Mixing
What's wrong with my proof of quotient rings?
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I seem to have arrived at a contradiction by applying what I know about quotient rings. I can't figure out where the mistake is.
Let $f(x)in mathbb{Z}[x]$, $deg f(x)ge 1$, and $p$ a prime number. Then
begin{align*}
mathbb{Z}[x]/f(x)mathbb{Z}[x]&cong (mathbb{Z}[x]/pmathbb{Z}[x])/((f(x)mathbb{Z}[x])/pmathbb{Z}[x])tag{Third Iso. Thm}\
&cong(mathbb{Z}[x]/pmathbb{Z}[x])/(hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x]))tag{$*$}\
&cong mathbb{Z}_p[x]/hat{f}(x)mathbb{Z}_p[x].
end{align*}
where $(*)$ comes from:
begin{align*}
(f(x)mathbb{Z}[x])/pmathbb{Z}[x]&={g(x)+pmathbb{Z}[x]:g(x)in f(x)mathbb{Z}[x]}\
&={f(x)g(x)+pmathbb{Z}[x]:g(x)inmathbb{Z}[x]}\
&={(f(x)+pmathbb{Z}[x])(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&={hat{f}(x)(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&=hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x])
end{align*}
For example, this would seem to show that $$mathbb{Z}[i]cong mathbb{Z}[x]/(x^2+1)mathbb{Z}[x]cong mathbb{Z}_2[x]/(x^2+1)mathbb{Z}_2[x]$$ The first ring is infinite and the last ring is finite. What's going on here?
abstract-algebra ring-theory fake-proofs
asked Jan 19 at 0:18
RileyRiley
2,1551420
$endgroup$
|
show 3 more comments
$begingroup$
I seem to have arrived at a contradiction by applying what I know about quotient rings. I can't figure out where the mistake is.
Let $f(x)in mathbb{Z}[x]$, $deg f(x)ge 1$, and $p$ a prime number. Then
begin{align*}
mathbb{Z}[x]/f(x)mathbb{Z}[x]&cong (mathbb{Z}[x]/pmathbb{Z}[x])/((f(x)mathbb{Z}[x])/pmathbb{Z}[x])tag{Third Iso. Thm}\
&cong(mathbb{Z}[x]/pmathbb{Z}[x])/(hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x]))tag{$*$}\
&cong mathbb{Z}_p[x]/hat{f}(x)mathbb{Z}_p[x].
end{align*}
where $(*)$ comes from:
begin{align*}
(f(x)mathbb{Z}[x])/pmathbb{Z}[x]&={g(x)+pmathbb{Z}[x]:g(x)in f(x)mathbb{Z}[x]}\
&={f(x)g(x)+pmathbb{Z}[x]:g(x)inmathbb{Z}[x]}\
&={(f(x)+pmathbb{Z}[x])(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&={hat{f}(x)(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&=hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x])
end{align*}
For example, this would seem to show that $$mathbb{Z}[i]cong mathbb{Z}[x]/(x^2+1)mathbb{Z}[x]cong mathbb{Z}_2[x]/(x^2+1)mathbb{Z}_2[x]$$ The first ring is infinite and the last ring is finite. What's going on here?
abstract-algebra ring-theory fake-proofs
asked Jan 19 at 0:18
RileyRiley
2,1551420
$endgroup$
$begingroup$
The problem is that $(Bbb Z[x])/(pBbb Z[x])$ is not equal to $(Bbb Z/p Bbb Z)[x]$
$endgroup$
– Omnomnomnom
Jan 19 at 0:24
$begingroup$
@Omnomnomnom Are you sure about that? Could you prove it in an answer form?
$endgroup$
– Riley
Jan 19 at 0:32
$begingroup$
@Omnomnomnom If you define $f:mathbb{Z}[x]to mathbb{Z}_p[x]$ by modding the coefficients by $p$, then isn't $text{Ker}(f)$ just $pmathbb{Z}[x]$, proving the desired isomorphism?
$endgroup$
– Riley
Jan 19 at 0:42
$begingroup$
@Riley An element in the ring $;left(Bbb Z/pBbb Zright)[x];$ is a polynomial, whereas an element in the ring $;Bbb Z[x]/pBbb Z[x];$ is not a polynomial but in fact an equivalence class...
$endgroup$
– DonAntonio
Jan 19 at 0:43
2
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]subset f(x)Bbb Z[x];$ , as required by the third isomorphism theorem.
$endgroup$
– DonAntonio
Jan 19 at 0:49
|
show 3 more comments
$begingroup$
I seem to have arrived at a contradiction by applying what I know about quotient rings. I can't figure out where the mistake is.
