Let a belong to a ring R. Let S={x in R | ax=0} . Show that S is a subring of R.

Question

Let a belong to a ring R. Let S={x in R | ax=0} . Show that S is a subring of R.

Question

Answers (1)

Relevant Questions

asked 2020-12-21

Let R be a commutative ring. Show that R[x] has a subring isomorphic to R.

See answers (1)

asked 2020-11-22

Suppose that R and S are commutative rings with unites, Let PSJphiZSK be a ring homomorphism from R onto S and let A be an ideal of S.
a. If A is prime in S, show that \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}={\left\lbrace{x}\in{R}{\mid}\phi{\left({x}\right)}\in{A}\right\rbrace}\) is prime \(\displaystyle\in{R}\).
b. If A is maximal in S, show that \(\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}\) is maximal \(\displaystyle\in{R}\).

See answers (1)

asked 2021-03-07

Let R be a commutative ring with unit element .if f(x) is a prime ideal of R[x] then show that R is an integral domain.

See answers (1)

asked 2020-11-08

Suppose that a and b belong to a commutative ring and ab is a zero-divisor. Show that either a or b is a zero-divisor.

See answers (1)

asked 2020-11-23

If R is a commutative ring, show that the characteristic of R[x] is the same as the characteristic of R.

See answers (1)

asked 2021-01-31

Suppose that R is a ring and that \(\displaystyle{a}^{{2}}={a}\) for all \(\displaystyle{a}\in{R}{Z}\). Show that R is commutative.

See answers (1)

asked 2020-12-14

Let R and S be commutative rings. Prove that (a, b) is a zero-divisor in \(\displaystyle{R}\oplus{S}\) if and only if a or b is a zero-divisor or exactly one of a or b is 0.

See answers (1)

asked 2021-02-11

Suppose that R is a commutative ring without zero-divisors. Show that all the nonzero elements of R have the same additive order.

See answers (1)

asked 2020-12-17

If R is a commutative ring with unity and A is a proper ideal of R, show that \(\displaystyle\frac{{R}}{{A}}\) is a commutative ring with unity.

See answers (1)

asked 2020-11-02

Let R be a commutative ring with identity and let I be a proper ideal of R. proe that \(\displaystyle\frac{{R}}{{I}}\) is a commutative ring with identity.

See answers (1)

Answer 1

To show a non-empty subset S of R is a subring if \(\displaystyle{a},{b}\in{S}\Rightarrow{a}-{b}\in{S}{\quad\text{and}\quad}{a}{b}\in{S}\)
Clearly S is non-empty as 0 belongs to Ring R so, ax = 0.
let \(\displaystyle{x},{y}\in{S}\Rightarrow\) ax=0, ay=0
is is needed to show that \(\displaystyle{x}-{y}\in{S}\)
consider, \(\displaystyle{a}{x}-{a}{y}={a}{\left({x}-{y}\right)}\in{S}\)
as \(\displaystyle{x}-{y}\in{R}\)
Now show \(\displaystyle{x}{y}\in{S}\)
cnsider, \(\displaystyle{\left({a}{x}\right)}{\left({a}{y}\right)}={0}\cdot{0}={0}\)
\(\displaystyle{a}^{{2}}{\left({x}{y}\right)}={b}{\left({x}{y}\right)}={0}{\left[\begin{array}{c} {a}^{=}{b}\\{x}{y}\in{R}\end{array}\right]}\)
thus, \(\displaystyle{x}{y}\in{S}\)
Therefore, S is a subring of ring R.

Didn’t find what you are looking for?

Let a belong to a ring R. Let S={x in R | ax=0} . Show that S is a subring of R.

Let a belong to a ring R. Let S={x in R | ax=0} . Show that S is a subring of R.

Answers (1)

Relevant Questions