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The problem:
Let $(\mathbb{Z}[x,y,z],+,\cdot)$ be the ring of polynomials with coefficients in $\mathbb{Z}$ in the variables $x$, $y$ and $z$ and the obvious operations $+$ and $\cdot$. Let $(x-y, x^3-z)$ be the ideal generated by $x-y$ and $x^3-z$. Find the ring to which $\mathbb{Z}[x,y,z]/(x-y,x^3-z)$ is isomorphic.

A similar question has been asked before: Which ring is $R[X,Y,Z,T]/(X-Y^2,T-Y^4,T^3-Z)$ isomorphic to?. Using the same approach, we set $x=y$ and $x^3=z$ and substitute for $y$ and $z$. Then we find $\mathbb{Z}[x,x,x^3]$ which is just $\mathbb{Z}[x]$. I'm trying to find out why this approach would work, i.e., formalise it in a proper proof. My idea is to use the first isomorphism theorem, but I get stuck near the end. I would very much appreciate if someone could help fill in this gap, check my proof for mistakes or provide additional (intuitive) information as to why the above method works.

My proof:
Consider the map $\varphi : \mathbb{Z}[x,y,z] \to \mathbb{Z}[x] : p(x,y,z) \mapsto p(x,x,x^3)$.

This is a homomorphism, since it obeys the relations $\varphi(p) + \varphi(q) = \varphi(p+q)$ and $\varphi(p) \cdot \varphi(q) = \varphi(p \cdot q)$ for all $p,q \in \mathbb{Z}[x,y,z]$ (verifying these is quite trivial).

Now take any $q \in \mathbb{Z}[x]$ and define a $p \in \mathbb{Z}[x,y,z]$ such that $p(x,y,z) = q(x)$. Since $\varphi(p(x,y,z)) = p(x,x,x^3) = q(x)$, $\varphi$ is surjective and it follows that $\mathbb{Z}[x] \subseteq \varphi(\mathbb{Z}[x,y,z])$. It is obvious that $\varphi(\mathbb{Z}[x,y,z]) \subseteq \mathbb{Z}[x]$ and, therefore, we must have $\varphi(\mathbb{Z}[x,y,z]) = \mathbb{Z}[x]$.

Lastly, we must show that $\ker(\varphi) = (x-y, x^3-z)$. For brevity's sake I will denote $I := (x-y,x^3-z).$
Let $p \in I$. Then there must exist $q_1, q_2 \in \mathbb{Z}[x,y,z]$ such that $p = q_1(x-y) + q_2(x^3-z)$. It follows that: \begin{align*}\varphi(p(x,y,z)) &= \varphi(q_1(x,y,z)(x-y) + q_2(x,y,z)(x^3-z))\\ &= q_1(x,x,x^3)(x-x) + q_2(x,x,x^3)(x^3-x^3) = 0.\end{align*} Thus $p \in \ker(\varphi)$ and $I \subseteq \ker(\varphi)$, since $p$ was chosen arbitrarily in $I$.
Now, take a $q \in \ker(\varphi)$. Then $\varphi(q(x,y,z)) = q(x,x,x^3) = 0$. Somehow we must get that $q \in I$ so that we can conclude $\ker(\varphi) \subseteq I$ and thus $I = \ker(\varphi)$, but I seem to get stuck on every path. Proof by contradiction or by contraposition also seems fruitless. What am I missing?

With all of the above, we may apply the first isomorphism theorem to conclude that: $$\mathbb{Z}[x,y,z]/I = \mathbb{Z}[x,y,z]/\ker(\varphi) \cong \varphi(\mathbb{Z}[x,y,z]) = \mathbb{Z}[x]$$

user26857
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beertje00
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2 Answers2

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Hint: Note that $q$ is congruent to $q(x,x,x^3)$ modulo $I$.

Thorgott
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    Nice hint! Maybe it also helps the OP to write $$q(x,y,z)= q(x,x-(x-y), x^3-(x^3-z)).$$ – Severin Schraven Dec 25 '20 at 02:15
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    Your suggestion, @SeverinSchraven, was the shortcut I was hoping to find. Never did, though! – Lubin Dec 25 '20 at 17:38
  • @Lubin I wouldn't know how to do it otherwise. Could you give me a hint how you would do it? – Severin Schraven Dec 25 '20 at 22:59
  • Sorry, @SeverinSchraven , I’m not going to embarrass myself by tell you the cracked-up strategy I was working on. – Lubin Dec 26 '20 at 03:15
  • @Lubin i think we can learn something from every approach (I found also the deleted answer quite interesting). In any case, I wish you happy holidays :) – Severin Schraven Dec 26 '20 at 10:43
  • Continuing from that hint is relatively obvious. We must have $q(x,y,z) + I = q(x,x,x^3) + I = I$, hence $q \in I$. But how do you get to that point? I suppose @SeverinSchraven 's hint was supposed to help there. I can intuitively understand that the hint is true, though. – beertje00 Dec 26 '20 at 10:54
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    You can write $q(x,y,z)=\sum_{i,j,k=1}^n a_{ijk} x^i y^j z^k$ and then use my hint. Note that for $t>0$ we have, for example, $y^t = (x-(x-y))^t = x^t + g$ with $g\in (y-x)$. – Severin Schraven Dec 26 '20 at 11:03
  • I just figured it out as well, using the binomial theorem. Thanks! – beertje00 Dec 26 '20 at 11:19
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Write $\mathbb{Z}[X]= R$ ring ( comm, $1$). If $a_1$, $\ldots$, $a_n$ are in $R$ then the evaluation map at $(a_1, \ldots, a_n)$, $$f(X_1, \ldots, X_n) \mapsto f(a_1, \ldots, a_n)$$ from $R[X_1, \ldots, X_n]$ to $R$ is surjective. Clearly the kernel contains the ideal $I= \langle X_1-a_1, \ldots, X_n - a_n\rangle$. In fact it equals $I$. Indeed, for every $f\in R[X_1, \ldots, X_n]$ we have $f(X_1, \ldots, X_n) \equiv f(a_1, \ldots, a_n) \mod I$, since $X_i \equiv a_i \mod I$ (@Thorgott:'s observation)- so $f(a_1, \ldots, a_n) = 0$ iff $f(X_1, \ldots, X_n) \in \langle X_1-a_1, \ldots, X_n - a_n\rangle$, a multidimensional Bézout's theorem. We conclude that the evaluation map induces the isomorphism $$R[X_1, \ldots, X_n]/\langle X_1-a_1, \ldots, X_n - a_n\rangle \stackrel{\sim}{\rightarrow} R$$

orangeskid
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