$R$ ring without identity generated by an element
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Let $R$ a commutative ring without unity and we suppose that exist $ain R$ such that:
$$R={ar+na;|;rin R,ninmathbb{Z}}=(a).$$
We observe that the principal generated ideal $(a^2)={a^2r+na^2;|;rin R,ninmathbb{Z}}$ is a proper ideal of $R$, in fact $ain(a)$, but $anotin(a^2)$. Indedd, observe that, were $a$ in $(a^2)$, we could write $a=a^2overline{r}+overline{n}a^2$ and at this point, if we define $e=aoverline{r}+overline{n}a$, $e$ would a multiplicative identity for $R$. For to prove this I take $r_1in R$ and after the calculation I get that $$er=e^2r.$$
A this point, can I conclude that $er=r$ even if $R$ is not a integral domain?
Thanks!
abstract-algebra
asked Nov 28 at 16:40
Jack J.
4601419
add a comment |
up vote
1
down vote
favorite
Let $R$ a commutative ring without unity and we suppose that exist $ain R$ such that:
$$R={ar+na;|;rin R,ninmathbb{Z}}=(a).$$
We observe that the principal generated ideal $(a^2)={a^2r+na^2;|;rin R,ninmathbb{Z}}$ is a proper ideal of $R$, in fact $ain(a)$, but $anotin(a^2)$. Indedd, observe that, were $a$ in $(a^2)$, we could write $a=a^2overline{r}+overline{n}a^2$ and at this point, if we define $e=aoverline{r}+overline{n}a$, $e$ would a multiplicative identity for $R$. For to prove this I take $r_1in R$ and after the calculation I get that $$er=e^2r.$$
A this point, can I conclude that $er=r$ even if $R$ is not a integral domain?
Thanks!
abstract-algebra
asked Nov 28 at 16:40
Jack J.
4601419
Where did you want to use $r_1$? What are the bars in $bar r$ and $bar n$ supposed to mean? just $bar rin R$ and $bar n in mathbb Z$?
– rschwieb
Nov 28 at 16:57
I was wrong to write $e$, sorry
– Jack J.
Nov 28 at 17:02
$r_1$ is a specific element of $R$ and $overline{r}$ and $overline{n}$ are the specific element which exist in conseguence to the fact that $ain (a^2)$
– Jack J.
Nov 28 at 17:04
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $R$ a commutative ring without unity and we suppose that exist $ain R$ such that:
$$R={ar+na;|;rin R,ninmathbb{Z}}=(a).$$
We observe that the principal generated ideal $(a^2)={a^2r+na^2;|;rin R,ninmathbb{Z}}$ is a proper ideal of $R$, in fact $ain(a)$, but $anotin(a^2)$. Indedd, observe that, were $a$ in $(a^2)$, we could write $a=a^2overline{r}+overline{n}a^2$ and at this point, if we define $e=aoverline{r}+overline{n}a$, $e$ would a multiplicative identity for $R$. For to prove this I take $r_1in R$ and after the calculation I get that $$er=e^2r.$$
A this point, can I conclude that $er=r$ even if $R$ is not a integral domain?
Thanks!
abstract-algebra
asked Nov 28 at 16:40
Jack J.
4601419
Let $R$ a commutative ring without unity and we suppose that exist $ain R$ such that:
$$R={ar+na;|;rin R,ninmathbb{Z}}=(a).$$
We observe that the principal generated ideal $(a^2)={a^2r+na^2;|;rin R,ninmathbb{Z}}$ is a proper ideal of $R$, in fact $ain(a)$, but $anotin(a^2)$. Indedd, observe that, were $a$ in $(a^2)$, we could write $a=a^2overline{r}+overline{n}a^2$ and at this point, if we define $e=aoverline{r}+overline{n}a$, $e$ would a multiplicative identity for $R$. For to prove this I take $r_1in R$ and after the calculation I get that $$er=e^2r.$$
A this point, can I conclude that $er=r$ even if $R$ is not a integral domain?
Thanks!
abstract-algebra
abstract-algebra
asked Nov 28 at 16:40
Jack J.
4601419
asked Nov 28 at 16:40
Jack J.
4601419
edited Nov 28 at 17:01
asked Nov 28 at 16:40
Jack J.
4601419
asked Nov 28 at 16:40
Jack J.
4601419
asked Nov 28 at 16:40
Jack J.
4601419
4601419
Where did you want to use $r_1$? What are the bars in $bar r$ and $bar n$ supposed to mean? just $bar rin R$ and $bar n in mathbb Z$?
– rschwieb
Nov 28 at 16:57
I was wrong to write $e$, sorry
– Jack J.
Nov 28 at 17:02
$r_1$ is a specific element of $R$ and $overline{r}$ and $overline{n}$ are the specific element which exist in conseguence to the fact that $ain (a^2)$
– Jack J.
Nov 28 at 17:04
add a comment |
Where did you want to use $r_1$? What are the bars in $bar r$ and $bar n$ supposed to mean? just $bar rin R$ and $bar n in mathbb Z$?
