Mixing
Isn't $mathbb Q[X]/(X^2+1)cong mathbb Q[i]$ wrong and should be $mathbb Q[X]/(X^2+1)cong mathbb Q(i)$?
$begingroup$
Isn't $mathbb Q[X]/(X^2+1)cong mathbb Q[i]$ wrong and should be $mathbb Q[X]/(X^2+1)cong mathbb Q(i)$ ?
Indeed, $mathbb Q[X]/(X^2+1)$ is a field whereas $mathbb Q[i]$ is a ring (is the Fraction ring of $mathbb Q(i)$).
abstract-algebra
asked Jan 13 at 15:57
user623855user623855
1507
$endgroup$
|
show 11 more comments
$begingroup$
Isn't $mathbb Q[X]/(X^2+1)cong mathbb Q[i]$ wrong and should be $mathbb Q[X]/(X^2+1)cong mathbb Q(i)$ ?
Indeed, $mathbb Q[X]/(X^2+1)$ is a field whereas $mathbb Q[i]$ is a ring (is the Fraction ring of $mathbb Q(i)$).
abstract-algebra
asked Jan 13 at 15:57
user623855user623855
1507
$endgroup$
$begingroup$
$mathbb{Q}[i]$ is in fact a field, since $i^2 = -1$.
$endgroup$
– Paul K
Jan 13 at 15:58
1
$begingroup$
Actually $Bbb Q [i]$ happens to be a field, thus it coincides with $Bbb Q (i)$
$endgroup$
– Crostul
Jan 13 at 15:58
4
$begingroup$
For any $;ainBbb C;$ , we have that $Bbb Q[a]=Bbb Q(a)iff a;$ is algebraic over $;Bbb Q;$ . You can also change $;Bbb Q;$ and use your favourite field and its algebraic closure.
$endgroup$
– DonAntonio
Jan 13 at 16:00
$begingroup$
A field is a ring, in case that is worrying you for some reason.
$endgroup$
– rschwieb
Jan 13 at 16:00
$begingroup$
Fields are also rings
$endgroup$
– Adam Higgins
Jan 13 at 16:00
|
show 11 more comments
$begingroup$
Isn't $mathbb Q[X]/(X^2+1)cong mathbb Q[i]$ wrong and should be $mathbb Q[X]/(X^2+1)cong mathbb Q(i)$ ?
Indeed, $mathbb Q[X]/(X^2+1)$ is a field whereas $mathbb Q[i]$ is a ring (is the Fraction ring of $mathbb Q(i)$).
abstract-algebra
asked Jan 13 at 15:57
user623855user623855
1507
$endgroup$
Isn't $mathbb Q[X]/(X^2+1)cong mathbb Q[i]$ wrong and should be $mathbb Q[X]/(X^2+1)cong mathbb Q(i)$ ?
Indeed, $mathbb Q[X]/(X^2+1)$ is a field whereas $mathbb Q[i]$ is a ring (is the Fraction ring of $mathbb Q(i)$).
abstract-algebra
abstract-algebra
asked Jan 13 at 15:57
user623855user623855
1507
asked Jan 13 at 15:57
user623855user623855
1507
asked Jan 13 at 15:57
user623855user623855
1507
asked Jan 13 at 15:57
user623855user623855
1507
asked Jan 13 at 15:57
user623855user623855
1507
1507
$begingroup$
$mathbb{Q}[i]$ is in fact a field, since $i^2 = -1$.
$endgroup$
– Paul K
Jan 13 at 15:58
1
$begingroup$
Actually $Bbb Q [i]$ happens to be a field, thus it coincides with $Bbb Q (i)$
$endgroup$
– Crostul
Jan 13 at 15:58
4
$begingroup$
For any $;ainBbb C;$ , we have that $Bbb Q[a]=Bbb Q(a)iff a;$ is algebraic over $;Bbb Q;$ . You can also change $;Bbb Q;$ and use your favourite field and its algebraic closure.
$endgroup$
– DonAntonio
Jan 13 at 16:00
$begingroup$
A field is a ring, in case that is worrying you for some reason.
$endgroup$
– rschwieb
Jan 13 at 16:00
$begingroup$
Fields are also rings
$endgroup$
– Adam Higgins
Jan 13 at 16:00
|
show 11 more comments
$begingroup$
$mathbb{Q}[i]$ is in fact a field, since $i^2 = -1$.
