# Alternative definitions of algebraically closed

Let $$k$$ be a field and $$k[x]$$ be the polynomial ring with coefficients from $$k$$. Prove whether the following are equivalent:

1. Every nonconstant polynomial in $$k[x]$$ has a root in $$k$$.
2. Every root of every nonconstant polynomial in $$k[x]$$ is in $$k$$.

It is obvious to me that 2 implies 1. I appreciate any help on this. Thanks!

• Hint: if $p(a)=0$, then $p(x)=(x-a)q(x)$. Do you see how you can use this to piece-by-piece to find all roots? – KReiser May 8 at 22:23
• If $a\in k$ is a root of $f\in k[x]$, then there exists $g\in k[x]$ such that $f(x)=(x-a)g(x)$. By induction on the degree one shows that every polynomial is then the product of linear factors which implies $2.)$ – Severin Schraven May 8 at 22:24

Condition 2 can be better stated as every nonconstant polynomial in $$k[x]$$ factors as a product of degree $$1$$ factors.

To see that 1 implies 2, note that this is obvious if the polynomial has degree $$1$$.

Suppose that if a polynomial has degree $$n$$, then it can be factored as a product of degree $$1$$ factors and let $$f(x)\in k[x]$$ have degree $$n+1$$. By condition 1 it has a root $$r\in k$$.

Can you finish without looking at the spoiler?

Then $$f(x)=(x-r)g(x)$$ and $$g(x)$$ has degree $$n$$.

Let $$p\in K[x]$$ and let's say $$deg(p)=n>0$$. By assumption it has a root $$\alpha_1\in K$$. Then $$x-\alpha_1|p$$. So there is a polynomial $$p_1\in K[x]$$ of degree $$n-1$$ such that $$p(x)=(x-\alpha_1)p_1(x)$$. If $$n-1>0$$ then $$p_1$$ is not constant and then it has a root $$\alpha_2\in K$$. Hence $$x-\alpha_2|p_1$$ and then we can write $$p_1(x)=(x-\alpha_2)p_2(x)$$ where $$p_2\in K[x]$$ has degree $$n-2$$. Then $$p(x)=(x-\alpha_1)(x-\alpha_2)p_2(x)$$. Now if $$p_2$$ is not constant then take a root of $$p_2$$ in $$K$$ and continue by induction. At the end you will get the following equality:

$$p(x)=c(x-\alpha_1)(x-\alpha_2)...(x-\alpha_n)$$

Where $$c,\alpha_1,\alpha_2,...,\alpha_n\in K$$. The roots of $$p$$ are $$\alpha_1,\alpha_2,...,\alpha_n$$ and they are all in $$K$$.