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Prove that the image of an ideal under a ring homomorphism $mathbb Z[x]/Ito M_2(mathbb Z)$ is not an ideal
Clash Royale CLAN TAG#URR8PPP
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Let $R=mathbb Z[x]/I$ where $I=(x^2-5x-2)$ and $S=M_2(mathbb Z)$. Let $B$ be the matrix with rows $(1,2), (3,4)$. The kernel of the ring homomorphism $mathbb Z[x]to S$ that sends $p(x)$ to $p(B)$ contains $x^2-5x-2$, so this homo factors through $R$ by the universal property of ring homomorphism, and there is a unique ring homomorphism $f:Rto S$ with $f(x+I)=B$ defined by $p(x)+Imapsto p(B)$.
I want to show that the image of the ideal $J=((x-1)+I)subset R$ under $f$ is not an ideal in $S$.
My thoughts are: the matrix with rows $(0,2), (3,3)$ lies in the image. If the image were an ideal, then multiplication by an elementary matrix would result in a matrix that lies in the image of $f$. E.g. the matrix with rows $(3,3),(0,2)$ would be in the image of $f$. But I'm not sure how to show that it isn't in the image of $f$. The image of $f$ is all elements of the form $f(overline p(x)overline(x-1))=p(B)(B-I)$ if I understand correctly.
abstract-algebra ring-theory ideals
asked Aug 6 at 18:35
user437309
556212
add a comment |Â
up vote
3
down vote
favorite
Let $R=mathbb Z[x]/I$ where $I=(x^2-5x-2)$ and $S=M_2(mathbb Z)$. Let $B$ be the matrix with rows $(1,2), (3,4)$. The kernel of the ring homomorphism $mathbb Z[x]to S$ that sends $p(x)$ to $p(B)$ contains $x^2-5x-2$, so this homo factors through $R$ by the universal property of ring homomorphism, and there is a unique ring homomorphism $f:Rto S$ with $f(x+I)=B$ defined by $p(x)+Imapsto p(B)$.
I want to show that the image of the ideal $J=((x-1)+I)subset R$ under $f$ is not an ideal in $S$.
My thoughts are: the matrix with rows $(0,2), (3,3)$ lies in the image. If the image were an ideal, then multiplication by an elementary matrix would result in a matrix that lies in the image of $f$. E.g. the matrix with rows $(3,3),(0,2)$ would be in the image of $f$. But I'm not sure how to show that it isn't in the image of $f$. The image of $f$ is all elements of the form $f(overline p(x)overline(x-1))=p(B)(B-I)$ if I understand correctly.
abstract-algebra ring-theory ideals
asked Aug 6 at 18:35
user437309
556212
What do you mean by $p(B)$?
– Javi
Aug 6 at 18:59
@Javi $p(x)$ evaluated at $x=B$
– user437309
Aug 6 at 19:00
Sorry, I don't know what you mean by evaluating a polynomial at a matrix. You said that $x^2-5x-2$ is contained in the kernel, so it's not evaluation at each of the entries, since $1$ is not a root of that polynomial.
– Javi
Aug 6 at 19:03
4
@Javi If $p(x)=x^2-5x-2$, then $p(B)=B^2-5B-2I=0$ ($0$ is the zero matrix in this case).
– user437309
Aug 6 at 19:04
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $R=mathbb Z[x]/I$ where $I=(x^2-5x-2)$ and $S=M_2(mathbb Z)$. Let $B$ be the matrix with rows $(1,2), (3,4)$. The kernel of the ring homomorphism $mathbb Z[x]to S$ that sends $p(x)$ to $p(B)$ contains $x^2-5x-2$, so this homo factors through $R$ by the universal property of ring homomorphism, and there is a unique ring homomorphism $f:Rto S$ with $f(x+I)=B$ defined by $p(x)+Imapsto p(B)$.
I want to show that the image of the ideal $J=((x-1)+I)subset R$ under $f$ is not an ideal in $S$.
