How Does Algebraic Closure Extend to Multivariable Polynomials?

[Math Amateur]

Dummit and Foote in their book Abstract Algebra give the following definition of an algebraically closed field ... ...View attachment 5679From the remarks following the definition it appears that the definition only applies to \(\displaystyle K[x]\) ...

Does it also apply to \(\displaystyle K[x_1, x_2], K[x_1, x_2, x_3], \ ... \ ... \ , K[x_1, x_2, \ ... \ ... \ , x_n] , \ ... \ ... \ ...\) ?

That is ... when we say \(\displaystyle K\) is an algebraically closed field does it imply that every polynomial in \(\displaystyle K[x_1, x_2, \ ... \ ... \ , x_n]\) has a root in \(\displaystyle K\) ... ... ?

... ... or maybe it is better if I say ... how does the definition of algebraically closed generalise to \(\displaystyle K[x_1, x_2, \ ... \ ... \ , x_n]\) ... ... ?

Hope someone can clarify this issue ... ...

Peter

 

[Deveno]

I think you are asking the wrong question: algebraic closure is a property of a field $K$, not the ring $K[x]$.

But it is trivially true that if $K$ is algebraically closed, any $p \in K[x_1,\dots,x_n]$ has a solution point in $K^n$, for, let:

$(a_2,\dots,a_n)$ be any $n-1$-tuple in $K^{n-1}$. We then obtain a polynomial in $K[x_1]$, which by dint of the algebraic closure of $K$ has a root in $K$.

In fact, what one typically sees is not a collection of "points" (a 0-manifold), but rather regions of solutions. For example, with $K = \Bbb R$, and:

$p(x,y) = xy - 1$

(a quadratic in two indeterminates)

the zero-set is a hyperbola, and it really doesn't make sense to speak of "roots" as *numbers*. A pair of numbers either lies in the zero-set, or it does not. If $K$ is algebraically closed, the zero-set of $p(x,y) \in K[x,y]$ will of necessity be non-empty, but if $K$ is not algebraically closed, it may not be, for example, the zero-set of $f(x,y) \in \Bbb R[x,y]$ given by:

$f(x,y) = x^2 + y^2 + 1$

is null.

 

[Math Amateur]

I think you are asking the wrong question: algebraic closure is a property of a field $K$, not the ring $K[x]$.

But it is trivially true that if $K$ is algebraically closed, any $p \in K[x_1,\dots,x_n]$ has a solution point in $K^n$, for, let:

$(a_2,\dots,a_n)$ be any $n-1$-tuple in $K^{n-1}$. We then obtain a polynomial in $K[x_1]$, which by dint of the algebraic closure of $K$ has a root in $K$.

In fact, what one typically sees is not a collection of "points" (a 0-manifold), but rather regions of solutions. For example, with $K = \Bbb R$, and:

$p(x,y) = xy - 1$

(a quadratic in two indeterminates)

the zero-set is a hyperbola, and it really doesn't make sense to speak of "roots" as *numbers*. A pair of numbers either lies in the zero-set, or it does not. If $K$ is algebraically closed, the zero-set of $p(x,y) \in K[x,y]$ will of necessity be non-empty, but if $K$ is not algebraically closed, it may not be, for example, the zero-set of $f(x,y) \in \Bbb R[x,y]$ given by:

$f(x,y) = x^2 + y^2 + 1$

is null.

Thanks for clarifying the issue, Deveno ...

What you said regarding the hyperbole was most helpful ...

Thanks again,

Peter