Primeness of an ideal

Question

Let $k$ be a field and $x_1, x_2, ...,x_n$ are independent over $k$. Is there a simple argument to prove that the ideal $\langle x_1-x_2, x_2-x_3,\dots,x_{n-1}-x_n\rangle$ is a prime ideal of $k[x_1, x_2, \dots, x_n]$? (Note that it can be checked easily by Macaulay2)

Is the result true when $n$ is infinite?

Answer 1

$$ k[x_1,\dots,x_n]/(x_1 - x_2, \dots, x_{n-1} - x_n) \cong k[x_1] $$

and since $k[x_1]$ is an integral domain, our ideal is indeed prime. The same is true for $n$ infinite.

Answer 2

If $R$ is any commutative ring, a ring homomorphism $\varphi\colon k[x_1,x_2,\dotsc]\to R$ (any number of indeterminates, even infinite) is determined by the action on $k$ (so the restriction should be a ring homomorphism $k\to R$) and by the action on each indeterminate.

Consider the homomorphism $\varphi\colon k[x_1,x_2,\dotsc]\to k[y]$ defined to be the identity on $k$ and such that $\varphi(x_i)=y$.

Clearly, $x_i-x_{i-1}\in\ker\varphi$, for each $i>1$, and $\varphi$ is surjective. If we show that $$ \ker\varphi=\langle x_{i}-x_{i-1}:1<i\le n\rangle $$ (the $\le n$ condition to be omitted in the infinite case), the ideal is prime because $k[y]$ is a domain.

The inclusion $\langle x_{i}-x_{i-1}:i>1\rangle\subseteq\ker\varphi$ has already been proved. Suppose $f\in\ker\varphi$. Since $f$ depends only on finitely many indeterminates, we can assume $f\in k[x_1,\dots,x_m]$, where $m$ is the largest index that actually appears in $f$ (if $f$ is constant, then $f=0$ and there's nothing to prove).

If $m=1$, then $\varphi(f)=0$ is a contradiction. Thus $m>1$.

We can write $f=(x_m-x_{m-1})g+r$, where $r$ only depends possibly on $x_1,\dots,x_{m-1}$. Applying $\varphi$ gives $$ 0=\varphi(f)=(y-y)\varphi(g)+\varphi(r) $$ and so induction says $r\in\langle x_{i}-x_{i-1}:i>1\rangle$, so also $f\in\langle x_{i}-x_{i-1}:i>1\rangle$.

Answer 3

As the ideal is also $\;(x_2-x_1,x_3-x_1,\dots, x_n-x_1)$, the quotient is clearly isomorphic to $\;k[x_1]$, hence the ideal is prime. The isomorphism is induced by the homorphism \begin{align*} k[x_1,x_2,\dots,x_n]&\longrightarrow k[x_1],\\ p(x_1,x_2,\dots,x_n]&\longmapsto p(x_1,0,\dots,0). \end{align*}

If ‘$n$ is infinite’, i.e. if we consider the ring $\;k[x_n]_{n\in\mathbf N}$, the same argument works, since each polynomial involves only a finite number of indeterminates.