My thoughts are: the matrix with rows $(0,2), (3,3)$ lies in the image. If the image were an ideal, then multiplication by an elementary matrix would result in a matrix that lies in the image of $f$. E.g. the matrix with rows $(3,3),(0,2)$ would be in the image of $f$. But I'm not sure how to show that it isn't in the image of $f$. The image of $f$ is all elements of the form $f(overline p(x)overline(x-1))=p(B)(B-I)$ if I understand correctly.
abstract-algebra ring-theory ideals
asked Aug 6 at 18:35
user437309
556212
Let $R=mathbb Z[x]/I$ where $I=(x^2-5x-2)$ and $S=M_2(mathbb Z)$. Let $B$ be the matrix with rows $(1,2), (3,4)$. The kernel of the ring homomorphism $mathbb Z[x]to S$ that sends $p(x)$ to $p(B)$ contains $x^2-5x-2$, so this homo factors through $R$ by the universal property of ring homomorphism, and there is a unique ring homomorphism $f:Rto S$ with $f(x+I)=B$ defined by $p(x)+Imapsto p(B)$.
I want to show that the image of the ideal $J=((x-1)+I)subset R$ under $f$ is not an ideal in $S$.
My thoughts are: the matrix with rows $(0,2), (3,3)$ lies in the image. If the image were an ideal, then multiplication by an elementary matrix would result in a matrix that lies in the image of $f$. E.g. the matrix with rows $(3,3),(0,2)$ would be in the image of $f$. But I'm not sure how to show that it isn't in the image of $f$. The image of $f$ is all elements of the form $f(overline p(x)overline(x-1))=p(B)(B-I)$ if I understand correctly.
abstract-algebra ring-theory ideals
asked Aug 6 at 18:35
user437309
556212
asked Aug 6 at 18:35
user437309
556212
asked Aug 6 at 18:35
user437309
556212
asked Aug 6 at 18:35
user437309
556212
556212
What do you mean by $p(B)$?
– Javi
Aug 6 at 18:59
@Javi $p(x)$ evaluated at $x=B$
– user437309
Aug 6 at 19:00
Sorry, I don't know what you mean by evaluating a polynomial at a matrix. You said that $x^2-5x-2$ is contained in the kernel, so it's not evaluation at each of the entries, since $1$ is not a root of that polynomial.
– Javi
Aug 6 at 19:03
4
@Javi If $p(x)=x^2-5x-2$, then $p(B)=B^2-5B-2I=0$ ($0$ is the zero matrix in this case).
– user437309
Aug 6 at 19:04
add a comment |Â
What do you mean by $p(B)$?
– Javi
Aug 6 at 18:59
@Javi $p(x)$ evaluated at $x=B$
– user437309
Aug 6 at 19:00
Sorry, I don't know what you mean by evaluating a polynomial at a matrix. You said that $x^2-5x-2$ is contained in the kernel, so it's not evaluation at each of the entries, since $1$ is not a root of that polynomial.
– Javi
Aug 6 at 19:03
4
@Javi If $p(x)=x^2-5x-2$, then $p(B)=B^2-5B-2I=0$ ($0$ is the zero matrix in this case).
– user437309
Aug 6 at 19:04
What do you mean by $p(B)$?
– Javi
Aug 6 at 18:59
What do you mean by $p(B)$?
– Javi
Aug 6 at 18:59
@Javi $p(x)$ evaluated at $x=B$
– user437309
Aug 6 at 19:00
@Javi $p(x)$ evaluated at $x=B$
– user437309
Aug 6 at 19:00
Sorry, I don't know what you mean by evaluating a polynomial at a matrix. You said that $x^2-5x-2$ is contained in the kernel, so it's not evaluation at each of the entries, since $1$ is not a root of that polynomial.
– Javi
Aug 6 at 19:03
Sorry, I don't know what you mean by evaluating a polynomial at a matrix. You said that $x^2-5x-2$ is contained in the kernel, so it's not evaluation at each of the entries, since $1$ is not a root of that polynomial.
– Javi
Aug 6 at 19:03
4
4
@Javi If $p(x)=x^2-5x-2$, then $p(B)=B^2-5B-2I=0$ ($0$ is the zero matrix in this case).
