Krdytkyu

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 $ein H$ such that $xcirc e=x$

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

Please guide me. Thank you.

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

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

Distributivity is obvious.

Note. If $n$ is an integer and $xin H$, $nx$ 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 $ecirc x=2ex=mathbf{1}x=x$.

Thus the ring $(H,+circ)$ is unital if and only if $2mathbf{1}=mathbf{1}+mathbf{1}$ 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 $kmathbf{1}=(kmathbf{1})circ e=2ke=0$ 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 $H=mathbb{Q}$.

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.