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Let $L$ be a field, $K$ a subfield of $L$ and $x\in L$.

I want to prove that $K[x]$ is an integral domain, a subring of $L$ but not always a field.

An integral domain is a commutative ring without zero-divisors. We have that $K[x]$ is commutative, since when $f(x), g(x)\in K[x]$ then $f(x)*g(x)=g(x)*f(x)$, right?

Suppose that $f(x)=a_0+a_1x+\dots a_nx^n$ and $g(x)=b_0+b_1x+\dots +b_nx^n$. We have that $f(x)\cdot g(x)=0\Rightarrow (a_0+a_1x+\dots +a_nx^n)(b_0+b_1x+\dots +b_nx^n)=0 \\ \Rightarrow \sum_{i=0}^{2n}c_ix^i=0, \text{ where } c_i=a_0b_i+a_1b_{i-1}+\dots +a_{i-1}b_1+a_ib_0$

From the relation $\sum_{i=0}^{2n}c_ix^i=0$ we get that $c_i=0, \forall i=0, \dots , 2n$.

So we have the following: $$c_0=0 : a_0b_0=0 \\ c_1=0 : a_0b_1+a_1b_0=0 \\ \dots \\ a_0b_n+a_1b_{n-1}+\dots a_{n-1}b_1+a_nb_0=0$$ But how do we conclude that one of the polynomials $f$ or $g$ must be the zero polynomials?

Also, how can we show that $K[x]$ is a ring and a subring?

$K[x]$ is not always a field because the quotients of polynomials are not contained in $K[x]$, right?

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    $\begingroup$ It's waaaay easier: since $\;K\subset L,\,x\in L\implies K[x]\subset L\;$, and since $\;L\;$ is a field $\;K[x]\;$ cannot have non-trivial zero divisors. $\endgroup$ – DonAntonio Sep 15 '16 at 17:23
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I already commented why it is almost immediate: $\;K[x]\;$ is an integral domain. Now, you may want to prove the following very important classic lemma from field theory:

Lemma: with the given data, $\;K[x]\;$ is a field iff $\;x\;$ is algebraic over $\;K\;$, and this happens iff $\;K[x]=K(x)\;$.

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  • $\begingroup$ To show that $K[x]$ is a subring of $L$ do we use this lemma? $\endgroup$ – Mary Star Sep 15 '16 at 17:52
  • $\begingroup$ @MaryStar No. It is a ring because it is a homomorphic image of the ring of polynomials over $\;K\;$ , namely $\;K[t]\;$ . $\endgroup$ – DonAntonio Sep 15 '16 at 18:11
  • $\begingroup$ What do you mean? I got stuck right now... $\endgroup$ – Mary Star Sep 15 '16 at 18:19

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