Exercise 9.1.14 in Dummit and Foote's Abstract Algebra

Question

Exercise 9.1.14 in Dummit and Foote's Abstract Algebra

Let $R$ be an integral domain and let $i, j$ be relatively prime integers. Prove that the ideal $\langle x^i-y^j\rangle$ is a prime ideal in $R[x, y]$.

[Hint. Consider the ring homomorphism $\varphi$ from $R[x, y]$ to $R[t]$ defined by mapping $x$ to $t^j$ and mapping $y$ to $t^i$. Show that an element of $R[x, y]$ differs from an element in $\langle x^i-y^j\rangle$ by a polynomial $f(x)$ of degree at most $j-1$ in $y$ and observe that the exponents of $\varphi(x^r y^s)$ are distinct for $0\leq s<j$.]

Question. Is there another elegant proof?

My Proof. Since $\varphi(x^i-y^j)=0$, we have $\langle x^i-y^j\rangle\subseteq \ker{\varphi}$. We show that $\langle x^i-y^j\rangle\supseteq \ker{\varphi}$.

Consider $\varphi(f(x, y))=\sum_{k, l}c_{kl}t^{jk+il}$. \begin{eqnarray} &\text{If}& jk_1+il_1=jk_2+il_2=\cdots=jk_n+il_n\nonumber\\ &\Rightarrow& \text{for }1\leq m\leq n, ~j(k_1-k_m)=i(l_m-l_1)...(1)\\ &\stackrel{\gcd{(i, j)}=1}{\Rightarrow}& i\mid k_1-k_m, ~j\mid l_m-l_1\nonumber\\ &\Rightarrow& k_1-k_m=is_m, ~l_m-l_1=jt_m...(2)\\ &\stackrel{\text{substitute (2) to (1)}}{\Rightarrow}& s_m=t_m...(3)\\ &\stackrel{\text{substitute (3) to (2)}}{\Rightarrow}& k_m=k_1-is_m, ~l_m=js_m+l_1...(4)\\ &\Rightarrow& c_{k_1 l_1}x^{k_1}y^{l_1}+c_{k_2 l_2}x^{k_2}y^{l_2}+\cdots+c_{k_n l_n}x^{k_n}y^{l_n}\nonumber\\ &\stackrel{(4)}{=}& c_{k_1 l_1}x^{k_1}y^{l_1}+c_{k_2 l_2}x^{k_1-is_2}y^{js_2+l_1}+\cdots+c_{k_n l_n}x^{k_1-is_n}y^{js_n+l_1}....(5) \end{eqnarray}

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\begin{eqnarray} &\text{If}& f(x, y)\in \ker{\varphi}\nonumber\\ &\Rightarrow& \varphi(f(x, y))=\sum_{k, l}c_{kl}t^{jk+il}=0\nonumber\\ &\Rightarrow& c_{k_1 l_1}t^{jk_1+il_1}+c_{k_2 l_2}t^{jk_2+il_2}+\cdots+c_{k_n l_n}t^{jk_n+il_n}=0\nonumber\\ &\stackrel{(4)}{\Rightarrow}& c_{k_1 l_1}t^{jk_1+il_1}+c_{k_2 l_2}t^{jk_1-{ijs_2}+{ijs_2}+il_1}+\cdots+c_{k_n l_n}t^{jk_1-{ijs_n}+{ijs_n}+il_1}=0\nonumber\\ &\Rightarrow& (c_{k_1 l_1}+c_{k_2 l_2}+\cdots+c_{k_n l_n})t^{jk_1+il_1}=0\nonumber\\ &\Rightarrow& c_{k_1 l_1}+c_{k_2 l_2}+\cdots+c_{k_n l_n}=0...(6) \end{eqnarray}

