Prove that $(H,+,\circ)$ is unital commutative ring.

Question

Suppose that $(H,+,.)$ is a unital commutative ring. Now for all $x,y \in H$ define $$x \circ y= xy+yx.$$

With this operation we have another ring which is also commutative. $(H,+,\circ)$

Now I have to show that this ring is also unital under $\circ$ operation.

My observation:

I have to find $e\in H$ such that $x\circ e=x$

$xe+ex=x$ which gives $e=1/2$ but we don't know the elements of $H$ so we can't say $1/2\in H$, can we?

Please guide me. Thank you.

Answer 1

Since $H$ is commutative, $x\circ y=xy+yx=2xy$. This new multiplication is obviously commutative, but need not have a neutral element. For instance, if $H$ has characteristic $2$, we have $x\circ y=0$ for every $x,y\in H$.

Is the operation associative? $(x\circ y)\circ z=(2xy)\circ z=4xyz$ and $x\circ(y\circ z)=2x(y\circ z)=2x(2yz)=4xyz$.

Distributivity is obvious.

Note. If $n$ is an integer and $x\in H$, $nx$ is the standard multiple.

Under what condition does the circle operation have a neutral element $e$? We need it satisfies $2ex=x$, for every $x$, in particular for $\mathbf{1}$ (the neutral element of $H$), so $2e=\mathbf{1}$. This condition is also easily seen to be sufficient: indeed, if $2e=\mathbf{1}$, then $e\circ x=2ex=\mathbf{1}x=x$.

Thus the ring $(H,+\circ)$ is unital if and only if $2\mathbf{1}=\mathbf{1}+\mathbf{1}$ is invertible in $(H,+,\cdot)$. This is granted if the characteristic of $H$ is finite and coprime with $2$. It is false whenever the characteristic is divisible by $2$: if the characteristic is $2k$, then $k\mathbf{1}=(k\mathbf{1})\circ e=2ke=0$ gives a contradiction.

If the characteristic is $0$, it may happen or not: the identity doesn't exist for $H=\mathbb{Z}$, it obviously exists for $H=\mathbb{Q}$.

Answer 2

As is pointed out in the comments, the construction which replaces $xy$ with

$x \circ y = xy + yx \tag 1$

will leave the ring $\Bbb Z$ unitless under the multiplication operation "$\circ$"; first we note that since $H$ is commutative,

$x \circ y = xy + yx = xy + xy = 2xy; \tag 2$

then if

$\exists e \in \Bbb Z, \; x \circ e = e \circ x = x, \tag 3$

we would also have, via (2),

$x \circ e = 2xe; \tag 4$

but then from (3),

$2xe = x \Longrightarrow 2xe - x = 0 \Longrightarrow x(2e - 1) = 0; \tag 5$

since $\Bbb Z$ is an integral domain, if $x \ne 0$ we obtain

$2e - 1 = 0, \tag 6$

which has no solution in $\Bbb Z$.

Will this construction work in $\Bbb Q$? In this case, we still find that (2) binds; therefore so do (4)-(6), and taking

$e = \dfrac{1}{2}, \tag 7$

we see that

$x \circ e = x \circ \dfrac{1}{2} = 2 \dfrac{1}{2} x = x, \tag 8$

so apparently $e = 1/2$ is indeed the unit of $(H, +, \circ) = (\Bbb Q, +, \circ)$. As a check, we observe that

$e \circ e = 2 \left ( \dfrac{1}{2} \right )^2 = 2 \dfrac{1}{4} = \dfrac{1}{2} = e, \tag 9$

so $e$ is idempotent with respect to $\circ$, as should be true of any unit.

Nota Bene: Caveat Emptor!!! I have confined my remarks here to the specific questions surrounding $e$; I have granted our OP Hitman's (perhaps tacit) assertion that indeed $(H, +, \circ)$ does indeed satisfy all the ring axioms, such as associativity, distributivity, etc, without checking them myself. End of Note.