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Let $k$ be a field and $k[X_1,...,X_n]$ a ring of polynomials in $n$ variables. In which case is $k[X_1,...,X_n]$ an integral domain?

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  • $\begingroup$ For any integral domain $R$, the polynomial ring $R[X_1,\dots,X_n]$ is an integral domain. $\endgroup$ – Bernard Feb 10 '20 at 12:46
  • $\begingroup$ Sorry for a easy questions (I am very beginner in algebra). Can you write a prove of that fact? $\endgroup$ – VDGG Feb 10 '20 at 12:48
  • $\begingroup$ See this post and apply recursively. $\endgroup$ – Dietrich Burde Feb 10 '20 at 12:55
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Always.

For a given ring $R$ we have: $R[X_1,\ldots,X_n]$ is an integral domain if and only if $R$ is. And in your case $R=k$ is a field, which is an integral domain.

Proof. Step 1. We will show that $R[X]$ is an integral domain if and only if $R$ is.

"$\Rightarrow$" Note that $R$ is a subring of $R[X]$. And subrings of integral domains are integral domains.

"$\Leftarrow$" Assume that $W=\sum w_nX^n$ and $U=\sum u_nX^n$ and $WU=0$. Assume that both $W,U\neq 0$ and note that the coefficient of $WU$ at $\deg(WU)$ is $w_{\deg{W}}\cdot u_{\deg{U}}$. But since $WU=0$ then $w_{\deg{W}}\cdot u_{\deg{U}}=0$ and since $R$ is an integral domain then $w_{\deg{W}}=0=u_{\deg{u}}$ which contradicts the definition of $\deg$ (which is the highest index with nonzero coefficient).

Step 2. $R[X_1,\ldots,X_n]$ is isomorphic to $R[X_1][X_2]\cdots[X_n]$ and therfore by Step 1 and induction on $n$ we conclude that $R[X_1,\ldots, X_n]$ is an integral domain if and only if $R$ is.

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  • $\begingroup$ Thank a lot! and how I can prove that? $\endgroup$ – VDGG Feb 10 '20 at 12:48
  • $\begingroup$ Erekle, see the above duplicate for a proof. $\endgroup$ – Dietrich Burde Feb 10 '20 at 12:58
  • $\begingroup$ @ErekleKhurodze I've updated the answer. $\endgroup$ – freakish Feb 10 '20 at 13:00
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A proof that $R[X_1,\dots,X_n]$ is an integral domain if $R$ is this:

By a trivial induction, it is enough to show it for a single indeterminate. Consider two polynomials $$p(X)=a_0+a_1X+\dots+a_n X^d\enspace(a_d\ne 0),\quad q(X)=b_0+b_1X+\dots+b_e X^e\enspace(b_e\ne 0).$$ The leading term of $p(X)q(X)$ is $a_db_e X^{d+e}$, by the definition of multiplication in $R[X]$, and it is non-zero since $a_db_e\ne 0$.

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