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\begin{eqnarray*} &\text{consider}&(c_{k_1 l_1}x^{k_1}y^{l_1}+c_{k_2 l_2}x^{k_2}y^{l_2}+\cdots+c_{k_n l_n}x^{k_n}y^{l_n}){+\langle x^i-y^j\rangle}\\ &\stackrel{(5)}{=}& (c_{k_1 l_1}x^{k_1}y^{l_1}+c_{k_2 l_2}x^{k_1-is_2}y^{js_2+l_1}+\cdots+c_{k_n l_n}x^{k_1-is_n}y^{js_n+l_1})+\langle x^i-y^j\rangle\\ &\stackrel{x^i+\langle x^i-y^j\rangle=y^j+\langle x^i-y^j\rangle}{=}& (c_{k_1 l_1}x^{k_1}y^{l_1}+c_{k_2 l_2}x^{k_1-{is_2}}x^{{is_2}} y^{l_1}+\cdots+c_{k_n l_n}x^{k_1-{is_n}}x^{{is_n}}y^{l_1})+\langle x^i-y^j\rangle\\ &=& (c_{k_1 l_1}+c_{k_2 l_2}+\cdots+c_{k_n l_n})x^{k_1}y^{l_1}+\langle x^i-y^j\rangle\\ &\stackrel{(6)}{=}& 0+\langle x^i-y^j\rangle\\ &\Rightarrow& x^i-y^j\mid c_{k_1 l_1}x^{k_1}y^{l_1}+c_{k_2 l_2}x^{k_2}y^{l_2}+\cdots+c_{k_n l_n}x^{k_n}y^{l_n}\\ &\Rightarrow& x^i-y^j\mid f(x, y)\\ &\Rightarrow& f(x, y)\in \langle x^i-y^j\rangle\\ &\Rightarrow& \ker{\varphi}\subseteq \langle x^i-y^j\rangle. \end{eqnarray*}

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Therefore, $$R[x, y]/\langle x^i-y^j\rangle =R[x, y]/\ker{\varphi} \cong \text{Im }{\varphi}\leq R[t].$$ Since $R$ is an integral domain, so is $R[t]$. The subring $R[x, y]/\langle x^i-y^j\rangle\cong \text{Im }{\varphi}$ of $R[t]$ contains the unity of $R[t]$, hence, $R[x, y]/\langle x^i-y^j\rangle$ is also an integral domain and $\langle x^i-y^j\rangle$ is a prime ideal in $R[x, y]$ by Proposition 7.4.13.

Answer 1

Following the hint provided in Exercise 9.1.14 in Dummit and Foote's Abstract Algebra and cited above, I came up with the following proof. Not sure it is much more elegant than the one provided above, though.

