# Ideal generated by $x_1 - a_1, \dots, x_k - a_k$ is prime in polynomial ring over integral domain

I would like to show the ideal $P_k = \langle x_1 - a_1, \dots, x_k - a_k \rangle$ of $R = F[x_1,\dots,x_n]$, where $F$ is a field, $k \leq n$ and $a_i \in F$, is a prime ideal.

In the case where all $a_i = 0$, I think the natural isomorphism $R/P_k \rightarrow F[x_{k+1},\dots,x_n]$ allows us to conclude $R/P_k$ is an integral domain completing the argument. But in the case of general $a_i \neq 0$, I can't see how to set up a similar isomorphism neatly.

Also, am I correct in thinking the result holds if we replace $F$ with any integral domain?

The result does indeed hold; the ideal $P_k$ is the kernel of the map $$\varepsilon_a:\ F[x_1,\ldots,x_n]\ \longrightarrow\ F[x_{k+1},\ldots,x_n]:\ x_i\ \longmapsto\ a_i.$$ Verify that for every polynomial $Q\in F[x_1,\ldots,x_n]$ we have $$\varepsilon_a(Q)=Q(a_1,\ldots,a_k,x_{k+1},\ldots,x_n),$$ from which it easily follows that $\varepsilon_a$ is a ring homomorphism with kernel $P_k$. Conclude that $P_k$ is prime because $F[x_{k+1},\ldots,x_n]$ is an integral domain (because $F$ is).
• Thanks! That looks like a much nicer way of setting it up. Although intuitively clear, how can you neatly show that if $Q(a_1,\dots,a_k,x_{k+1},\dots,x_n) = 0$ then one can write it in the form $\sum_{i=1}^k (x_i-a_i)g_i$ for some $g_i \in R$? – dstivd May 12 '18 at 22:01
• Note that it suffices to show this for $n=k=1$, as $F[x]$ is an integral domain whenever $F$ is an integral domain. And in this case you can simply perform long division. – Servaes May 12 '18 at 22:06
• More concretely, you can perform long division on $Q(x_1,\ldots,x_n)$ first by $a_1$, then by $a_2$, etc. – Servaes May 12 '18 at 22:12