# Cosets in rings?

Does cosets exist in rings?

i.e R = Ring, a in R

set {a*R}

or

set {a+R}

The above two sets looks very similar to cosets in groups but there are two operations in rings so potentially two different cosets both involving the same ring R and element a. If the above two sets are not cosets than what are they called?

matt grime
Homework Helper
The former is the right ideal generated by R, the latter is the right coset of a in R (and since addition is commutative, the left coset).

I is a right ideal if Ix is in x for all x in R. Clearly aR is a right ideal. Similarly Ra is a left ideal, and RaR is an ideal (that is a two sided ideal).

In this case aR is also a left ideal? Because x(aR) with x and a in R means xa is in R. (xa)R is a subset of R.

Similarly Ra is also a right ideal.

StatusX
Homework Helper
No, because (xa)R is not necessarily a subset of aR.

Good point. In fact x(Ra)=Ra, for x and a in R which is the ring.

Hurkyl
Staff Emeritus
Gold Member
Also, a + R is a rather boring coset, since a + R = R. a + I is more interesting, for ideals I.

matt grime
Homework Helper
Good point. In fact x(Ra)=Ra, for x and a in R which is the ring.

No. x(Ra) is a subset of Ra. It is most definitely not equal to it in general (x=0, for instance).

matt grime
Homework Helper
Also, a + R is a rather boring coset, since a + R = R. a + I is more interesting, for ideals I.

Doh. Can't believe I missed this. Elements don't have cosets, sugroups/ideals, have cosets

Ra gives R. xR gives R again. so x(Ra)=xR=R=Ra. In this way x(Ra)=Ra. I agree that it is quite boring but at least it's correct? You think not Matt? R is the ring and a and x are elements in R. All elements in R are closed under multiplication.

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StatusX
Homework Helper
Closure under multiplication means xR is a subset of R. To get the other inclusion, you need the existence of inverses. Thus x+R=R, or xR=R when x is a unit, but in general this isn't true. If the example x=0 isn't enough to convince yourself that xR is not always R, you need to go back to the definitions.

Closure under multiplication means xR is a subset of R. To get the other inclusion, you need the existence of inverses. Thus x+R=R, or xR=R when x is a unit, but in general this isn't true. If the example x=0 isn't enough to convince yourself that xR is not always R, you need to go back to the definitions.

I should have said x must be non zero.

If x is a unit then x cannot be a zero divisor. This latter point is the important thing because if x was a zero divisor than some non zero entities in R may not be produced from xR. If x is not a zero divisor and is not a unit than xR=R if R does not contain the multiplicative identity, 1?

x+R=R will work all the time with x in R wouldn't it? Without any limitations on x?

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matt grime