Kernel of ring homomorphisms and subring test

Question

Let $R$ be a ring and we adjoin an element "$a$" to the ring $R$ with some relation $f(a) = 0$. The resulting ring is $R[x]/\langle f\rangle= R'$ (say). Now latest consider the inclusion map $i$ from $R$ to $R[x]$ and then consider the canonical projection $\pi$ from $R[x]$ to $R'$. Clearly both the maps $i$, $\pi$ are homomorphisms, so their composition $\pi\circ i = g$ (say) is also a homomorphism.

Then I am very interested to see what is $\ker(g)$ but I couldn't find it. Help me.

Also it is written that of $g$ is not injective, then we can't identify $R$ with a subring of $R'$.

I couldn't understand why?

Any help would be appreciated.

Answer 1

The kernel of $g$ is $\{r\in\mathcal R\mid \pi(r)=0\}=\{r\in\mathcal R\mid r\in(f(x))\}$.
If, and only if, this kernel is zero, we then have an embedding of $\mathcal R$ into $\mathcal R'$ (in which case we can identify it with a subring of $\mathcal R'$).

Answer 2

The kernel is of course the intersection of the ideal $(f(x))$ of $R[x]$ with (the image of) $R$.

When $R$ is an integral domain, if $f(x)$ has degree at least 1 then this intersection consists only of $0$; indeed, the elements of $(f(x))$ are multiples of $f(x)$, and because $R$ is an integral domain, every nonzero multiple of $f(x)$ has degree greater than or equal to $1$. Thus, $R\cap(f(x))=\{0\}$. So the kernel of the composite map $R\to R[x]\to R[x]/(f(x))$ is trivial in that case.

But if $R$ is not an integral domain, then the ideal $(f(x))$ may contain nonzero constants.

For example, take $R=\mathbb{Z}/4\mathbb{Z}$, and suppose you try to adjoin a multiplicative inverse to $2$, which is nilpotent in this ring. That would mean "adjoining" an $a$ that satisfies $f(x) = 2x-1$.

Now, $2(2x-1)$ lies in the ideal $(2x-1)$ in $R[x]$. But $2(2x-1) = 2$ (since $(2)(2)=0$ and $-2=2$ in $R$). So $2$ lies in the kernel of the map.

So $\ker(g) = R\cap(f(x))$ in $R[x]$; under some circumstances you can prove that this is trivial: among them, if $R$ is an integral domain, or more generally if $\deg(f)\gt 0$ and the leading coefficient of $f$ is not a zero divisor (but this is not the full set of circumstances under which you could show that). But for more general rings, you can certainly have nontrivial kernel.