Let \(R\) be a simple rng (a nonzero ring which is not assume to have a 1, whose only two-sided ideals are \((0)\) and \(R\)) satisfying the following two conditions:
- \(R\) has no zero divisors, and
- If \(x\in R\) with \(x\neq 0\) then \(2x\neq 0\), where \(2x\definedas x+x\).
Prove the following:
For each \(x\in R\) there is one and only one element \(y\in R\) such that \(x = 2y\).
Suppose \(x,y\in R\) such that \(x\neq 0\) and \(2(xy) = x\), then \(yz = zy\) for all \(z\in R\).
You can get partial credit for (b) by showing it in the case \(R\) has a 1.
A general opinion is that this is not a great qual problem! Possibly worth skipping.
- \(R\) has no left zero divisors iff \(R\) has the left cancellation property: \(xa=xb \implies a=b\).
- \(R\) has no right zero divisors iff \(R\) has the right cancellation property: \(ax=bx \implies a=b\).
Solution
Note: solutions borrowed from folks on Math twitter!
(part 1)
- Existence: the claim is that \(2R \da
\ts{2y \st y\in R}\) is a nontrivial two-sided ideal of \(R\), forcing \(2R
= R\) by simpleness.
- That \(2R \neq 0\) follows from condition (1): Provided \(y\neq 0\), we have \(2y\neq 0\), and so if \(R\neq 0\) then there exists some nonzero \(a\in R\), in which case \(2a\neq 0\) and \(2a\in 2R\).
- That \(2R\) is a right ideal: clear, since \((2y)\cdot r = 2(yr)\in 2R\).
- That \(2R\) is a left ideal: use that multiplication is distributive: \[ r\cdot 2y \da r(y+y) = ry + ry \da 2(ry) \in 2R .\]
- So \(2R = R\) by simpleness.
- Uniqueness:
- Use the contrapositive of condition (1), so that \(2x = 0 \implies x=0\).
- Suppose toward a contradiction that \(x=2y_1 = 2y_2\), then \[ 0 = x-x = 2y_1 - 2y_2 = 2(y_1 - y_2) \implies y_1 - y_2 = 0 \implies y_1 = y_2 .\]
(part 2)
First we’ll show \(z=2(yz)\): \[ xy + xy &= x \\ \implies xy + xy - x &= 0 \\ \implies xyz + xyz - xz &= 0 \\ \implies x(yz + yz - z) &= 0 \\ \implies yz + yz - z &= 0 && \text{since } x\neq 0 \text{ and no zero divisors }\\ \implies 2(yz) &= z .\]
Now we’ll show \(z=2(zy)\): \[ yz + yz &= z \\ \implies zyz + zyz &= zz \\ \implies zyz + zyz - zz &= 0 \\ \implies (zy + zy - z)z &= 0\\ \implies z=0 \text{ or } zy+zy-z &= 0 && \text{ no zero divisors } .\]
Then if \(z=0\), we have \(yz = 0 = zy\) and we’re done.
Otherwise, \(2(zy) = z\), and thus \[ 2(zy) = z = 2(yz) \implies 2(zy - yz) = 0 \implies zy-yz = 0 ,\] so \(zy=yz\).
(of 2, if $R$ is unital)
- If \(1\in R\), \[ 2xy &= x \\ \implies 2xy-x &= 0 \\ \implies x(2y-1) &= 0 \\ \implies 2y-1 &= 0 && x\neq 0 \text{ and no zero divisors}\\ \implies 2y &= 1 .\]
- Now use \[ 1\cdot z &= z\cdot 1 \\ \implies (2y)z &= z(2y) \\ \implies (y+y)z &= z(y+y) \\ \implies yz+yz &= zy+zy \\ \implies 2(yz) &= 2(zy) \\ \implies 2(yz-zy) &= 0 \\ \implies yz-zy &= 0 \\ ,\] using condition (2).