Ring Theory: Is the ring $S$ a prime ring? Ref. Page no. 8 (Section 2) Neal H. McCoy “The Theory of...
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Let $R$ be a non commutative prime ring without unity and let $S$ be the set of all ordered pair $(a,i)$, where $ain R$ and $iin I$, where $I$ is the ring of integers. On $S$ we define addition and multiplication as follows:
$(a,i)+(b,j)=(a+b,i+j)$,
$(a,i)(b,j)=(ab+ib+ja, ij)$.
Then $S$ is a ring with unity $(0,1)$.
My question-
Is the ring $S$ a prime ring?
In case $R$ is a commutative prime ring with unity, I have shown $S$ is not a prime ring.
But I am stuck in the case $R$ is non commutative prime ring without unity.
ring-theory
$endgroup$
add a comment |
$begingroup$
Let $R$ be a non commutative prime ring without unity and let $S$ be the set of all ordered pair $(a,i)$, where $ain R$ and $iin I$, where $I$ is the ring of integers. On $S$ we define addition and multiplication as follows:
$(a,i)+(b,j)=(a+b,i+j)$,
$(a,i)(b,j)=(ab+ib+ja, ij)$.
Then $S$ is a ring with unity $(0,1)$.
My question-
Is the ring $S$ a prime ring?
In case $R$ is a commutative prime ring with unity, I have shown $S$ is not a prime ring.
But I am stuck in the case $R$ is non commutative prime ring without unity.
ring-theory
$endgroup$
1
$begingroup$
If you have already shown the statement to be false for commutative ring with identity, what is there left to do? You have not been asked to prove it is false for all categories of rings.
$endgroup$
– rschwieb
Jan 3 at 14:49
$begingroup$
My question is that if R is non commutative prime ring without unity, then what we can say about S, whether S is a prime ring or not?
$endgroup$
– nazim khan
Jan 5 at 8:08
$begingroup$
Then I submit to you that you are making a mistake of logic. Subtracting hypotheses from a disproven statement cannot cause it to become true. And if you are simply asking for a new counterexample that isn't commutative and doesn't have an identity, you are effectively asking people to work harder to re-disprove the statement, because every hypothesis you forbid adds a requirement to the needed counterexample. My guess is that users will be more annoyed than compelled by this activity.
$endgroup$
– rschwieb
Jan 5 at 13:46
add a comment |
$begingroup$
Let $R$ be a non commutative prime ring without unity and let $S$ be the set of all ordered pair $(a,i)$, where $ain R$ and $iin I$, where $I$ is the ring of integers. On $S$ we define addition and multiplication as follows:
$(a,i)+(b,j)=(a+b,i+j)$,
$(a,i)(b,j)=(ab+ib+ja, ij)$.
Then $S$ is a ring with unity $(0,1)$.
My question-
Is the ring $S$ a prime ring?
In case $R$ is a commutative prime ring with unity, I have shown $S$ is not a prime ring.
But I am stuck in the case $R$ is non commutative prime ring without unity.
ring-theory
$endgroup$
Let $R$ be a non commutative prime ring without unity and let $S$ be the set of all ordered pair $(a,i)$, where $ain R$ and $iin I$, where $I$ is the ring of integers. On $S$ we define addition and multiplication as follows:
$(a,i)+(b,j)=(a+b,i+j)$,
$(a,i)(b,j)=(ab+ib+ja, ij)$.
Then $S$ is a ring with unity $(0,1)$.
My question-
Is the ring $S$ a prime ring?
In case $R$ is a commutative prime ring with unity, I have shown $S$ is not a prime ring.
But I am stuck in the case $R$ is non commutative prime ring without unity.
ring-theory
ring-theory
asked Jan 3 at 8:03
nazim khannazim khan
555
555
1
$begingroup$
If you have already shown the statement to be false for commutative ring with identity, what is there left to do? You have not been asked to prove it is false for all categories of rings.
$endgroup$
– rschwieb
Jan 3 at 14:49
$begingroup$
My question is that if R is non commutative prime ring without unity, then what we can say about S, whether S is a prime ring or not?
