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I want to prove that when $F:K\rightarrow K[X]/\langle f\rangle $ is a map such that $F(a)=a+\langle f \rangle$, then $F$ is an embedding from $K$ to $K[X]/\langle f \rangle$, when $f\in K[X]\backslash K$.

To show the function is an embedding do I need to check that, for $a,b\in K$:

$$F(a+b)=F(a)+F(b)$$

Which seems straightforward in this case: $$F(a+b)=I+a+b=I+a+I+b=F(a)+F(b)$$

And the same for multiplication... And in addition check that the map is an injection?

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  • $\begingroup$ In what sense are you using embedding here, i.e. what's your definition for it? $\endgroup$ – Guido A. Feb 27 at 3:47
  • $\begingroup$ You need to assume that $K$ is a field and that $\deg f \geq 1$. $\endgroup$ – darij grinberg Feb 27 at 3:58
  • $\begingroup$ Hint: If an element $a \in K$ would map to $0$ under $f$, then $a$ would be divisible by $f$ as a (constant) polynomial in $K\left[x\right]$. Why would that be absurd? $\endgroup$ – darij grinberg Feb 27 at 3:59
  • $\begingroup$ @darijgrinberg it's in the post that $f \in k[X] \setminus k$. $\endgroup$ – Guido A. Feb 27 at 3:59
  • $\begingroup$ @darijgrinberg the question says that $f \in K[X]\setminus K$. So, $\deg(f) \geq 1$, I guess. $\endgroup$ – stressed out Feb 27 at 4:00
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A quick way of seeing that $F$ is a morphism is to note that $F = \pi \iota$ with $\pi : k[X] \to k[X]/(f)$ the canonical projection and $\iota : k \hookrightarrow k[X]$ the inclusion. Both are morphisms, hence their composition $F$ will be.

As for injectivity, we can equivalently prove that $\ker F = 0$. Take $a \in k$ such that $$ a + (f) = [a] = F(a) = (f) = [0]. $$ Then $a - 0 \in (f)$, which implies $a = fg$ for some $g \in k[X]$. By degree considerations, it must be that $g = 0$ because otherwise we would have

$$ 0 = \deg a = \deg fg = \deg f + \deg g \geq \deg f \geq 1. $$

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I think you are trying to prove Kronecker's theorem for field extensions. So, by embedding, I assume that you're trying to show that $F$ is an injective homomorphism.

Your map is the composition of two homomorphisms $i: K \hookrightarrow K[X]$ and $\pi: K[X] \to K[X]/\langle f \rangle$, i.e. $F=\pi \circ i$. So, it's a homomorphism.

Now let's find its kernel to show that it's injective. Suppose that $a \neq0 \in \ker(F)$ $$F(a) = a + \langle f \rangle = \langle f \rangle$$ Then $a \in \langle f \rangle$. So, $a = p(x)f(x)$. Since $a \in K \setminus \{0\}$, $p(x)$ cannot be the zero polynomial and also, $\deg{a}=0$ while $\deg{p(x)q(x)}=\deg{p(x)}+\deg{f(x)} \geq 1$.

So, $a=p(x)f(x)$ is impossible. Hence, $a \not\in \langle f\rangle$. This shows that $\ker(F) = 0$. Hence, the map is injective.

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