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There is a theorem in my book which says the following:

Let $K$ be a field and let $f(X) \in K[X]$ be irreducible over $K$. Then there exists a field extension $L/K$, such that $\exists u \in L$: $f(u) = 0$.

Proof: Notice that $K\subseteq K[X]/(f(x))$. Clearly, $K[X]/(f(x))$ is a field since $(f(x))$ is maximum ideal. If we take $u = \overline{X}$, then $f(u) = f(\overline{X})= \overline{f(X)} = 0$.

Question: I really don't get if $u = ...$ in the proof, since I don't know what $\overline{X}$ means in this context, does anyone know?

Thanks in advance.

Unfortunately I cannot provide a link to the book as it is my dutch (non pdf) syllabus for my algebra course.

EDIT: Seems like I get it now: $f(\overline{X}) = \overline{f(X)} = f(x)+ (f(x)) = (f(x))$ which is $0$ for $K\subseteq K[X]/(f(x))$.

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This is not really related to your question, but the proof you have stated is a very nice proof, and the one usually given in textbooks. But you can strictly construct a field containing $K$ which has a root of your polynomial, and not just a field containing an isomorphic copy of $K$ where your polynomial has root. Take a look at Proposition 2.3 on page 231 of Lang's Algebra (Third Edition). This is really a minor point, since isomorphic fields are essentially the same, but to me Lang's proof is also very illuminating. – Rankeya Dec 14 '11 at 2:08
Also, disregard my comment if it confuses you at all. – Rankeya Dec 14 '11 at 2:08
@Rankeya, thanks, i'll check it out! – sxd Dec 14 '11 at 2:15
@Rankeya: It's looks like it's not quite an alternate proof, but rather a nicer-looking (and more satisfying) statement of the result, as the proof of Proposition 2.3 does use the "standard" method (mod out by the ideal...) as a starting point. – Riley E Dec 14 '11 at 2:44
@Riley E: Sorry, if my comment gave the impression that it is an alternate proof (in my defense, I did not mention it was an alternate proof. I just said it was more illuminating to me). You are right. It does use the standard method, but then expands on it to give a more 'satisfying' result. – Rankeya Dec 14 '11 at 2:48
up vote 5 down vote accepted

$\overline{X}$ means the class of $X$ mod $f(X)$ in the quotient ring $K[X]/(f(x))$.

Now, given $g\in K[X]$, consider the associated polynomial function in $K[X]/(f(x))$. Then, by the definition of the ring operations in $K[X]/(f(x))$, we have $g(u) = g(\overline{X}) = \overline{g(X)}$. When $g=f$, we get $g(u)=\overline{0}$.

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Thanks, could you also help me clear up why $f(\overline{X})= \overline{f(X)} = 0$. – sxd Dec 14 '11 at 1:52
@DimitriSurinx, I've edited my answer. – lhf Dec 14 '11 at 1:56
I don't see how this follows from the ring operations, could you elaborate a little more? Thanks in advance. – sxd Dec 14 '11 at 2:13
@DimitriSurinx, for instance $u^2 = (\overline{X})^2 = \overline{X} \cdot \overline{X} = \overline{X^2}$, where the definition of multiplication in the quotient ring was used in the last step. – lhf Dec 14 '11 at 2:17
Write out the polynomial $f(X)$ explicitly as $f(X) = a_nX^{n} + ... + a_0$ and then substitute $X + (f(X))$ for $X$. That should help. – Rankeya Dec 14 '11 at 2:18

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