$endgroup$
– Paul K
Jan 13 at 15:58
1
$begingroup$
Actually $Bbb Q [i]$ happens to be a field, thus it coincides with $Bbb Q (i)$
$endgroup$
– Crostul
Jan 13 at 15:58
4
$begingroup$
For any $;ainBbb C;$ , we have that $Bbb Q[a]=Bbb Q(a)iff a;$ is algebraic over $;Bbb Q;$ . You can also change $;Bbb Q;$ and use your favourite field and its algebraic closure.
$endgroup$
– DonAntonio
Jan 13 at 16:00
$begingroup$
A field is a ring, in case that is worrying you for some reason.
$endgroup$
– rschwieb
Jan 13 at 16:00
$begingroup$
Fields are also rings
$endgroup$
– Adam Higgins
Jan 13 at 16:00
$begingroup$
$mathbb{Q}[i]$ is in fact a field, since $i^2 = -1$.
$endgroup$
– Paul K
Jan 13 at 15:58
$begingroup$
$mathbb{Q}[i]$ is in fact a field, since $i^2 = -1$.
$endgroup$
– Paul K
Jan 13 at 15:58
1
1
$begingroup$
Actually $Bbb Q [i]$ happens to be a field, thus it coincides with $Bbb Q (i)$
$endgroup$
– Crostul
Jan 13 at 15:58
$begingroup$
Actually $Bbb Q [i]$ happens to be a field, thus it coincides with $Bbb Q (i)$
$endgroup$
– Crostul
Jan 13 at 15:58
4
4
$begingroup$
For any $;ainBbb C;$ , we have that $Bbb Q[a]=Bbb Q(a)iff a;$ is algebraic over $;Bbb Q;$ . You can also change $;Bbb Q;$ and use your favourite field and its algebraic closure.
$endgroup$
– DonAntonio
Jan 13 at 16:00
$begingroup$
For any $;ainBbb C;$ , we have that $Bbb Q[a]=Bbb Q(a)iff a;$ is algebraic over $;Bbb Q;$ . You can also change $;Bbb Q;$ and use your favourite field and its algebraic closure.
$endgroup$
– DonAntonio
Jan 13 at 16:00
$begingroup$
A field is a ring, in case that is worrying you for some reason.
$endgroup$
– rschwieb
Jan 13 at 16:00
$begingroup$
A field is a ring, in case that is worrying you for some reason.
$endgroup$
– rschwieb
Jan 13 at 16:00
$begingroup$
Fields are also rings
$endgroup$
– Adam Higgins
Jan 13 at 16:00
$begingroup$
Fields are also rings
$endgroup$
– Adam Higgins
Jan 13 at 16:00
|
show 11 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If $alpha$ is a root of a monic polynomial $rm:f(x):$ over a domain $rm:D:$ then $rm:D[alpha]:$ is a field iff $rm:D:$ is a field. Ditto for arbitrary integral extensions of domains. Here is a detailed treament of the quadratic case.
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
add a comment |
$begingroup$
Addressing your question in the comments: observe that in $;Bbb F:=Bbb Q[X]/langle X^2+1rangle;$ , we have that if $;x:=Z+langle x^2+1rangle;$ , then $;x^2+1=0implies x^2=-1;$ , so in $;Bbb F;$ we get
$$frac1{x+2}=frac{2-x}{4-x^2}=frac{2-x}{4+1}=frac15(-x+2)$$
so
$$frac{x+1}{x+2}=(x+1)frac1{x+2}=(x+1)frac15(-x+2)=frac15(-x^2+x-2)=frac15(x-1)$$
In similar form you can do other calculation. Observe the is just hte usual operations inthe complex $;Bbb C;$ , just "in disguise".
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $alpha$ is a root of a monic polynomial $rm:f(x):$ over a domain $rm:D:$ then $rm:D[alpha]:$ is a field iff $rm:D:$ is a field. Ditto for arbitrary integral extensions of domains. Here is a detailed treament of the quadratic case.
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
add a comment |
$begingroup$
If $alpha$ is a root of a monic polynomial $rm:f(x):$ over a domain $rm:D:$ then $rm:D[alpha]:$ is a field iff $rm:D:$ is a field. Ditto for arbitrary integral extensions of domains. Here is a detailed treament of the quadratic case.
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
add a comment |
$begingroup$
If $alpha$ is a root of a monic polynomial $rm:f(x):$ over a domain $rm:D:$ then $rm:D[alpha]:$ is a field iff $rm:D:$ is a field. Ditto for arbitrary integral extensions of domains. Here is a detailed treament of the quadratic case.
