This question follows from this basic theorem:

Let $I$ be an ideal of ring $R$. The map $\phi:R\to R/I$ defined as $\phi(r)=r+I$ is a ring homomorphism of $R$ onto $R/I$ with kernel $I$.

I've tried to prove the last part of the theorem, but stumbled in understanding the basic thing on the identity element of $R/I$.

I suspect that, since the definition of multiplication of two cosets in $R/I$ defined as $$(r+I)(s+I)=rs+I$$

then the identity element should be $1+I$?

But, when trying to prove that last part of the theorem I got:

$\ker{\phi}=\{x\in R:\phi(x)=1+I\}$, can I claim from here that $\ker{\phi}=I$?

Thank you.

  • $\begingroup$ The point of the quotient map is that the kernel is exactly the quotiented ideal. So the identity is $1+I$, and you can claim and easily prove that $I$ is the kernel. Attempt it here so that we can help you. $\endgroup$ – астон вілла олоф мэллбэрг Aug 7 '16 at 1:30
  • $\begingroup$ It might help to think of the elements of a quotient ring not being cosets, but rather being the same elements as the original ring, subject to a system of relations. Namely, if x is in I, then x = 0 in the quotient ring. For instance, take $\mathbb{R}[x]/<x^2 + 1>$. This is just the polynomials, but with $x^2 + 1 = 0$. Put another way, $x^2 = -1$. So it is the complex numbers in disguise. $\endgroup$ – Tac-Tics Aug 7 '16 at 1:34
  • $\begingroup$ Also, you can use \ker in LaTeX to display $\ker \phi$ correctly. $\endgroup$ – Tac-Tics Aug 7 '16 at 1:35
  • 1
    $\begingroup$ Rings have both multiplication and addition, and each operation has an identity element. The coset $1+I$ is the multiplicative identity of $R/I$, but the kernel of a ring homomorphism is defined to be the stuff mapping to the additive identity. $\endgroup$ – Julian Rosen Aug 7 '16 at 1:35

There are two identity elements at play: the additive identity $0_{R/I} = 0+ I = I$ and the multiplicative identity $1_{R/I} = 1+I$. The kernel of $\phi$ is the set of elements of $R$ that map to the additive identity. Therefore it is precisely $I$.


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