0
$\begingroup$

I'm reading "A First Course in Abstract Algebra, John B. Fraleigh, 7th edition", and I'm having trouble seeing the connection between the following two theorems:

14.9 Theorem Let $H$ be a normal subgroup of $G$. Then $\gamma: G \to G/H$ given by $\gamma(x) = xH$ is a homomorphism with kernel $H$.

26.16 Theorem (Analogue of Theorem 14.9) Let $N$ be an ideal of a ring $R$. Then $\gamma: R \to R/N$ given by $\gamma(x) = x + N$ is a ring homomorphism with kernel $N$.

Now, in the proof of Theorem 26.16 he prefaces by saying:

The additive part is done in Theorem 14.9

So my question is this, does $xH$ in Theorem 14.9 really denote $x + H$ where the operation is implicit, or are they different expressions all together?

$\endgroup$
2
$\begingroup$

A ring has a group structure for its addition. This means that viewing a ring $R$ with just its addition and ignoring the multiplication will leave just a group where $+$ is the group operation. So Yes,

$$xH\qquad\text{means}\qquad x+H$$

in the ring.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.