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
$endgroup$
If $alpha$ is a root of a monic polynomial $rm:f(x):$ over a domain $rm:D:$ then $rm:D[alpha]:$ is a field iff $rm:D:$ is a field. Ditto for arbitrary integral extensions of domains. Here is a detailed treament of the quadratic case.
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
answered Jan 13 at 16:37
Bill DubuqueBill Dubuque
210k29192645
210k29192645
add a comment |
add a comment |
$begingroup$
Addressing your question in the comments: observe that in $;Bbb F:=Bbb Q[X]/langle X^2+1rangle;$ , we have that if $;x:=Z+langle x^2+1rangle;$ , then $;x^2+1=0implies x^2=-1;$ , so in $;Bbb F;$ we get
$$frac1{x+2}=frac{2-x}{4-x^2}=frac{2-x}{4+1}=frac15(-x+2)$$
so
$$frac{x+1}{x+2}=(x+1)frac1{x+2}=(x+1)frac15(-x+2)=frac15(-x^2+x-2)=frac15(x-1)$$
In similar form you can do other calculation. Observe the is just hte usual operations inthe complex $;Bbb C;$ , just "in disguise".
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
$begingroup$
Addressing your question in the comments: observe that in $;Bbb F:=Bbb Q[X]/langle X^2+1rangle;$ , we have that if $;x:=Z+langle x^2+1rangle;$ , then $;x^2+1=0implies x^2=-1;$ , so in $;Bbb F;$ we get
$$frac1{x+2}=frac{2-x}{4-x^2}=frac{2-x}{4+1}=frac15(-x+2)$$
so
$$frac{x+1}{x+2}=(x+1)frac1{x+2}=(x+1)frac15(-x+2)=frac15(-x^2+x-2)=frac15(x-1)$$
In similar form you can do other calculation. Observe the is just hte usual operations inthe complex $;Bbb C;$ , just "in disguise".
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
$begingroup$
Addressing your question in the comments: observe that in $;Bbb F:=Bbb Q[X]/langle X^2+1rangle;$ , we have that if $;x:=Z+langle x^2+1rangle;$ , then $;x^2+1=0implies x^2=-1;$ , so in $;Bbb F;$ we get
$$frac1{x+2}=frac{2-x}{4-x^2}=frac{2-x}{4+1}=frac15(-x+2)$$
so
$$frac{x+1}{x+2}=(x+1)frac1{x+2}=(x+1)frac15(-x+2)=frac15(-x^2+x-2)=frac15(x-1)$$
In similar form you can do other calculation. Observe the is just hte usual operations inthe complex $;Bbb C;$ , just "in disguise".
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
$endgroup$
Addressing your question in the comments: observe that in $;Bbb F:=Bbb Q[X]/langle X^2+1rangle;$ , we have that if $;x:=Z+langle x^2+1rangle;$ , then $;x^2+1=0implies x^2=-1;$ , so in $;Bbb F;$ we get
$$frac1{x+2}=frac{2-x}{4-x^2}=frac{2-x}{4+1}=frac15(-x+2)$$
so
$$frac{x+1}{x+2}=(x+1)frac1{x+2}=(x+1)frac15(-x+2)=frac15(-x^2+x-2)=frac15(x-1)$$
In similar form you can do other calculation. Observe the is just hte usual operations inthe complex $;Bbb C;$ , just "in disguise".
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 17:18
DonAntonioDonAntonio
178k1494230
178k1494230
add a comment |
add a comment |
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$begingroup$
$mathbb{Q}[i]$ is in fact a field, since $i^2 = -1$.
$endgroup$
– Paul K
Jan 13 at 15:58
1
$begingroup$
Actually $Bbb Q [i]$ happens to be a field, thus it coincides with $Bbb Q (i)$
$endgroup$
– Crostul
Jan 13 at 15:58
4
$begingroup$
For any $;ainBbb C;$ , we have that $Bbb Q[a]=Bbb Q(a)iff a;$ is algebraic over $;Bbb Q;$ . You can also change $;Bbb Q;$ and use your favourite field and its algebraic closure.
$endgroup$
– DonAntonio
Jan 13 at 16:00
$begingroup$
A field is a ring, in case that is worrying you for some reason.
$endgroup$
– rschwieb
Jan 13 at 16:00
$begingroup$
Fields are also rings
$endgroup$
– Adam Higgins
Jan 13 at 16:00