I have been told that if we have a ring $R$ and $a$ is an element of $R$ and if we adjoin an element $b$ with the relation $b=a$ then we will get a ring isomorphic to $R$. However, I have not seen a formal proof for this. Is there a textbook or nice proof of this anyone knows of? I can't seem to see exactly how this is true.
Asked
Active
Viewed 233 times
1 Answers
0
Formally, adjoining $b=a$ is forming a quotient: $R[x]/(x-a)$. Then $b$ is the class of $x$.
Now $p(x) \mapsto p(a)$ is a surjective homomorphism $R[x] \to R$ whose kernel is the ideal $(x-a)$ because $x-a$ is monic and we can divide $p(x)$ by $x-a$, obtaining $p(x)=(x-a)q(x)+p(a)$.
Therefore, $R[x]/(x-a) \cong R$.
lhf
- 216,483
-
Did you mean $R[x]/(x)$? – David Wheeler Apr 12 '16 at 01:52
-
No need to thank me, I take cash... – David Wheeler Apr 12 '16 at 02:04
-
How did you go from R[x]≅R[x−a] to R[x]/(x−a)≅R[x]/(x)≅R? – RobKard Apr 12 '16 at 13:17
-
hmm i'm not seeing that completely can you add an edit with some more explanation. I've been looking at it for a while and am confused by that last part – RobKard Apr 12 '16 at 13:21
-
Thanks one last question what does $b$ of class $x$ mean? – RobKard Apr 13 '16 at 01:50
-
@RobKard, the coset of $x \bmod (x-a)$. – lhf Apr 13 '16 at 01:54