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Show that $k[x,y]/\langle y+1\rangle \cong k[x]$.

Define $\varphi : k[x,y] \to k[x]$ such that $x \longmapsto 0$ and $y \longmapsto -1$, then $\ker \varphi=\{p(x,y) \in k[x,y] \mid \varphi(p)=p(0,-1)=0\}$.

I'm now trying to show that $\ker \varphi=\langle y+1 \rangle$ in order to invoke the first isomorphism theorem to conclude that $k[x,y]/\langle y + 1 \rangle = k[x,y]/\ker \varphi \cong k[x].$

If $g \in \langle y+1 \rangle$, then $g(x,y)=q(x,y)(y+1)$ for $q \in k[x,y]$ so $$\varphi(g)=g(0,-1)=q(0,-1)(-1+1)= 0 \implies g \in \ker \varphi.$$

How should I approach the direction $\ker \varphi \subset \langle y +1 \rangle$? Usually with the one variable approach I'm able to use the division algorithm and conlcude from the remainder's degree the result, but it doesn't apply to the multivariate case?

Jaipur
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    $\varphi$ is not surjective. You should reevaluate the image of $x$. – Oiler Apr 06 '22 at 14:00
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    As Oiler notes, you need to map $x$ somewhere else. Once that is fixed, this reduces to the univariate case if you use the natural isomorphism $k[x][y]\cong k[x,y]$ (more generally, the same argument shows $R[y]/(y+1)\cong R$ for any ring $R$ and this is the case $R=k[x]$). Alternatively, take $f\in\ker\varphi$ and write $f=f(x,y)=f(x,(y+1)-1)$ and then expand the terms $((y+1)-1)^k$ out using the binomial formula. You will get that $f$ differs from the constant $f(-1)$ by an element of $(y+1)$. – Thorgott Apr 06 '22 at 14:01
  • @Oiler Aha! If I send $x \longmapsto x$ I should get in the right direction? Is the map still not yet surjective though? – Jaipur Apr 06 '22 at 14:06
  • @Thorgott May you elaborate on how to use $k[x][y]\cong k[x,y]$ here? – Jaipur Apr 06 '22 at 18:08
  • If I take $h \in \ker \varphi$, then I would like to put $h(x,y)=q(x,y)(y+1) + r(x,y)$ and conclude that since the degree of $y+1$ is $1$ in $y$ the degree of $r(x,y)$ must only depend on $x$ so $$h(x,y)=q(x,y)(y+1) + r(x)$$ – Jaipur Apr 06 '22 at 18:14
  • Now since $h \in \ker \varphi$ we have that $$0 = \varphi(h) = r(x)$$ so $r$ is identically zero implying that $h(x,y) = q(x,y)(y+1) \in \langle y+1 \rangle$. – Jaipur Apr 06 '22 at 18:25
  • For the details to the outline in Thorgott's comment, see this post: https://math.stackexchange.com/questions/2593642/fr-0-for-r-in-r-implies-that-x-r-divides-fx – Viktor Vaughn Apr 06 '22 at 20:32

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