– user437309
Aug 6 at 19:04
@Javi If $p(x)=x^2-5x-2$, then $p(B)=B^2-5B-2I=0$ ($0$ is the zero matrix in this case).
– user437309
Aug 6 at 19:04
add a comment |Â
2 Answers
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active
oldest
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up vote
2
down vote
accepted
Let $T:=textim(f)$. Clearly, $T$ is not the whole $S$, as elements of $T$ commute with $B$. Observe that
$$fbig((x-1)+Ibig)=f(x)-1_S=B-1_S=beginbmatrix1&2\3&4endbmatrix-beginbmatrix1&0\0&1endbmatrix=beginbmatrix0&2\3&3endbmatrix=:C,.$$
Thus, the image of the ideal $J$ of $R$ under $f$ is $TC$. It is strictly contained in the left ideal $SC$ of $S$. This is because everything in $TC$ commutes with $B$, whereas the element
$$beginbmatrix0&1\1&0endbmatrix,C=beginbmatrix3&3\0&2endbmatrix$$
of $SC$ does not commute with $B$, as we can readily check
$$beginbmatrix3&3\0&2endbmatrix,B=beginbmatrix3&3\0&2endbmatrix,beginbmatrix1&2\3&4endbmatrix=beginbmatrix12&18\6&8endbmatrixneqbeginbmatrix3&7\9&17endbmatrix=beginbmatrix1&2\3&4endbmatrix,beginbmatrix3&3\0&2endbmatrix=B,beginbmatrix3&3\0&2endbmatrix,.$$
Thus, $TC$ is not even a left ideal of $S$.
Note that the left ideal $SC$ of $S$ consists of matrices $beginbmatrixa&b\c&dendbmatrix$,
where $(a,b)$ and $(c,d)$ are in the $mathbbZ$-span of $(0,2)$ and $(3,3)$. The right ideal $CS$ of $S$ is composed by matrices $beginbmatrixa&b\c&dendbmatrix$, where $(a,c)$ and $(b,d)$ lie within the $mathbbZ$-span of $(0,3)$ and $(2,0)$. Finally, the two-sided ideal $SCS$ of $S$ equals $S$ itself.
answered Aug 6 at 19:24
Batominovski
23.4k22779
What do you mean by $TC$ (which is $operatornameim(f) C$)?
– user437309
Aug 6 at 19:38
This is just the set of $tcdot C$, where $tin T=textim(f)$. Note that $T$ is a commutative subalgebra of $S$.
– Batominovski
Aug 6 at 19:39
It seems I no longer understand this: why does every matrix in the image commute with $B$?
– user437309
Aug 7 at 22:16
Because everything in $T$ is a polynomial in $B$. Every matrix $X$ commutes with $p(X)$ for any polynomial $p$.
– Batominovski
Aug 7 at 22:18
1
But $B$ and $C$ do commute. $C$ is a polynomial in $B$.
– Batominovski
Aug 7 at 22:33
 |Â
show 1 more comment
up vote
0
down vote
If $f(R/I)$ were an ideal, $C(B-I)$ should be an element of $f(R/I)$ for any $C in M_2(mathbbZ)$ and thus there should exist $p in mathbbZ[X]$ such that $p(B)(B-I) = C(B-I)$, that is,
$$
p(B) = C
$$
since $B-I$ is invertible. In particular, the eigenvalues of $C$ have to be those of $p(B)$ which in turn are the eigenvalues of $B$ via $p$. However, $B$ has real eigenvalues and therefore this would imply that any matrix with integer coefficients has real eigenvalues, which is not the case: take for example
$$
left[ beginarraycc
0 & 1 \
-1 & 0 \
endarray right]
$$
with characteristic polynomial $X^2 + 1$.
answered Aug 6 at 19:41
Guido A.
