Therefore, $m$ should be irreducible over $mathbbF$.
– user578878
Jul 26 at 2:22
Hint: $mu_T(x)$ is irreducible, so $(mu_T(x))$ is prime.
– Ashwin Iyengar
Jul 26 at 2:22
Of course, if $mu_T$ is irreducible, then $mathbbF[T]$ is a field.
– Dylan_Carlo_Beck
Jul 26 at 2:44
But is $T$ invertible enough to guarantee that $mathbbF[T]$ is a field?
– Dylan_Carlo_Beck
Jul 26 at 2:44
Consider the linear operator $T : mathbbC^2 to mathbbC^2$ that is represented by the matrix $A = beginpmatrix 1 & phantom- 0 \ 0 & -1 endpmatrix$ with respect to the standard basis of $mathbbC^2$ over $mathbbC.$ We have that $I + A = beginpmatrix 2 & 0 \ 0 & 0 endpmatrix$ is not a unit even though $T$ is invertible. We conclude that $T$ invertible is not sufficient for $mathbbF[T]$ a field.
– Dylan_Carlo_Beck
Jul 26 at 8:06
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up vote
3
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favorite
Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $mathbbF,$ consider the ring $mathbbF[T]$ of polynomials in the linear operator $T.$ Under what conditions is $mathbbF[T]$ a field? One sufficient condition with which I am already familiar is that $T$ maps no nonzero subspace of $V$ into itself.
Proof. Given an element $p(T) in mathbbF[T],$ let $v$ be a vector in $ker p(T).$ Considering that $T$ commutes with $p(T),$ we have that $p(T)(T(v)) = T(p(T)(v)) = T(0) = 0,$ from which it follows that $T(ker p(T)) subseteq ker p(T).$ Our assumption that $T$ maps no nonzero subspace of $V$ into itself implies that $ker p(T) = 0,$ i.e., $p(T)$ is injective. Of course, $p(T)$ is a linear operator on $V,$ hence $p(T)$ is invertible. We conclude that every element of $mathbbF[T]$ is invertible. Furthermore, $mathbbF[T]$ is an integral domain since we have that $p(T) q(T) = 0$ if and only if $p(T) q(T)(v) = 0$ for every nonzero vector $v$ in $V$ if and only if $p(T) = 0$ or $q(T) = 0.$ So, $mathbbF[T]$ is a field.
Lately, I have been thinking about whether it is sufficient that $T$ is invertible. Using the evaluation map $varphi_T : mathbbF[x] to mathbbF[T]$ defined by $varphi_T(p(x)) = p(T)$ in tandem with the fact that $mathbbF[x]$ is a PID, it is clear that $mathbbF[T] cong mathbbF[x]/(mu_T(x)),$ where $mu_T(x)$ is the minimal polynomial of $T.$ We have therefore that $mathbbF[T]$ is a finite-dimensional vector space over $mathbbF$ with a basis $I, T, T^2, dots, T^d-1 ,$ where $d = deg mu_T.$ We could conclude that $mathbbF[T]$ is a field if we could show that $mathbbF[T]$ is an integral domain since every integral domain that is a finite-dimensional vector space over a field is itself a field. But I am not entirely sure how to prove that $mathbbF[T]$ is an integral domain in this case.
Generally, are there any other conditions than the one I initially stated for which $mathbbF[T]$ is a field?
vector-spaces field-theory linear-transformations minimal-polynomials
asked Jul 26 at 2:14
Dylan_Carlo_Beck
361211
Given a linear operator $T$ on a finite-dimensional vector space $V$ over a field $mathbbF,$ consider the ring $mathbbF[T]$ of polynomials in the linear operator $T.$ Under what conditions is $mathbbF[T]$ a field? One sufficient condition with which I am already familiar is that $T$ maps no nonzero subspace of $V$ into itself.