Let $f(x)in mathbb{Z}[x]$, $deg f(x)ge 1$, and $p$ a prime number. Then
begin{align*}
mathbb{Z}[x]/f(x)mathbb{Z}[x]&cong (mathbb{Z}[x]/pmathbb{Z}[x])/((f(x)mathbb{Z}[x])/pmathbb{Z}[x])tag{Third Iso. Thm}\
&cong(mathbb{Z}[x]/pmathbb{Z}[x])/(hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x]))tag{$*$}\
&cong mathbb{Z}_p[x]/hat{f}(x)mathbb{Z}_p[x].
end{align*}
where $(*)$ comes from:
begin{align*}
(f(x)mathbb{Z}[x])/pmathbb{Z}[x]&={g(x)+pmathbb{Z}[x]:g(x)in f(x)mathbb{Z}[x]}\
&={f(x)g(x)+pmathbb{Z}[x]:g(x)inmathbb{Z}[x]}\
&={(f(x)+pmathbb{Z}[x])(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&={hat{f}(x)(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&=hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x])
end{align*}
For example, this would seem to show that $$mathbb{Z}[i]cong mathbb{Z}[x]/(x^2+1)mathbb{Z}[x]cong mathbb{Z}_2[x]/(x^2+1)mathbb{Z}_2[x]$$ The first ring is infinite and the last ring is finite. What's going on here?
abstract-algebra ring-theory fake-proofs
asked Jan 19 at 0:18
RileyRiley
2,1551420
$endgroup$
I seem to have arrived at a contradiction by applying what I know about quotient rings. I can't figure out where the mistake is.
Let $f(x)in mathbb{Z}[x]$, $deg f(x)ge 1$, and $p$ a prime number. Then
begin{align*}
mathbb{Z}[x]/f(x)mathbb{Z}[x]&cong (mathbb{Z}[x]/pmathbb{Z}[x])/((f(x)mathbb{Z}[x])/pmathbb{Z}[x])tag{Third Iso. Thm}\
&cong(mathbb{Z}[x]/pmathbb{Z}[x])/(hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x]))tag{$*$}\
&cong mathbb{Z}_p[x]/hat{f}(x)mathbb{Z}_p[x].
end{align*}
where $(*)$ comes from:
begin{align*}
(f(x)mathbb{Z}[x])/pmathbb{Z}[x]&={g(x)+pmathbb{Z}[x]:g(x)in f(x)mathbb{Z}[x]}\
&={f(x)g(x)+pmathbb{Z}[x]:g(x)inmathbb{Z}[x]}\
&={(f(x)+pmathbb{Z}[x])(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&={hat{f}(x)(g(x)+pmathbb{Z}[x]):g(x)inmathbb{Z}[x]}\
&=hat{f}(x)(mathbb{Z}[x]/pmathbb{Z}[x])
end{align*}
For example, this would seem to show that $$mathbb{Z}[i]cong mathbb{Z}[x]/(x^2+1)mathbb{Z}[x]cong mathbb{Z}_2[x]/(x^2+1)mathbb{Z}_2[x]$$ The first ring is infinite and the last ring is finite. What's going on here?
abstract-algebra ring-theory fake-proofs
abstract-algebra ring-theory fake-proofs
asked Jan 19 at 0:18
RileyRiley
2,1551420
asked Jan 19 at 0:18
RileyRiley
2,1551420
asked Jan 19 at 0:18
RileyRiley
2,1551420
asked Jan 19 at 0:18
RileyRiley
2,1551420
asked Jan 19 at 0:18
RileyRiley
2,1551420
2,1551420
$begingroup$
The problem is that $(Bbb Z[x])/(pBbb Z[x])$ is not equal to $(Bbb Z/p Bbb Z)[x]$
$endgroup$
– Omnomnomnom
Jan 19 at 0:24
$begingroup$
@Omnomnomnom Are you sure about that? Could you prove it in an answer form?
$endgroup$
– Riley
Jan 19 at 0:32
$begingroup$
@Omnomnomnom If you define $f:mathbb{Z}[x]to mathbb{Z}_p[x]$ by modding the coefficients by $p$, then isn't $text{Ker}(f)$ just $pmathbb{Z}[x]$, proving the desired isomorphism?
$endgroup$
– Riley
Jan 19 at 0:42
$begingroup$
@Riley An element in the ring $;left(Bbb Z/pBbb Zright)[x];$ is a polynomial, whereas an element in the ring $;Bbb Z[x]/pBbb Z[x];$ is not a polynomial but in fact an equivalence class...
$endgroup$
– DonAntonio
Jan 19 at 0:43
2
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]subset f(x)Bbb Z[x];$ , as required by the third isomorphism theorem.
$endgroup$
– DonAntonio
Jan 19 at 0:49
|
show 3 more comments
$begingroup$
The problem is that $(Bbb Z[x])/(pBbb Z[x])$ is not equal to $(Bbb Z/p Bbb Z)[x]$
$endgroup$
– Omnomnomnom
Jan 19 at 0:24
$begingroup$
@Omnomnomnom Are you sure about that? Could you prove it in an answer form?
$endgroup$
– Riley
Jan 19 at 0:32
$begingroup$
@Omnomnomnom If you define $f:mathbb{Z}[x]to mathbb{Z}_p[x]$ by modding the coefficients by $p$, then isn't $text{Ker}(f)$ just $pmathbb{Z}[x]$, proving the desired isomorphism?
