If $x$ is algebraic over $K$, show that $K[x]$ is a field.

Let $L/K$ an extension field (and $K$ a domain). $x$ is algebraic over $K$ if there is a polynomial $P(X)\in K[X]\backslash \{0\}$ such that $P(x)=0$. I want to show that $K[x]$ is a field.

I know that $P(x)=\prod_i P_i(x)^{v_i}=0$ where $P_i(X)$ are irreducible and thus, since $K$ is a domain, there is an $i$ such that $P_i(x)=0$.

I don't understand what follow

$K[x]$ is the range of the ring homomorphism

\begin{align*} \text{eval}_x:K[X]&\longrightarrow L\\ P(X)&\longmapsto P(x), \end{align*}

therefore it's kernel is a principal ideal that contain $(P_i)$ which is an ideal maximal.

Q1) Why $K[x]=\text{Im}(\text{eval}_x)$ ? To me we only have $\text{Im}(\text{eval}_x)\subset K[x]$.

Q2) Why $\ker(\text{eval}_x)$ is a principal ideal ? (I know that it's an ideal, but why is it principal ?)

Q3) Why $(P_i)$ is maximal ? And why $\ker(\text{eval}_x)\supset (P_i)$

May be my question are obvious for the most part of you, but I'm starting this course, and I didn't do algebra before, that's why it's complicate to me.

• thank you, I corrected it :-) – idm Oct 14 '15 at 11:00

Q1: Elements of $K[x]$ are of the form $a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n = f(x)$ for $f \in K[X]$.

Q2: The kernel surely is an ideal. Take as a generator the minimal polynomial (i.e. the monic polynomial s.t. $p(x) =0$ of least degree). Then surely $(p) \subseteq \ker(\text{eval}_x)$, and the other inclusion follows since the minimal polynomial must divide any polynomial which is zero at $x$.

There is actually an argument behind this: Suppose $q(x) = 0$. By division with remainder, $$q = hp + r, \deg r < \deg p.$$ Then $q(x) = r(x)$. Since $p$ is of minimal degree, $r = 0$.

Q3: The second part was answered in my comment regarding Q2. Further, $(p)$ is maximal since $K[X]$ is a principal ideal domain (since it's Euclidean), and if $(q) \supsetneq (p)$ we obtain $(q) = K[X]$ since $p$ is irreducible as follows: If $p = fg$, then $g(x) = 0$ or $f(x) = 0$, in contradiction to the minimality of the degree of $p$.

I denote $e_x$ for $eval_x$. $$P(X)\in\ker(e_x)\iff e_x(P(X))=0\iff P(x)=0\iff P_i(x)\mid P(x)\iff P(x)=P_i(x)Q(x)\iff P(x)\in (P_i)$$ therefore $\ker(e_x)=(P_i)$.

The maximality of $P_i$ is obvious since every polynomial $Q(X)\in K[X]$ such that $Q(x)=0$ is in $(P_i)$.

• Your third equivalence requires some work though, doesn't it? – Cloudscape Oct 14 '15 at 11:50
• I don't think so... What would you add ? – Surb Oct 14 '15 at 15:15
• You would have to choose the right $P_i$, wouldn't you? – Cloudscape Oct 14 '15 at 15:19
• Not here since the $P_i$ is already fixed by the OP. – Surb Oct 14 '15 at 15:52
• OK then you take $P_i$ to be several polynomials. But how do you deduce maximality then? – Cloudscape Oct 14 '15 at 17:58