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I am asked to show that if f is a ring homomorphism from R to R' then kernel of f is an ideal of R.

According to definition of ideal : A non empty subset of R is an ideal for any two elements of ideal their substraction must be in that ideal and the product of any element of R and an element of ideal must be in ideal, I am not able to prove second condition. Please help

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    $\begingroup$ Let $f \colon R \to R'$ be a ring homomorphism, and let $x \in \ker(f)$. Then for any $r \in R$, $f(rx) = f(r)f(x) = ...$ $\endgroup$ – Alex Wertheim Apr 20 '16 at 7:35
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Let ϕ:(R1,+1,∘1)→(R2,+2,∘2) be a ring homomorphism.

Then the kernel of ϕ is an ideal of R1.

Proof By Kernel of Ring Homomorphism is Subring, ker(ϕ) is a subring of R1.

Let s∈ker(ϕ), so ϕ(s)=0 in R2.

Suppose x∈R1. Then:

ϕ(x∘1s) = ϕ(x)∘2ϕ(s) Definition of Morphism Property
= ϕ(x)∘20R2 as s∈ker(ϕ)
= 0R2 Properties of 0R2

and similarly for ϕ(s∘1x).

The result follows.

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