3,851624
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $T:=textim(f)$. Clearly, $T$ is not the whole $S$, as elements of $T$ commute with $B$. Observe that
$$fbig((x-1)+Ibig)=f(x)-1_S=B-1_S=beginbmatrix1&2\3&4endbmatrix-beginbmatrix1&0\0&1endbmatrix=beginbmatrix0&2\3&3endbmatrix=:C,.$$
Thus, the image of the ideal $J$ of $R$ under $f$ is $TC$. It is strictly contained in the left ideal $SC$ of $S$. This is because everything in $TC$ commutes with $B$, whereas the element
$$beginbmatrix0&1\1&0endbmatrix,C=beginbmatrix3&3\0&2endbmatrix$$
of $SC$ does not commute with $B$, as we can readily check
$$beginbmatrix3&3\0&2endbmatrix,B=beginbmatrix3&3\0&2endbmatrix,beginbmatrix1&2\3&4endbmatrix=beginbmatrix12&18\6&8endbmatrixneqbeginbmatrix3&7\9&17endbmatrix=beginbmatrix1&2\3&4endbmatrix,beginbmatrix3&3\0&2endbmatrix=B,beginbmatrix3&3\0&2endbmatrix,.$$
Thus, $TC$ is not even a left ideal of $S$.
Note that the left ideal $SC$ of $S$ consists of matrices $beginbmatrixa&b\c&dendbmatrix$,
where $(a,b)$ and $(c,d)$ are in the $mathbbZ$-span of $(0,2)$ and $(3,3)$. The right ideal $CS$ of $S$ is composed by matrices $beginbmatrixa&b\c&dendbmatrix$, where $(a,c)$ and $(b,d)$ lie within the $mathbbZ$-span of $(0,3)$ and $(2,0)$. Finally, the two-sided ideal $SCS$ of $S$ equals $S$ itself.
answered Aug 6 at 19:24
Batominovski
23.4k22779
What do you mean by $TC$ (which is $operatornameim(f) C$)?
– user437309
Aug 6 at 19:38
This is just the set of $tcdot C$, where $tin T=textim(f)$. Note that $T$ is a commutative subalgebra of $S$.
– Batominovski
Aug 6 at 19:39
It seems I no longer understand this: why does every matrix in the image commute with $B$?
– user437309
Aug 7 at 22:16
Because everything in $T$ is a polynomial in $B$. Every matrix $X$ commutes with $p(X)$ for any polynomial $p$.
– Batominovski
Aug 7 at 22:18
1
But $B$ and $C$ do commute. $C$ is a polynomial in $B$.
– Batominovski
Aug 7 at 22:33
 |Â
show 1 more comment
up vote
2
down vote
accepted
Let $T:=textim(f)$. Clearly, $T$ is not the whole $S$, as elements of $T$ commute with $B$. Observe that
$$fbig((x-1)+Ibig)=f(x)-1_S=B-1_S=beginbmatrix1&2\3&4endbmatrix-beginbmatrix1&0\0&1endbmatrix=beginbmatrix0&2\3&3endbmatrix=:C,.$$
Thus, the image of the ideal $J$ of $R$ under $f$ is $TC$. It is strictly contained in the left ideal $SC$ of $S$. This is because everything in $TC$ commutes with $B$, whereas the element
$$beginbmatrix0&1\1&0endbmatrix,C=beginbmatrix3&3\0&2endbmatrix$$
of $SC$ does not commute with $B$, as we can readily check
$$beginbmatrix3&3\0&2endbmatrix,B=beginbmatrix3&3\0&2endbmatrix,beginbmatrix1&2\3&4endbmatrix=beginbmatrix12&18\6&8endbmatrixneqbeginbmatrix3&7\9&17endbmatrix=beginbmatrix1&2\3&4endbmatrix,beginbmatrix3&3\0&2endbmatrix=B,beginbmatrix3&3\0&2endbmatrix,.$$
Thus, $TC$ is not even a left ideal of $S$.