Proof. Given an element $p(T) in mathbbF[T],$ let $v$ be a vector in $ker p(T).$ Considering that $T$ commutes with $p(T),$ we have that $p(T)(T(v)) = T(p(T)(v)) = T(0) = 0,$ from which it follows that $T(ker p(T)) subseteq ker p(T).$ Our assumption that $T$ maps no nonzero subspace of $V$ into itself implies that $ker p(T) = 0,$ i.e., $p(T)$ is injective. Of course, $p(T)$ is a linear operator on $V,$ hence $p(T)$ is invertible. We conclude that every element of $mathbbF[T]$ is invertible. Furthermore, $mathbbF[T]$ is an integral domain since we have that $p(T) q(T) = 0$ if and only if $p(T) q(T)(v) = 0$ for every nonzero vector $v$ in $V$ if and only if $p(T) = 0$ or $q(T) = 0.$ So, $mathbbF[T]$ is a field.
Lately, I have been thinking about whether it is sufficient that $T$ is invertible. Using the evaluation map $varphi_T : mathbbF[x] to mathbbF[T]$ defined by $varphi_T(p(x)) = p(T)$ in tandem with the fact that $mathbbF[x]$ is a PID, it is clear that $mathbbF[T] cong mathbbF[x]/(mu_T(x)),$ where $mu_T(x)$ is the minimal polynomial of $T.$ We have therefore that $mathbbF[T]$ is a finite-dimensional vector space over $mathbbF$ with a basis $I, T, T^2, dots, T^d-1 ,$ where $d = deg mu_T.$ We could conclude that $mathbbF[T]$ is a field if we could show that $mathbbF[T]$ is an integral domain since every integral domain that is a finite-dimensional vector space over a field is itself a field. But I am not entirely sure how to prove that $mathbbF[T]$ is an integral domain in this case.
Generally, are there any other conditions than the one I initially stated for which $mathbbF[T]$ is a field?
vector-spaces field-theory linear-transformations minimal-polynomials
asked Jul 26 at 2:14
Dylan_Carlo_Beck
361211
edited Jul 26 at 2:19
asked Jul 26 at 2:14
Dylan_Carlo_Beck
361211
asked Jul 26 at 2:14
Dylan_Carlo_Beck
361211
asked Jul 26 at 2:14
Dylan_Carlo_Beck
361211
361211
3
$mathbbF[T]$ is isomorphic to $mathbbF[x]/(m(x))$, where $m(x)$ is the minimal polynomial of $T$, by sending $T$ to the class of $x$. Therefore, $m$ should be irreducible over $mathbbF$.
– user578878
Jul 26 at 2:22
Hint: $mu_T(x)$ is irreducible, so $(mu_T(x))$ is prime.
– Ashwin Iyengar
Jul 26 at 2:22
Of course, if $mu_T$ is irreducible, then $mathbbF[T]$ is a field.
– Dylan_Carlo_Beck
Jul 26 at 2:44
But is $T$ invertible enough to guarantee that $mathbbF[T]$ is a field?
– Dylan_Carlo_Beck
Jul 26 at 2:44
Consider the linear operator $T : mathbbC^2 to mathbbC^2$ that is represented by the matrix $A = beginpmatrix 1 & phantom- 0 \ 0 & -1 endpmatrix$ with respect to the standard basis of $mathbbC^2$ over $mathbbC.$ We have that $I + A = beginpmatrix 2 & 0 \ 0 & 0 endpmatrix$ is not a unit even though $T$ is invertible. We conclude that $T$ invertible is not sufficient for $mathbbF[T]$ a field.
– Dylan_Carlo_Beck
Jul 26 at 8:06
add a comment |Â
3
$mathbbF[T]$ is isomorphic to $mathbbF[x]/(m(x))$, where $m(x)$ is the minimal polynomial of $T$, by sending $T$ to the class of $x$. Therefore, $m$ should be irreducible over $mathbbF$.
– user578878
Jul 26 at 2:22
Hint: $mu_T(x)$ is irreducible, so $(mu_T(x))$ is prime.
– Ashwin Iyengar
Jul 26 at 2:22
Of course, if $mu_T$ is irreducible, then $mathbbF[T]$ is a field.
– Dylan_Carlo_Beck
Jul 26 at 2:44
But is $T$ invertible enough to guarantee that $mathbbF[T]$ is a field?
– Dylan_Carlo_Beck
Jul 26 at 2:44
Consider the linear operator $T : mathbbC^2 to mathbbC^2$ that is represented by the matrix $A = beginpmatrix 1 & phantom- 0 \ 0 & -1 endpmatrix$ with respect to the standard basis of $mathbbC^2$ over $mathbbC.$ We have that $I + A = beginpmatrix 2 & 0 \ 0 & 0 endpmatrix$ is not a unit even though $T$ is invertible. We conclude that $T$ invertible is not sufficient for $mathbbF[T]$ a field.