$endgroup$
– Riley
Jan 19 at 0:42
$begingroup$
@Riley An element in the ring $;left(Bbb Z/pBbb Zright)[x];$ is a polynomial, whereas an element in the ring $;Bbb Z[x]/pBbb Z[x];$ is not a polynomial but in fact an equivalence class...
$endgroup$
– DonAntonio
Jan 19 at 0:43
2
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]subset f(x)Bbb Z[x];$ , as required by the third isomorphism theorem.
$endgroup$
– DonAntonio
Jan 19 at 0:49
$begingroup$
The problem is that $(Bbb Z[x])/(pBbb Z[x])$ is not equal to $(Bbb Z/p Bbb Z)[x]$
$endgroup$
– Omnomnomnom
Jan 19 at 0:24
$begingroup$
The problem is that $(Bbb Z[x])/(pBbb Z[x])$ is not equal to $(Bbb Z/p Bbb Z)[x]$
$endgroup$
– Omnomnomnom
Jan 19 at 0:24
$begingroup$
@Omnomnomnom Are you sure about that? Could you prove it in an answer form?
$endgroup$
– Riley
Jan 19 at 0:32
$begingroup$
@Omnomnomnom Are you sure about that? Could you prove it in an answer form?
$endgroup$
– Riley
Jan 19 at 0:32
$begingroup$
@Omnomnomnom If you define $f:mathbb{Z}[x]to mathbb{Z}_p[x]$ by modding the coefficients by $p$, then isn't $text{Ker}(f)$ just $pmathbb{Z}[x]$, proving the desired isomorphism?
$endgroup$
– Riley
Jan 19 at 0:42
$begingroup$
@Omnomnomnom If you define $f:mathbb{Z}[x]to mathbb{Z}_p[x]$ by modding the coefficients by $p$, then isn't $text{Ker}(f)$ just $pmathbb{Z}[x]$, proving the desired isomorphism?
$endgroup$
– Riley
Jan 19 at 0:42
$begingroup$
@Riley An element in the ring $;left(Bbb Z/pBbb Zright)[x];$ is a polynomial, whereas an element in the ring $;Bbb Z[x]/pBbb Z[x];$ is not a polynomial but in fact an equivalence class...
$endgroup$
– DonAntonio
Jan 19 at 0:43
$begingroup$
@Riley An element in the ring $;left(Bbb Z/pBbb Zright)[x];$ is a polynomial, whereas an element in the ring $;Bbb Z[x]/pBbb Z[x];$ is not a polynomial but in fact an equivalence class...
$endgroup$
– DonAntonio
Jan 19 at 0:43
2
2
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]subset f(x)Bbb Z[x];$ , as required by the third isomorphism theorem.
$endgroup$
– DonAntonio
Jan 19 at 0:49
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]subset f(x)Bbb Z[x];$ , as required by the third isomorphism theorem.
$endgroup$
– DonAntonio
Jan 19 at 0:49
|
show 3 more comments
1 Answer
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$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]⊂f(x)Bbb Z[x];$ , as required by the third isomorphism theorem
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
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add a comment |
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$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]⊂f(x)Bbb Z[x];$ , as required by the third isomorphism theorem
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
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add a comment |
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]⊂f(x)Bbb Z[x];$ , as required by the third isomorphism theorem
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]⊂f(x)Bbb Z[x];$ , as required by the third isomorphism theorem
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
$endgroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]⊂f(x)Bbb Z[x];$ , as required by the third isomorphism theorem
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
answered Jan 19 at 1:17
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
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$begingroup$
The problem is that $(Bbb Z[x])/(pBbb Z[x])$ is not equal to $(Bbb Z/p Bbb Z)[x]$
$endgroup$
– Omnomnomnom
Jan 19 at 0:24
$begingroup$
@Omnomnomnom Are you sure about that? Could you prove it in an answer form?
$endgroup$
– Riley
Jan 19 at 0:32
$begingroup$
@Omnomnomnom If you define $f:mathbb{Z}[x]to mathbb{Z}_p[x]$ by modding the coefficients by $p$, then isn't $text{Ker}(f)$ just $pmathbb{Z}[x]$, proving the desired isomorphism?
$endgroup$
– Riley
Jan 19 at 0:42
$begingroup$
@Riley An element in the ring $;left(Bbb Z/pBbb Zright)[x];$ is a polynomial, whereas an element in the ring $;Bbb Z[x]/pBbb Z[x];$ is not a polynomial but in fact an equivalence class...
$endgroup$
– DonAntonio
Jan 19 at 0:43
2
$begingroup$
Whatever you meant, the very first equality (isomorphism) is wrong unless $;pBbb Z[x]subset f(x)Bbb Z[x];$ , as required by the third isomorphism theorem.
$endgroup$
– DonAntonio
Jan 19 at 0:49