– rschwieb
Nov 28 at 16:57
I was wrong to write $e$, sorry
– Jack J.
Nov 28 at 17:02
$r_1$ is a specific element of $R$ and $overline{r}$ and $overline{n}$ are the specific element which exist in conseguence to the fact that $ain (a^2)$
– Jack J.
Nov 28 at 17:04
Where did you want to use $r_1$? What are the bars in $bar r$ and $bar n$ supposed to mean? just $bar rin R$ and $bar n in mathbb Z$?
– rschwieb
Nov 28 at 16:57
Where did you want to use $r_1$? What are the bars in $bar r$ and $bar n$ supposed to mean? just $bar rin R$ and $bar n in mathbb Z$?
– rschwieb
Nov 28 at 16:57
I was wrong to write $e$, sorry
– Jack J.
Nov 28 at 17:02
I was wrong to write $e$, sorry
– Jack J.
Nov 28 at 17:02
$r_1$ is a specific element of $R$ and $overline{r}$ and $overline{n}$ are the specific element which exist in conseguence to the fact that $ain (a^2)$
– Jack J.
Nov 28 at 17:04
$r_1$ is a specific element of $R$ and $overline{r}$ and $overline{n}$ are the specific element which exist in conseguence to the fact that $ain (a^2)$
– Jack J.
Nov 28 at 17:04
add a comment |
1 Answer
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up vote
1
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No, this logic is incorrect.
Take, for example, any boolean ring without identity. In that case, $(a)=(a^2)$ for every $a$, but there is no identity.
From $a=a^2r+na^2=e$, it simply does not follow that $e$ is an identity for the ring. (Edit: this has since been corrected in the post. The following comment confirms what the OP says.)
It is true, though, if you assume $R$ is a domain that the existence of a single $aneq 0$ such that $ain (a^2)$ does imply $ar+an$ is the identity for the ring. It's not true otherwise, as the example I gave shows.
answered Nov 28 at 16:56
rschwieb
104k1299241
Sorry! I was wrong to write $e$....
– Jack J.
Nov 28 at 17:02
Now, I correct.
– Jack J.
Nov 28 at 17:03
Ok, but my book conclude saying that $e$ is a identity for $R$, but it does not specify if $R$ is a integral domain....
– Jack J.
Nov 28 at 17:07
Perhaps we can show otherwise that $e$ is identity without arriving at that equality?
– Jack J.
Nov 28 at 17:08
1
@JackJ. The counterexample I gave shows that it is possible for $(a)=(a^2)$ for all $a$, and yet there is no identity. So it would be a waste of time to attempt a proof... The situation is as I say in the last paragraph.
– rschwieb
Nov 28 at 17:25
add a comment |
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1 Answer
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1 Answer
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up vote
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accepted
No, this logic is incorrect.
Take, for example, any boolean ring without identity. In that case, $(a)=(a^2)$ for every $a$, but there is no identity.
From $a=a^2r+na^2=e$, it simply does not follow that $e$ is an identity for the ring. (Edit: this has since been corrected in the post. The following comment confirms what the OP says.)
It is true, though, if you assume $R$ is a domain that the existence of a single $aneq 0$ such that $ain (a^2)$ does imply $ar+an$ is the identity for the ring. It's not true otherwise, as the example I gave shows.
answered Nov 28 at 16:56
rschwieb
104k1299241
Sorry! I was wrong to write $e$....
– Jack J.
Nov 28 at 17:02
Now, I correct.
– Jack J.
Nov 28 at 17:03
Ok, but my book conclude saying that $e$ is a identity for $R$, but it does not specify if $R$ is a integral domain....
– Jack J.
Nov 28 at 17:07
Perhaps we can show otherwise that $e$ is identity without arriving at that equality?
– Jack J.
Nov 28 at 17:08
1
@JackJ. The counterexample I gave shows that it is possible for $(a)=(a^2)$ for all $a$, and yet there is no identity. So it would be a waste of time to attempt a proof... The situation is as I say in the last paragraph.
– rschwieb
Nov 28 at 17:25
add a comment |
up vote
1
down vote
accepted
No, this logic is incorrect.
Take, for example, any boolean ring without identity. In that case, $(a)=(a^2)$ for every $a$, but there is no identity.
From $a=a^2r+na^2=e$, it simply does not follow that $e$ is an identity for the ring. (Edit: this has since been corrected in the post. The following comment confirms what the OP says.)
It is true, though, if you assume $R$ is a domain that the existence of a single $aneq 0$ such that $ain (a^2)$ does imply $ar+an$ is the identity for the ring. It's not true otherwise, as the example I gave shows.
answered Nov 28 at 16:56
rschwieb
104k1299241
Sorry! I was wrong to write $e$....
– Jack J.
Nov 28 at 17:02
Now, I correct.
– Jack J.
Nov 28 at 17:03
Ok, but my book conclude saying that $e$ is a identity for $R$, but it does not specify if $R$ is a integral domain....
– Jack J.
Nov 28 at 17:07
Perhaps we can show otherwise that $e$ is identity without arriving at that equality?
– Jack J.