– Dylan_Carlo_Beck
Jul 26 at 8:06
3
3
$mathbbF[T]$ is isomorphic to $mathbbF[x]/(m(x))$, where $m(x)$ is the minimal polynomial of $T$, by sending $T$ to the class of $x$. Therefore, $m$ should be irreducible over $mathbbF$.
– user578878
Jul 26 at 2:22
$mathbbF[T]$ is isomorphic to $mathbbF[x]/(m(x))$, where $m(x)$ is the minimal polynomial of $T$, by sending $T$ to the class of $x$. Therefore, $m$ should be irreducible over $mathbbF$.
– user578878
Jul 26 at 2:22
Hint: $mu_T(x)$ is irreducible, so $(mu_T(x))$ is prime.
– Ashwin Iyengar
Jul 26 at 2:22
Hint: $mu_T(x)$ is irreducible, so $(mu_T(x))$ is prime.
– Ashwin Iyengar
Jul 26 at 2:22
Of course, if $mu_T$ is irreducible, then $mathbbF[T]$ is a field.
– Dylan_Carlo_Beck
Jul 26 at 2:44
Of course, if $mu_T$ is irreducible, then $mathbbF[T]$ is a field.
– Dylan_Carlo_Beck
Jul 26 at 2:44
But is $T$ invertible enough to guarantee that $mathbbF[T]$ is a field?
– Dylan_Carlo_Beck
Jul 26 at 2:44
But is $T$ invertible enough to guarantee that $mathbbF[T]$ is a field?
– Dylan_Carlo_Beck
Jul 26 at 2:44
Consider the linear operator $T : mathbbC^2 to mathbbC^2$ that is represented by the matrix $A = beginpmatrix 1 & phantom- 0 \ 0 & -1 endpmatrix$ with respect to the standard basis of $mathbbC^2$ over $mathbbC.$ We have that $I + A = beginpmatrix 2 & 0 \ 0 & 0 endpmatrix$ is not a unit even though $T$ is invertible. We conclude that $T$ invertible is not sufficient for $mathbbF[T]$ a field.
– Dylan_Carlo_Beck
Jul 26 at 8:06
Consider the linear operator $T : mathbbC^2 to mathbbC^2$ that is represented by the matrix $A = beginpmatrix 1 & phantom- 0 \ 0 & -1 endpmatrix$ with respect to the standard basis of $mathbbC^2$ over $mathbbC.$ We have that $I + A = beginpmatrix 2 & 0 \ 0 & 0 endpmatrix$ is not a unit even though $T$ is invertible. We conclude that $T$ invertible is not sufficient for $mathbbF[T]$ a field.
– Dylan_Carlo_Beck
Jul 26 at 8:06
add a comment |Â
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$mathbbF[T]$ is isomorphic to $mathbbF[x]/(m(x))$, where $m(x)$ is the minimal polynomial of $T$, by sending $T$ to the class of $x$. Therefore, $m$ should be irreducible over $mathbbF$.
– user578878
Jul 26 at 2:22
Hint: $mu_T(x)$ is irreducible, so $(mu_T(x))$ is prime.
– Ashwin Iyengar
Jul 26 at 2:22
Of course, if $mu_T$ is irreducible, then $mathbbF[T]$ is a field.
– Dylan_Carlo_Beck
Jul 26 at 2:44
But is $T$ invertible enough to guarantee that $mathbbF[T]$ is a field?
– Dylan_Carlo_Beck
Jul 26 at 2:44
Consider the linear operator $T : mathbbC^2 to mathbbC^2$ that is represented by the matrix $A = beginpmatrix 1 & phantom- 0 \ 0 & -1 endpmatrix$ with respect to the standard basis of $mathbbC^2$ over $mathbbC.$ We have that $I + A = beginpmatrix 2 & 0 \ 0 & 0 endpmatrix$ is not a unit even though $T$ is invertible. We conclude that $T$ invertible is not sufficient for $mathbbF[T]$ a field.
– Dylan_Carlo_Beck
Jul 26 at 8:06