# Polynomials of several variable is integral domain [closed]

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?

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

## 2 Answers

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.

• Thank a lot! and how I can prove that? – VDGG Feb 10 '20 at 12:48
• Erekle, see the above duplicate for a proof. – Dietrich Burde Feb 10 '20 at 12:58
• @ErekleKhurodze I've updated the answer. – freakish Feb 10 '20 at 13:00

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 $$a_db_e\ne 0$$.