If a field K is algebraically closed does it implies that any multivariate polinomial has a root?

Question

Does any of these two hold?:

1. if K an algebraically closed field then any multivariate polinomial in K has a root in K.
2. if any multivariate polinomial in K has a root in K then K is algebraically closed.

Answer 1

1. Yes. Pick any non-constant $f(x,y) \in K[x,y]$. You may consider $$f(x,y) = g_0(x) + g_1(x)y + g_2(x)y^2 + \cdots + g_n(x)y^n$$ where $g_i(x)\in K[x]$ for $i=0,1,\dots,n$. Substituting $y=0$ you get $$f(x,0) = g_0(x) $$ so as $g_0(x)\in K[x]$ it either has some root $x_0\in K$, or it is a constant. In the former case, $f(x_0,0) = g_0(x_0) = 0$ so $(x_0,0)$ is a root of $f(x,y)$ as desired. In the latter case, let $g_0(x) = a \in K$.
   Similarly, if there exists $x_0 \in K$ such $$f(x_0,y) = g_0(x_0)+g_1(x_0)y+\cdots+g_n(x_0)y^n\in K[y]$$ is not a constant, then $f(x_0,y)$ must have some root $y_0 \in K$ so that $f(x_0,y_0) = 0$ i.e. $(x_0,y_0)$ is a root of $f(x,y)$.
   Otherwise $g_1(x_0)=g_2(x_0)=g_3(x_0)=\cdots = g_n(x_0) = 0$ for all $x_0$, i.e. $g_1,g_2,\dots, g_n$ are all the zero polynomial. Then $f(x,y) = g_0(x) = a$, a contradiction.
   You may proceed by induction.

2. Yes. $K[x] \subseteq K[x,y] \subseteq K[x,y,z] \subseteq \cdots$. So if all multivariate polynomials have a root in $K$ so do all polynomials with only one variable i.e. $K$ is algebraically closed.

Answer 2

Claim: If $K$ is algebraically closed, and $p\in K[x_1,\dots,x_n]$ is non-constant, then $$p(k_1,\dots,k_n)=0,$$ for some $k_i\in K,$ and $$p(k_1’,\dots,k_n’)\neq 0$$ for some $k_1’,\dots,k_n’\in K.$

Proof: Induction on $n.$ $n=1$ means we have a root of $p(k)=0$ by definition of algebraically closed, and $p(k’)\neq 0$ because a polynomial can only have finitely many roots and algebraically closed fields are infinite.

Now, assume true for $n.$

Every $p\in K[x_1,x_2,\dots,x_n,x_{n+1}]$ can be written as:

$$p=\sum_{k=0}^m a_kx_{n+1}^k$$ where the $a_k\in K[x_1,\dots,a_{n}].$

$p$ non-constant either means $a_0$ is non-constant, or at least one of $a_i\neq 0,$ for $i>0.$

If $a_0$ is not constant, then by the induction hypothesis, we can find $k_1,\dots,k_n$ and $k_1’,\dots,k_n’$ so that:

$$p(k_1,\dots,k_n,0)=a_0(k_1,\dots,k_n)=0\\ p(k_1’,\dots,k_n’,0)=a_0(k_1’,\dots,k_n’)\neq0$$

So assume $a_0$ is constant.

If all the $a_i$ are constants, then we are back to the case of $n=1,$ and we can find any $k,k’$ and use any values for $k_1,\dots,k_n.$

So assume $a_k$ is non-constant for $k>0.$ Then find $k_1,\dots,k_n$ so that $$a_k\neq 0$$ Then $p(k_1,\cdots,k_n,x_{n+1})$ is a single-variable non-constant polynomial, because the coefficient of $x_{n+1}^k$ is non-zero. So by the case $n=1$ we can find $k_{n+1},k_{n+1}’$ so that:

$$p(k_1,\dots,k_n,k_{n+1})=0\\ p(k_1,\dots,k_n,k_{n+1}’)\neq0$$

This answers part 1.

This actually shows a stronger result, that we can find the values such that $k_i=k_i’$ for all $i$ except $1.$

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For part $2$ any non-constant polynomial in $1$ variable is a polynomial of $n$ variables.

I suppose you could make “non-constant” mean something like $p(x,y)\neq p(x,0),p(0,y).$ That would be “not constant in any variable.” That makes it harder - I’m not even sure it is true with that meaning. But that is not the usual definition.