Say I create an extension $K$ over a field $F$ obtained by adjoining an element $\alpha$, i.e. $K = F[\alpha]$ ($\alpha$ does not necessarily have to be the root of a polynomial with coefficients in $F$). In this process I might need to adjoin some powers of $\alpha$. Suppose I have a basis $B$ for $K$ and want to show that every element in $K$ has an inverse - that $K$ is indeed a field. If every element of $B$ has an inverse, is that enough to guarantee that every element in $K$ have an inverse, and is there a nice way to compute it? If not, is it true if the extension is finite-dimensional i.e. $|B|$ is finite?

An example of what I'm talking about: take (letting $\alpha = \sqrt[3]{2}$) $K = \mathbb{Q}(\alpha)$. I have my basis $B = \left\{1, \alpha, \alpha^2\right\}$, and clearly $\alpha^{-1} = \frac{\alpha^2}{2}$ and $\alpha^{2^{-1}} = \frac{\alpha}{2}$. Any element in $K$ is of form $a + b\alpha + c\alpha^2$ and has some inverse of the same form. If I were asked to prove that adding the two elements $\alpha, \alpha^2$ is enough to make $K$ into a field, i.e. we do not need to add more powers, by showing that everything with just the adjunction of those two elements is multiplicatively closed and has inverses, would the existence of inverses of $\alpha, \alpha^2$ guarantee the existence of inverses of general elements?

This is in trying to prove properties about the extension. This is indirectly homework, but this exact question is not so I would appreciate a full answer or a link to one.

  • $\begingroup$ The definition of $F(\alpha)$ is the smallest field containing $F$ and $\alpha$. $\endgroup$ – JSchlather Jan 29 '13 at 0:40
  • $\begingroup$ @JacobSchlather But if you want to characterize a field by its basis, which is what the homework question I am asking deals with, is there some criterion by which you can know that your basis at least induces the right field, other than (in the example I gave) just chugging through and manually finding an inverse for an arbitrary element? In an extension of degree four that would be very tedious. $\endgroup$ – Julien Clancy Jan 29 '13 at 0:42
  • $\begingroup$ @JulienClancy Jacob's comment is correct, in that $F(\alpha)$ is the usual notation for a field. What you are interested in is some algebra over $F$ obtained by adjoining an element $\alpha$, and that is usually denoted $F[\alpha]$. Can you edit your question to use that notation instead? $\endgroup$ – PatrickR Jan 29 '13 at 6:53

I think you're looking at the problem the wrong way. Let $F$ be a field and let $\alpha$ be algebraic of degree $n$ over $F$. I claim that $F[\alpha]=F(\alpha)$ and that $1,\alpha,\dots,\alpha^{n-1}$ is a basis of $F(\alpha)$ over $F$. If $1,\dots,\alpha^{n-1}$ were linearly dependent over $F$ then $\alpha$ would satisfy a polynomial of degree less than $n$, so they are linearly independent. Let $p(t)$ be the minimal polynomial of $\alpha$ over $F$. Any element of $F[\alpha]$ is of the form $f(\alpha)$ for some polynomial $f \in F[t]$. We may divide $f(t)$ by $p(t)$ so that


where $\deg r(t) <n$ evaluating at $\alpha$ we deduce that


Since $r(t)$ is of degree less than $n$ it follows that $r(\alpha)$ can be expressed as a linear combination of $1,\alpha,\dots,\alpha^{n-1}$. Thereby $1,\dots,\alpha^{n-1}$ form a basis for $F[\alpha]$. Now to see that $F[\alpha]$ is indeed a field we have a map $\varphi: F[t] \rightarrow F[\alpha]$ generated by sending $\varphi(t)=\alpha$. By definition of the minimal polynomial $\ker \phi = \langle p(t) \rangle$. Since $\varphi$ is surjective we have by the first isomorphism theorem that $F[\alpha] \cong F[t]/\langle p(t) \rangle$. Finally $p(t)$ is irreducible so it generates a maximal ideal in $F[t]$ and we conclude that $F[t]/\langle p(t) \rangle$ is a field.


Consider the extension $K$ of the field $F$, adjoining a root $\alpha$ of $x^2 - 1$. Clearly $1,\alpha$ is a basis, and both elements are invertible, but $\alpha-1$ is a zero-divisor, and thus not invertible.

PS I am assuming you only want $F$ to be a field.

  • $\begingroup$ What are $K$ and $F$ here? Are you not adjoining $\alpha = 1$, which is already in the field? $\endgroup$ – Julien Clancy Jan 29 '13 at 0:25
  • $\begingroup$ Now that you have rephrased your question, I see that I gave an answer to a different one. Let me explain, though. Given a field $F$ and any polynomial $f(x) \in F[x]$, you can construct the ring $K = F/(f(x))$ and regard it as an extension of $F$. In $K$ the class of $x$ will be a root of $f(x)$. [follows in next comment] $\endgroup$ – Andreas Caranti Jan 29 '13 at 7:18
  • $\begingroup$ [see previous comment] As noted above by @JacobSchlater, $K$ will be a field if and only if $f(x)$ is irreducible in $F[x]$. But to see it more concretely in the example I gave, identify $F$ with the ring of matrices $\{ \begin{bmatrix} a & 0\\0&b \end {bmatrix} : a \in F\}$, and consider $\alpha = \begin{bmatrix} 0 & 1\\1 & 0 \end {bmatrix}$. Then $\alpha^2 = 1 = \begin{bmatrix} 1 & 0\\0&1 \end {bmatrix}$ is invertible, but $0 = \alpha^2 - 1 = (\alpha - 1)(\alpha + 1)$, so $\alpha - 1, \alpha + 1$ are not. $\endgroup$ – