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Let $F$ a field and $p(x)$ an irreducible polynomial in $F[x]$.

I'm trying to prove that $F[x]/(p(x))$ is an extension of $F$. I know there are two approaches. Either we can prove that $F$ is a subset of $F[x]/(p(x))$ (because $F[x]/(p(x))$ is a field) or we can show that there is an embedding between $F$ and $F[x]/(p(x))$.

I've tried the both cases, it seems to be easy, but I didn't get to solve it. Maybe because I'm a really beginner in this subject. I need help...

Thanks

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Since $F$ is a field, any homomorphism $F\to K$ is an embedding (because its kernel is a proper ideal of $F$ and so must be zero). It then suffices to give any homomorphism $F\to F[x]/(p(x))$. Do you see how this can be done? –  Brad Oct 21 '12 at 14:38
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3 Answers 3

up vote 2 down vote accepted

It is not true that $F$ is a subset of $F[x]/\langle p(x)\rangle$

However, you have a monomorphism $\Psi:F\to F[x]/\langle p(x)\rangle$ defined by $$\Psi(a)=\bar{a}=a+\langle p(x) \rangle\in F[x]/\langle p(x)\rangle$$

So we have $\Psi(F)\subset F[x]/\langle p(x)\rangle$ and $\Psi(F)\cong F$ .

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you meant: $\overline a = a + (p(x))$? p(x) have brackets, no? –  user42912 Oct 21 '12 at 14:52
    
@user42912 corrected. –  Belgi Oct 21 '12 at 14:54
    
It is not an isomorphism. It is a monomorphism. –  Martin Brandenburg Oct 21 '12 at 16:41
    
@MartinBrandenburg - thanks for the note, corrected –  Belgi Oct 21 '12 at 16:46
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Hint: Consider the map $\varphi: F \to F[x]/(p(x))$, $f \mapsto f + (p(x))$ and show that it is injective and a ring homomorphism.

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Showing it it homomorphism and $\not\equiv 0$ is sufficient –  Belgi Oct 21 '12 at 14:41
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The obvious map $\iota\colon F\to F[x]/(p(x))$, $a\mapsto a+(p(x))$ is a homomorphism at least of rings (i.e. compatible with addition and multiplication). Since $1\notin(p(x))$, we have $\iota(1)\ne 0$, i.e. $\iota$ is not the zero map. The only ideals of a field $F$ are $F$ and $\{0\}$, hence $\iota$ is an embedding.

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