Obviously, the ideal $(x^i-y^j)⊆\ker ⁡φ$.
Suppose there is a polynomial $p∈R[x,y]$ such that $p∈\ker⁡ φ$, but $p∉(x^i-y^j)$. Then $φ(p)=0$ and for all polynomials $k∈(x^i-y^j)$ [so that $φ(k)=0$] we have $φ(p)-φ(k)=0 ⇔ φ(p-k)=0$.
The ring $R[x,y]$ can be viewed as $R[x,y]=R[x][y]$ (i.e., as a ring of polynomials over the indeterminate $y$ with coefficients taken from the ring $R[x]$).
"->" Then $p=p(x,y)$ can be written as $p=v(x)_ny^n+v(x)_{n-1}y^{n-1}+⋯+v(x)_{j+1}y^{j+1}+v(x)_jy^j+v(x)_{j-1}y^{j-1}+⋯+v(x)_2y^2+v(x)_1y+v(x)_0$ where $v(x)_l∈R[x]$ is a polynomial-coefficient before $y^l$ with $l∈[0,n]$.
Since $(x^i-y^j )$ is an ideal in $R[x,y]$, it contains all the polynomials: $-(x^i-y^j)v(x)_j=-v(x)_j x^i+v(x)_j y^j$,
$-(x^i-y^j)v(x)_{j+1}y=-v(x)_{j+1}x^iy+v(x)_{j+1}y^{j+1}$,
$-(x^i-y^j)v(x)_{j+2}y^2=-v(x)_{j+2}x^iy^2+v(x)_{j+2}y^{j+2}$,
... ,
$-(x^i-y^j)v(x)_ny^{n-j}=-v(x)_nx^iy^{n-j}+v(x)_ny^n$.
Hence the ideal $(x^i-y^j)$ also contains their sum $S_1=-(v(x)_j x^i+v(x)_{j+1}x^i y+⋯+v(x)_nx^iy^{n-j} )+(v(x)_jy^j+v(x)_{j+1}y^{j+1}+⋯+v(x)_n y^n)$
Now, the difference $$p-S_1=(v(x)_{j-1} y^{j-1}+⋯+v(x)_2 y^2+v(x)_1 y+v(x)_0)+(v(x)_j x^i+v(x)_{j+1} x^i y+⋯+v(x)_n x^i y^{n-j})$$ If $n-j≤j-1$, then we stop here. If not, then $p-S_1$ is a polynomial of degree $n-j$ in $y$. Since $p-S_1∈\ker ⁡φ$, we can consider $p-S_1$ in place of $p$ starting at "->", so that we get a polynomial $S_2∈(x^i-y^j)$ and hence the polynomial $p-S_1-S_2$. If $n-2j>j-1$, then we consider $p-S_1-S_2∈\ker ⁡φ$ in place of $p$ starting at "->". And so on. Since $n$ is a finite integer, after $m$ iterations we’ll necessarily get a polynomial $p-S_1-S_2-…-S_m$ of degree $n-mj≤j-1$ in $y$. Since $S_1,S_2,…,S_m∈(x^i-y^j)$, it follows that their sum $k=S_1+S_2+⋯+S_m∈(x^i-y^j)$, so that $p-S_1-S_2-…-S_m=p-(S_1+S_2+⋯+S_m)=p-k$ is a polynomial of degree at most $j-1$ in $y$.
Thus, our $p∈R[x,y]$ differs from an element $k∈(x^i-y^j)$ by a polynomial $f(x,y)=p-k$ of degree at most $j-1$ in $y$ [as was noted in the hint.]

So we have $φ(p-k)=0$, where $p-k∈R[x,y]$ is a polynomial of degree at most $j-1$. Then the image of indeterminate part of each of the monomial terms of $p-k$ is $φ(x^ry^s)$ with $0≤s<j$. Suppose $φ(x^{r_1} y^{s_1})=φ(x^{r_2} y^{s_2} ) ⇔ t^{jr_1+is_1}=t^{jr_2+is_2} ⇔ jr_1+is_1=jr_2+is_2 ⇔ j(r_1-r_2)=i(s_2-s_1 )$, therefore, $j∣i(s_2-s_1)$. Since $i,j$ are relatively prime (by the statement), the latter implies $j∣s_2-s_1$. But, since $0≤s_1,s_2<j$, it follows that $|s_2-s_1|<j$, therefore, $j∤s_2-s_1$, a contradiction. Hence, all the images $φ(x^r y^s)=t^{jr+is}$ have distinct exponents $jr+is$ of $t$, therefore, $φ(p-k)≠0$. We’ve reached a contradiction to our initial assumption about the existence of the polynomial $p$. Thus, $\ker⁡ φ=(x^i-y^j)$.
Now, by the 1st Isomorphism Theorem, $R[x,y]/\ker ⁡φ≅φ(R[x,y]) ⇔ R[x,y]/(x^i-y^j)≅φ(R[x,y])$, where $φ(R[x,y])$ is an image of $R[x,y]$ in $R[t]$ under the ring homomorphism $φ$. Since $R$ is an integral domain (by the statement), $R[t]$ is also an integral domain, hence its subring $φ(R[x,y])$ is also an integral domain. Therefore, $(x^i-y^j)$ is a prime ideal.