How is this function injective?

Question

I'm currently studying Galois Theory and I came across this theorem.

Theorem Let $E$ be a field, $p(x)\in E[x]$ an irreducible polynomial of degree $d$ and $I = \langle p(x) \rangle$ the ideal generated by $p(x)$. Then $E[x]/I$ is an extension field of $E$.

The proof first uses the fact that since $E$ is a field, $E[x]$ is an euclidean domain so $p(x)$ being irreducible is also prime. Then as $E[x]$ is also a principal ideal domain, $I$ is a maximal ideal and thus $E[x]/I$ is a field. Finally proceeds to define the function $$ \varphi : E\to E[x]/I$$ given by $$\varphi (a) = \bar{a} = a + I $$ and affirms its injectivity to get an isomorphism $$ \varphi : E \to \varphi(E)\subseteq E[x]/I $$ Concluding that $E[x]/I$ is an extension field of $E$

I get the proof except for the fact that $\varphi$ is injective. I assume that we can think of this function as the composition $$ \varphi = \pi\circ f $$ where $$ f : E \to E[x] $$ assigns each element in $E$ its constant polynomial, and $$ \pi : E[x]\to E[x]/I $$ which assigns each element in $E[x]$ its left coset in $E[x]/I$

However, I know $f$ is injective but $\pi$ is not, so I'm having a hard time showing that $\varphi$ is injective.

How should I proceed? :(

Thank you.

Answer 1

Note $a$ and $b$ are constants (polynomials of degree $0$ if you like). Assume $a+I = b+I$. Then $a-b \in I = \langle p(x) \rangle$, so $a-b$ is a multiple of $p(x)$. But $a-b$ is a constant, and how can a constant be a multiple of a polynomial of degree $1$ or higher over a field? By taking degrees, only if that constant is $0$, so $a-b=0$ and $\varphi$ is injective. (You need not consider $\pi$ for this argument.)

Answer 2

Note that $\pi$ is injective on the constants.
If not, then you'd have that the difference of two constants belongs to $I$, that is, is divisible by $p(x)$. It is easy to see that this must force the constants to be equal. (Since $p(x)$ is not a constant polynomial.)

Answer 3

A rather general argument: A ring homomorphism $\varphi:K\to R$ from field to a commutative Ring (with 1) is always injective.

Proof: $\ker\varphi$ is an ideal in $K$. Therefore $\ker\varphi = (0)$ or $\ker\varphi =K$. Because a field has only two ideals. From $\varphi(0)\neq\varphi(1)$ it follows $\ker\varphi = (0)$ and therefore $\varphi$ injective.