Proof that $F(a)=a+\langle f\rangle$ is an embedding from $K$ to $K[X]/\langle f\rangle$

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?

• In what sense are you using embedding here, i.e. what's your definition for it? – Guido A. Feb 27 at 3:47
• You need to assume that $K$ is a field and that $\deg f \geq 1$. – darij grinberg Feb 27 at 3:58
• 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? – darij grinberg Feb 27 at 3:59
• @darijgrinberg it's in the post that $f \in k[X] \setminus k$. – Guido A. Feb 27 at 3:59
• @darijgrinberg the question says that $f \in K[X]\setminus K$. So, $\deg(f) \geq 1$, I guess. – stressed out Feb 27 at 4:00

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

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

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.