Isomorphisim of quotient ring: $\mathbb{Z}[x, y]/(y+1) \cong \mathbb{Z}[x]$

Question

How can I prove that ring $\mathbb{Z}[x, y]/(y+1)$ is isomorphic to $\mathbb{Z}[x]$?

I think these are isomorphic as there is a epimorphisim $T \colon \mathbb{Z}[x, y] \to \mathbb{Z}[x]$ defined by $T\big(p (x,y)\big) = p (x,-1)$ with kernel $(y+1)$. So by fundamental theorem both are isomorphic. Am I right?

Answer 1

You're right.

Perhaps it's clearer if you argue as follows:

If $D$ is a domain and $a \in D$, then $D[y] \to D$ given by $f \mapsto f(a$) is an epimorphism with kernel $y-a$ and so induces an isomorphism $D[y]/(y-a) \cong D$.

Then apply this to $D=\mathbb Z[x]$.

Answer 2

You're correct. However, you should justify why the kernel is $(y+1)$.

If $p(x,y)\in\mathbb{Z}[x,y]=\mathbb{Z}[x][y]$, then we can write $$ p(x,y)=(y+1)q(x,y)+r $$ because division with remainder by a monic polynomial (with coefficients in $\mathbb{Z}[x]$) is possible. The remainder $r$ is constant and, if $p\in\ker T$, $r=0$.

Note it is false that $\mathbb{Z}[x,y]/(2y+1)$ is isomorphic to $\mathbb{Z}[x]$.