$endgroup$
– nazim khan
Jan 5 at 8:08
$begingroup$
Then I submit to you that you are making a mistake of logic. Subtracting hypotheses from a disproven statement cannot cause it to become true. And if you are simply asking for a new counterexample that isn't commutative and doesn't have an identity, you are effectively asking people to work harder to re-disprove the statement, because every hypothesis you forbid adds a requirement to the needed counterexample. My guess is that users will be more annoyed than compelled by this activity.
$endgroup$
– rschwieb
Jan 5 at 13:46
add a comment |
1
$begingroup$
If you have already shown the statement to be false for commutative ring with identity, what is there left to do? You have not been asked to prove it is false for all categories of rings.
$endgroup$
– rschwieb
Jan 3 at 14:49
$begingroup$
My question is that if R is non commutative prime ring without unity, then what we can say about S, whether S is a prime ring or not?
$endgroup$
– nazim khan
Jan 5 at 8:08
$begingroup$
Then I submit to you that you are making a mistake of logic. Subtracting hypotheses from a disproven statement cannot cause it to become true. And if you are simply asking for a new counterexample that isn't commutative and doesn't have an identity, you are effectively asking people to work harder to re-disprove the statement, because every hypothesis you forbid adds a requirement to the needed counterexample. My guess is that users will be more annoyed than compelled by this activity.
$endgroup$
– rschwieb
Jan 5 at 13:46
1
1
$begingroup$
If you have already shown the statement to be false for commutative ring with identity, what is there left to do? You have not been asked to prove it is false for all categories of rings.
$endgroup$
– rschwieb
Jan 3 at 14:49
$begingroup$
If you have already shown the statement to be false for commutative ring with identity, what is there left to do? You have not been asked to prove it is false for all categories of rings.
$endgroup$
– rschwieb
Jan 3 at 14:49
$begingroup$
My question is that if R is non commutative prime ring without unity, then what we can say about S, whether S is a prime ring or not?
$endgroup$
– nazim khan
Jan 5 at 8:08
$begingroup$
My question is that if R is non commutative prime ring without unity, then what we can say about S, whether S is a prime ring or not?
$endgroup$
– nazim khan
Jan 5 at 8:08
$begingroup$
Then I submit to you that you are making a mistake of logic. Subtracting hypotheses from a disproven statement cannot cause it to become true. And if you are simply asking for a new counterexample that isn't commutative and doesn't have an identity, you are effectively asking people to work harder to re-disprove the statement, because every hypothesis you forbid adds a requirement to the needed counterexample. My guess is that users will be more annoyed than compelled by this activity.
$endgroup$
– rschwieb
Jan 5 at 13:46
$begingroup$
Then I submit to you that you are making a mistake of logic. Subtracting hypotheses from a disproven statement cannot cause it to become true. And if you are simply asking for a new counterexample that isn't commutative and doesn't have an identity, you are effectively asking people to work harder to re-disprove the statement, because every hypothesis you forbid adds a requirement to the needed counterexample. My guess is that users will be more annoyed than compelled by this activity.
$endgroup$
– rschwieb
Jan 5 at 13:46
add a comment |
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$begingroup$
If you have already shown the statement to be false for commutative ring with identity, what is there left to do? You have not been asked to prove it is false for all categories of rings.
$endgroup$
– rschwieb
Jan 3 at 14:49
$begingroup$
My question is that if R is non commutative prime ring without unity, then what we can say about S, whether S is a prime ring or not?
$endgroup$
– nazim khan
Jan 5 at 8:08
$begingroup$
Then I submit to you that you are making a mistake of logic. Subtracting hypotheses from a disproven statement cannot cause it to become true. And if you are simply asking for a new counterexample that isn't commutative and doesn't have an identity, you are effectively asking people to work harder to re-disprove the statement, because every hypothesis you forbid adds a requirement to the needed counterexample. My guess is that users will be more annoyed than compelled by this activity.
$endgroup$
– rschwieb
Jan 5 at 13:46