1
$\begingroup$

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

  • 2
    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?2012-10-21

3 Answers 3

2

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 $\Psi(F)\cong F$ .

  • 0
    @MartinBrandenburg - thanks for the note, corrected2012-10-21
2

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.

  • 1
    Showing it it homomorphism and $\not\equiv 0$ is sufficient2012-10-21
2

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.