$(x_1, ..., x_k)$ is prime in $R[x_1, ..., x_n]$ if $R$ is an integral domain Let $R$ be an integral domain. I need to prove that $\forall k = 1, ..., n \ \ \ (x_1, ..., x_k)$ is prime in $R[x_1, ..., x_n]$.
I managed to do it for $k = 1$.
Let $f, g \in R[x_1, ..., x_n]$. Then if $f \notin (x_1)$, then $\forall y \in R[x_1, ..., x_n] \ \  f \neq x_1y$. Same for $g:   \forall t \in R[x_1, ..., x_n] \ \  g \neq x_1t$. That means that both $f$ and $g$ and nonzero and they contain a summand $rx^{i_1}_1...x^{i_n}_n$ where $i_1 = 0$( and obviously, $r \neq 0$). So their product will also be nonzero and contain this kind of summand with a nonzero coefficient(since $R$ is an integral domain ), and hence it cannot be written as a product of some polynomial from $R[x_1, ...,x_n]$ and $x_1$, that is, $fg \notin (x_1)$.
I'm not sure if induction would actually help here. Assume
$f, g \notin (x_1, ..., x_k)$. So, $\forall y_1, ..., y_k \in R[x_1, ..., x_n] \ \ \ f \neq x_1y_1 + ... + x_ky_k$. But what does this actually mean?
 A: Thanks to the hints of Plankton and user26857 from the comments, here is what I'
ve come up with.
We construct a homomorphism of rings $\phi: R[x_1, ..., x_n] \to R[x_{k+1}, ..., x_n]$.
Define $\phi(r) = r$ if $r \in R, \phi(x_i) = x_i$ if $k+1 \leq i \leq n$, and $\phi(x_i) = 0$ if $1 \leq i \leq k$. Since we assume that $\phi$ is a homomorphism, this extends to any polynomials. As $\phi(R[1, ..., 1, x_{k+1}, ..., x_n]) = R[x_{k+1}, ..., x_n]$ , this homomorphism is surjective. Have to say I find it hard to check "by hand" that it is indeed a homomorphism, though it seems intuitively obvious.
It's obvious that $(x_1, ..., x_k) \subseteq \ker \phi$. Now I will prove that the converse also holds. If $f \in R[x_1, ..., x_n]$ has at least one nonzero summand "free" of variables $x_1, ..., x_k$, that is, of the form $r x^{a_1}_{k+1}...x^{a_{n-k}}_n, r \neq 0$, then $\phi(f) \neq 0$. So if $f \in \ker \phi$, then each nonzero summand  in $f$ must be of the form $rx^{i_1}_1...x^{i_k}_kx^{i_{k+1}}_{k+1}...x^{i_n}_n$, where there is at least one $j = 1, ..., k$ such that $i_j \neq 0$. Then each summand in $f$ will be of the form $gx_l, l = 1, ... ,k$ for some polynomial $g$ with one summand. Combining $yx_r's$ and $tx_d's$ where $d = r$, we get $f = b_1x_1 + ... + b_kx_k$, so $f \in (x_1, ..., x_k)$, and $\ker \phi \subseteq (x_1, ..., x_k)$.
Then $\ker \phi = (x_1, ..., x_k)$, and applying First Isomorphism Theorem, we get:
$$\frac{R[x_1, ..., x_n]}{(x_1, ..., x_k)} \cong R[x_{k+1}, ..., x_n].$$ And the latter is indeed an integral domain, since it can be realised as a subring of $R[x_1, ..., x_n]$.