Nov 28 at 17:08
1
@JackJ. The counterexample I gave shows that it is possible for $(a)=(a^2)$ for all $a$, and yet there is no identity. So it would be a waste of time to attempt a proof... The situation is as I say in the last paragraph.
– rschwieb
Nov 28 at 17:25
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
No, this logic is incorrect.
Take, for example, any boolean ring without identity. In that case, $(a)=(a^2)$ for every $a$, but there is no identity.
From $a=a^2r+na^2=e$, it simply does not follow that $e$ is an identity for the ring. (Edit: this has since been corrected in the post. The following comment confirms what the OP says.)
It is true, though, if you assume $R$ is a domain that the existence of a single $aneq 0$ such that $ain (a^2)$ does imply $ar+an$ is the identity for the ring. It's not true otherwise, as the example I gave shows.
answered Nov 28 at 16:56
rschwieb
104k1299241
No, this logic is incorrect.
Take, for example, any boolean ring without identity. In that case, $(a)=(a^2)$ for every $a$, but there is no identity.
From $a=a^2r+na^2=e$, it simply does not follow that $e$ is an identity for the ring. (Edit: this has since been corrected in the post. The following comment confirms what the OP says.)
It is true, though, if you assume $R$ is a domain that the existence of a single $aneq 0$ such that $ain (a^2)$ does imply $ar+an$ is the identity for the ring. It's not true otherwise, as the example I gave shows.
answered Nov 28 at 16:56
rschwieb
104k1299241
edited Nov 28 at 17:04
answered Nov 28 at 16:56
rschwieb
104k1299241
answered Nov 28 at 16:56
rschwieb
104k1299241
answered Nov 28 at 16:56
rschwieb
104k1299241
104k1299241
Sorry! I was wrong to write $e$....
– Jack J.
Nov 28 at 17:02
Now, I correct.
– Jack J.
Nov 28 at 17:03
Ok, but my book conclude saying that $e$ is a identity for $R$, but it does not specify if $R$ is a integral domain....
– Jack J.
Nov 28 at 17:07
Perhaps we can show otherwise that $e$ is identity without arriving at that equality?
– Jack J.
Nov 28 at 17:08
1
@JackJ. The counterexample I gave shows that it is possible for $(a)=(a^2)$ for all $a$, and yet there is no identity. So it would be a waste of time to attempt a proof... The situation is as I say in the last paragraph.
– rschwieb
Nov 28 at 17:25
add a comment |
Sorry! I was wrong to write $e$....
– Jack J.
Nov 28 at 17:02
Now, I correct.
– Jack J.
Nov 28 at 17:03
Ok, but my book conclude saying that $e$ is a identity for $R$, but it does not specify if $R$ is a integral domain....
– Jack J.
Nov 28 at 17:07
Perhaps we can show otherwise that $e$ is identity without arriving at that equality?
– Jack J.
Nov 28 at 17:08
1
@JackJ. The counterexample I gave shows that it is possible for $(a)=(a^2)$ for all $a$, and yet there is no identity. So it would be a waste of time to attempt a proof... The situation is as I say in the last paragraph.
– rschwieb
Nov 28 at 17:25
Sorry! I was wrong to write $e$....
– Jack J.
Nov 28 at 17:02
Sorry! I was wrong to write $e$....
– Jack J.
Nov 28 at 17:02
Now, I correct.
– Jack J.
Nov 28 at 17:03
Now, I correct.
– Jack J.
Nov 28 at 17:03
Ok, but my book conclude saying that $e$ is a identity for $R$, but it does not specify if $R$ is a integral domain....
– Jack J.
Nov 28 at 17:07
Ok, but my book conclude saying that $e$ is a identity for $R$, but it does not specify if $R$ is a integral domain....
– Jack J.
Nov 28 at 17:07
Perhaps we can show otherwise that $e$ is identity without arriving at that equality?
– Jack J.
Nov 28 at 17:08
Perhaps we can show otherwise that $e$ is identity without arriving at that equality?
– Jack J.
Nov 28 at 17:08
1
1
@JackJ. The counterexample I gave shows that it is possible for $(a)=(a^2)$ for all $a$, and yet there is no identity. So it would be a waste of time to attempt a proof... The situation is as I say in the last paragraph.
– rschwieb
Nov 28 at 17:25
@JackJ. The counterexample I gave shows that it is possible for $(a)=(a^2)$ for all $a$, and yet there is no identity. So it would be a waste of time to attempt a proof... The situation is as I say in the last paragraph.
– rschwieb
Nov 28 at 17:25
add a comment |
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Where did you want to use $r_1$? What are the bars in $bar r$ and $bar n$ supposed to mean? just $bar rin R$ and $bar n in mathbb Z$?
– rschwieb
Nov 28 at 16:57
I was wrong to write $e$, sorry
– Jack J.
Nov 28 at 17:02
$r_1$ is a specific element of $R$ and $overline{r}$ and $overline{n}$ are the specific element which exist in conseguence to the fact that $ain (a^2)$
– Jack J.
Nov 28 at 17:04