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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!

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    $\begingroup$ 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? $\endgroup$ – KReiser May 8 at 22:23
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    $\begingroup$ 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 $2.)$ $\endgroup$ – Severin Schraven May 8 at 22:24
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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 $n$.

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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 $K$.

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