Note that the left ideal $SC$ of $S$ consists of matrices $beginbmatrixa&b\c&dendbmatrix$,
where $(a,b)$ and $(c,d)$ are in the $mathbbZ$-span of $(0,2)$ and $(3,3)$. The right ideal $CS$ of $S$ is composed by matrices $beginbmatrixa&b\c&dendbmatrix$, where $(a,c)$ and $(b,d)$ lie within the $mathbbZ$-span of $(0,3)$ and $(2,0)$. Finally, the two-sided ideal $SCS$ of $S$ equals $S$ itself.
answered Aug 6 at 19:24
Batominovski
23.4k22779
What do you mean by $TC$ (which is $operatornameim(f) C$)?
– user437309
Aug 6 at 19:38
This is just the set of $tcdot C$, where $tin T=textim(f)$. Note that $T$ is a commutative subalgebra of $S$.
– Batominovski
Aug 6 at 19:39
It seems I no longer understand this: why does every matrix in the image commute with $B$?
– user437309
Aug 7 at 22:16
Because everything in $T$ is a polynomial in $B$. Every matrix $X$ commutes with $p(X)$ for any polynomial $p$.
– Batominovski
Aug 7 at 22:18
1
But $B$ and $C$ do commute. $C$ is a polynomial in $B$.
– Batominovski
Aug 7 at 22:33
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $T:=textim(f)$. Clearly, $T$ is not the whole $S$, as elements of $T$ commute with $B$. Observe that
$$fbig((x-1)+Ibig)=f(x)-1_S=B-1_S=beginbmatrix1&2\3&4endbmatrix-beginbmatrix1&0\0&1endbmatrix=beginbmatrix0&2\3&3endbmatrix=:C,.$$
Thus, the image of the ideal $J$ of $R$ under $f$ is $TC$. It is strictly contained in the left ideal $SC$ of $S$. This is because everything in $TC$ commutes with $B$, whereas the element
$$beginbmatrix0&1\1&0endbmatrix,C=beginbmatrix3&3\0&2endbmatrix$$
of $SC$ does not commute with $B$, as we can readily check
$$beginbmatrix3&3\0&2endbmatrix,B=beginbmatrix3&3\0&2endbmatrix,beginbmatrix1&2\3&4endbmatrix=beginbmatrix12&18\6&8endbmatrixneqbeginbmatrix3&7\9&17endbmatrix=beginbmatrix1&2\3&4endbmatrix,beginbmatrix3&3\0&2endbmatrix=B,beginbmatrix3&3\0&2endbmatrix,.$$
Thus, $TC$ is not even a left ideal of $S$.
Note that the left ideal $SC$ of $S$ consists of matrices $beginbmatrixa&b\c&dendbmatrix$,
where $(a,b)$ and $(c,d)$ are in the $mathbbZ$-span of $(0,2)$ and $(3,3)$. The right ideal $CS$ of $S$ is composed by matrices $beginbmatrixa&b\c&dendbmatrix$, where $(a,c)$ and $(b,d)$ lie within the $mathbbZ$-span of $(0,3)$ and $(2,0)$. Finally, the two-sided ideal $SCS$ of $S$ equals $S$ itself.
answered Aug 6 at 19:24
Batominovski
23.4k22779
Let $T:=textim(f)$. Clearly, $T$ is not the whole $S$, as elements of $T$ commute with $B$. Observe that
$$fbig((x-1)+Ibig)=f(x)-1_S=B-1_S=beginbmatrix1&2\3&4endbmatrix-beginbmatrix1&0\0&1endbmatrix=beginbmatrix0&2\3&3endbmatrix=:C,.$$
Thus, the image of the ideal $J$ of $R$ under $f$ is $TC$. It is strictly contained in the left ideal $SC$ of $S$. This is because everything in $TC$ commutes with $B$, whereas the element
$$beginbmatrix0&1\1&0endbmatrix,C=beginbmatrix3&3\0&2endbmatrix$$
of $SC$ does not commute with $B$, as we can readily check
$$beginbmatrix3&3\0&2endbmatrix,B=beginbmatrix3&3\0&2endbmatrix,beginbmatrix1&2\3&4endbmatrix=beginbmatrix12&18\6&8endbmatrixneqbeginbmatrix3&7\9&17endbmatrix=beginbmatrix1&2\3&4endbmatrix,beginbmatrix3&3\0&2endbmatrix=B,beginbmatrix3&3\0&2endbmatrix,.$$
Thus, $TC$ is not even a left ideal of $S$.
