Isomorphism between $K[X]/P(X)$ and $K(A)$ for $A$ a root of the irreducible polynomial $P(X)$ over $K$

Question

I'm afraid the answer to this question should be clear to anyone who knows a little bit of algebra, field theory, field extensions, and polynomials.

In his answer to a question about Galois theory and permutations of roots, Eric Wofsey writes:

Since the polynomial $x^2−4x+1$ is irreducible over $\mathbb{Q}$, there is an isomorphism $f:\mathbb{Q}[x]/(x^2−4x+1)\rightarrow\mathbb{Q}(A)$ sending $x$ to $A$

where $A$ is a root of $x^2−4x+1$.

My questions are:

1. By which argument or theorem does this follow, resp. more generally: For any irreducible polynomial P(X) with root $A$ over a field $K$ there is an isomorphism $f:K[X]/P(X)\rightarrow K(A)$ which sends $X$ to $A$?

2. With $X$ the monic polynomial of degree 1, $X$, is meant?

3. What does the isomorphism look like (which sends polynomials to numbers)?

Answer 1

Let $L$ be an extension field of $K$ and let $\alpha\in L$. The substitution mapping $\phi:K[x]\rightarrow L$ with $\phi:f\mapsto f(\alpha)$ is ring homomorphism, which is surjective if the mapping is restricted to the image $K(\alpha)$. The kernel of the mapping is the ideal $\langle g\rangle$, where $g\in K[x]$ is the minimal polynomial of $\alpha$ over $K$. By the homomorphism theorem, we obtain the isomorphism $\psi:K[x]/\langle g\rangle\rightarrow K(\alpha)$ given by $f+\langle g\rangle \mapsto f(\alpha)$.

Answer 2

Edit. The proof being very easy, the point is to phrase the statement as clearly as possible.

Let $K$ be a field, $X$ an indeterminate, and $P(X)\in K[X]$ an irreducible polynomial. Then

(a) $L:=K[X]/(P(X))$ is a field,

(b) if we write $\pi:K[X]\to L$ for the canonical projection and identify $\pi(K)$ to $K$ (noting that $\pi$ is injective on $K$), then we have $L=K[\pi(X)]$, and $\pi(X)$ is a root of $P(X)$ in $L$,

(c) if $M$ is any extension of $K$ containing a root $\alpha$ of $P(X)$, then there is a unique $K$-linear embedding of $L$ in $M$ mapping $\pi(X)$ to $\alpha$.

Proof. (a) and the fact that $\pi$ is injective on $K$ are clear. To prove (b) it suffices to observe that, for any polynomial $Q(X)=a_nX^n+\cdots+a_0$ in $K[X]$ we have $$ \pi(Q(X))=\pi(a_nX^n+\cdots+a_0)=\pi(a_n)\pi(X)^n+\cdots+\pi(a_0) $$ $$ =a_n\pi(X)^n+\cdots+a_0=Q(\pi(X)) $$ because we have identified $\pi(K)$ to $K$. To prove (c) note that there is a unique $K$-algebra morphism from $K[X]$ to $M$ mapping $X$ to $\alpha$, and that this morphism factors through $L$.

Previous answer

Let $\pi:K[X]\to K[X]/(P(X))$ be the canonical projection.

As $P(X)$ is irreducible, $L:=K[X]/(P(X))$ is a field.

Clearly $K$ embeds in $L$, and $L$ is generated, as a ring, by $K$ and $\pi(X)$.

The equalities $0=\pi(P(X))=P(\pi(X))$ show that $\pi(X)$ is a root of $P(X)$ in $L$.

Answer 3

^{Eric Wofsey in a comment to another question gave this answer:}

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By the universal property of $K[X]$ there exists a unique homomorphism $K[X]\rightarrow K(A)$ that is the identity on $K$ and sends [the polynomial] $X$ to [the number] $A$.

This homomorphism is surjective since $K$ and $A$ generate $K(A)$ as a ring.

The kernel is the ideal of polynomials that vanish when evaluated at $A$. Since $P(X)$ is irreducible, the ideal it generates is maximal, so the kernel must be generated by $P(X)$ (it cannot be larger).