What means $ℝ[i]$?
$begingroup$
From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?
complex-numbers
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
$endgroup$
add a comment |
$begingroup$
From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?
complex-numbers
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
$endgroup$
3
$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33
1
$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34
add a comment |
$begingroup$
From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?
complex-numbers
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
$endgroup$
From what i‘ve understood, a set, say $ℤ[i]$, denotes the set of all integers, in addition to i. ($i^2 = -1$) So for example, the complex number $1+i$ would be within that set.
In short, my question is, would then $ℝ[i]$ be the equivalent set as $ℂ$?
complex-numbers
complex-numbers
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
edited Dec 13 '18 at 21:49
Asaf Karagila♦
308k33441774
308k33441774
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
asked Dec 13 '18 at 21:32
Nils Phillip TalgöNils Phillip Talgö
659
659
3
$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33
1
$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34
add a comment |
3
$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33
1
$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34
3
3
$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33
$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33
1
1
$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34
$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In general, $S=R[a]$ is a ring extension
of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
add a comment |
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$begingroup$
In general, $S=R[a]$ is a ring extension
of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
add a comment |
$begingroup$
In general, $S=R[a]$ is a ring extension
of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
add a comment |
$begingroup$
In general, $S=R[a]$ is a ring extension
of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
In general, $S=R[a]$ is a ring extension
of $R$ by an element $a$ in a ring containing $R$, i.e., $S$ is the smallest ring which contains $R$ and $a$. Notice that all elements of $R[a]$ are necessarily in all rings containing $R$ and $a$, and it is itself a ring. Now apply this for $R=Bbb{Z}$ or $R=Bbb{R}$ in the ring $Bbb{C}$ and for $a=i$. This answers your question "what means $Bbb{R}[i]$?". Finally, $Bbb{R}[i]cong Bbb{C}$ as rings (and as fields).
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
edited Dec 13 '18 at 21:48
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
answered Dec 13 '18 at 21:43
Dietrich BurdeDietrich Burde
81.8k648106
81.8k648106
add a comment |
add a comment |
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3
$begingroup$
I see no reason it shouldn't be.
$endgroup$
– Arthur
Dec 13 '18 at 21:33
1
$begingroup$
yes, it is and it's exactly how you build $mathbb{C}$
$endgroup$
– qbert
Dec 13 '18 at 21:34