Note that the left ideal $SC$ of $S$ consists of matrices $beginbmatrixa&b\c&dendbmatrix$,
where $(a,b)$ and $(c,d)$ are in the $mathbbZ$-span of $(0,2)$ and $(3,3)$. The right ideal $CS$ of $S$ is composed by matrices $beginbmatrixa&b\c&dendbmatrix$, where $(a,c)$ and $(b,d)$ lie within the $mathbbZ$-span of $(0,3)$ and $(2,0)$. Finally, the two-sided ideal $SCS$ of $S$ equals $S$ itself.
answered Aug 6 at 19:24
Batominovski
23.4k22779
edited Aug 6 at 19:31
answered Aug 6 at 19:24
Batominovski
23.4k22779
answered Aug 6 at 19:24
Batominovski
23.4k22779
answered Aug 6 at 19:24
Batominovski
23.4k22779
23.4k22779
What do you mean by $TC$ (which is $operatornameim(f) C$)?
– user437309
Aug 6 at 19:38
This is just the set of $tcdot C$, where $tin T=textim(f)$. Note that $T$ is a commutative subalgebra of $S$.
– Batominovski
Aug 6 at 19:39
It seems I no longer understand this: why does every matrix in the image commute with $B$?
– user437309
Aug 7 at 22:16
Because everything in $T$ is a polynomial in $B$. Every matrix $X$ commutes with $p(X)$ for any polynomial $p$.
– Batominovski
Aug 7 at 22:18
1
But $B$ and $C$ do commute. $C$ is a polynomial in $B$.
– Batominovski
Aug 7 at 22:33
 |Â
show 1 more comment
What do you mean by $TC$ (which is $operatornameim(f) C$)?
– user437309
Aug 6 at 19:38
This is just the set of $tcdot C$, where $tin T=textim(f)$. Note that $T$ is a commutative subalgebra of $S$.
– Batominovski
Aug 6 at 19:39
It seems I no longer understand this: why does every matrix in the image commute with $B$?
– user437309
Aug 7 at 22:16
Because everything in $T$ is a polynomial in $B$. Every matrix $X$ commutes with $p(X)$ for any polynomial $p$.
– Batominovski
Aug 7 at 22:18
1
But $B$ and $C$ do commute. $C$ is a polynomial in $B$.
– Batominovski
Aug 7 at 22:33
What do you mean by $TC$ (which is $operatornameim(f) C$)?
– user437309
Aug 6 at 19:38
What do you mean by $TC$ (which is $operatornameim(f) C$)?
– user437309
Aug 6 at 19:38
This is just the set of $tcdot C$, where $tin T=textim(f)$. Note that $T$ is a commutative subalgebra of $S$.
– Batominovski
Aug 6 at 19:39
This is just the set of $tcdot C$, where $tin T=textim(f)$. Note that $T$ is a commutative subalgebra of $S$.
– Batominovski
Aug 6 at 19:39
It seems I no longer understand this: why does every matrix in the image commute with $B$?
– user437309
Aug 7 at 22:16
It seems I no longer understand this: why does every matrix in the image commute with $B$?
– user437309
Aug 7 at 22:16
Because everything in $T$ is a polynomial in $B$. Every matrix $X$ commutes with $p(X)$ for any polynomial $p$.
– Batominovski
Aug 7 at 22:18
Because everything in $T$ is a polynomial in $B$. Every matrix $X$ commutes with $p(X)$ for any polynomial $p$.
– Batominovski
Aug 7 at 22:18
1
1
But $B$ and $C$ do commute. $C$ is a polynomial in $B$.
– Batominovski
Aug 7 at 22:33
But $B$ and $C$ do commute. $C$ is a polynomial in $B$.
