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When checking if two rings are isomorphic, we check if mapping is homomorphism and then we check if it is bijective (injective and surjective). In some tasks when checking if isomorphisms, we checked if it is homomorphism, surjective and instead of injective we questioned the kernel (Ker) if it is ideal.

Do we need to question all of this or there are some cases? I think that if we have ring factor, we check homomorphism, surjective and Ker, but if we have just ring on the left side, we check if mapping is homomorphism, surjective and injective.

Can anyone explain? Thank you so much!

Bernard
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Haus
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  • Kernel is always an ideal. For injectivity, the thing you check for the kernel is whether or not it is trivial. – Anurag A Dec 06 '18 at 22:37
  • So when checking if isomorphism, we look if mapping is homomorphism, surjective, injective and we check kernel. All of this? – Haus Dec 06 '18 at 22:39
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    for isomorphism you need 3 things: homomorphic property, surjectivity and injectivity. Usually kernel helps with injectivity. – Anurag A Dec 06 '18 at 22:40
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    A homomorphism is injective if and only if its kernel is trivial (proof: one direction is obvious. For the other, if $f(a) = f(b)$, then $f(a - b) = f(0) = 0$, so $a - b$ lies in the kernel). – user3482749 Dec 06 '18 at 22:40
  • I got it now. Thank you! :) – Haus Dec 06 '18 at 22:40

2 Answers2

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A mapping is an isomorphism if and only if it is a homomorphism and a bijection. In particular, a homomorphism is injective if and only if its kernel is trivial, so these are exactly equivalent.

user3482749
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To check a map $f$ is isomorphism you need to do

(1) $f$ is a homomorphism.

(2) $f$ is surjective.

(3) $f$ is injective.

Equivalently, you can check

(1) $f$ is a homomorphism.

(2) $f$ is surjective.

(3') $\ker f=0$.

That is because (3) and (3') are equivalent once you prove (1).

Eclipse Sun
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