– Batominovski
Aug 7 at 22:33
 |Â
show 1 more comment
up vote
0
down vote
If $f(R/I)$ were an ideal, $C(B-I)$ should be an element of $f(R/I)$ for any $C in M_2(mathbbZ)$ and thus there should exist $p in mathbbZ[X]$ such that $p(B)(B-I) = C(B-I)$, that is,
$$
p(B) = C
$$
since $B-I$ is invertible. In particular, the eigenvalues of $C$ have to be those of $p(B)$ which in turn are the eigenvalues of $B$ via $p$. However, $B$ has real eigenvalues and therefore this would imply that any matrix with integer coefficients has real eigenvalues, which is not the case: take for example
$$
left[ beginarraycc
0 & 1 \
-1 & 0 \
endarray right]
$$
with characteristic polynomial $X^2 + 1$.
answered Aug 6 at 19:41
Guido A.
3,851624
add a comment |Â
up vote
0
down vote
If $f(R/I)$ were an ideal, $C(B-I)$ should be an element of $f(R/I)$ for any $C in M_2(mathbbZ)$ and thus there should exist $p in mathbbZ[X]$ such that $p(B)(B-I) = C(B-I)$, that is,
$$
p(B) = C
$$
since $B-I$ is invertible. In particular, the eigenvalues of $C$ have to be those of $p(B)$ which in turn are the eigenvalues of $B$ via $p$. However, $B$ has real eigenvalues and therefore this would imply that any matrix with integer coefficients has real eigenvalues, which is not the case: take for example
$$
left[ beginarraycc
0 & 1 \
-1 & 0 \
endarray right]
$$
with characteristic polynomial $X^2 + 1$.
answered Aug 6 at 19:41
Guido A.
3,851624
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If $f(R/I)$ were an ideal, $C(B-I)$ should be an element of $f(R/I)$ for any $C in M_2(mathbbZ)$ and thus there should exist $p in mathbbZ[X]$ such that $p(B)(B-I) = C(B-I)$, that is,
$$
p(B) = C
$$
since $B-I$ is invertible. In particular, the eigenvalues of $C$ have to be those of $p(B)$ which in turn are the eigenvalues of $B$ via $p$. However, $B$ has real eigenvalues and therefore this would imply that any matrix with integer coefficients has real eigenvalues, which is not the case: take for example
$$
left[ beginarraycc
0 & 1 \
-1 & 0 \
endarray right]
$$
with characteristic polynomial $X^2 + 1$.
answered Aug 6 at 19:41
Guido A.
3,851624
If $f(R/I)$ were an ideal, $C(B-I)$ should be an element of $f(R/I)$ for any $C in M_2(mathbbZ)$ and thus there should exist $p in mathbbZ[X]$ such that $p(B)(B-I) = C(B-I)$, that is,
$$
p(B) = C
$$
since $B-I$ is invertible. In particular, the eigenvalues of $C$ have to be those of $p(B)$ which in turn are the eigenvalues of $B$ via $p$. However, $B$ has real eigenvalues and therefore this would imply that any matrix with integer coefficients has real eigenvalues, which is not the case: take for example
$$
left[ beginarraycc
0 & 1 \
-1 & 0 \
endarray right]
$$
with characteristic polynomial $X^2 + 1$.
answered Aug 6 at 19:41
Guido A.
3,851624
answered Aug 6 at 19:41
Guido A.
3,851624
answered Aug 6 at 19:41
Guido A.
3,851624
answered Aug 6 at 19:41
Guido A.
3,851624
3,851624
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What do you mean by $p(B)$?
– Javi
Aug 6 at 18:59
@Javi $p(x)$ evaluated at $x=B$
– user437309
Aug 6 at 19:00
Sorry, I don't know what you mean by evaluating a polynomial at a matrix. You said that $x^2-5x-2$ is contained in the kernel, so it's not evaluation at each of the entries, since $1$ is not a root of that polynomial.
– Javi
Aug 6 at 19:03
4
@Javi If $p(x)=x^2-5x-2$, then $p(B)=B^2-5B-2I=0$ ($0$ is the zero matrix in this case).
– user437309
Aug 6 at 19:04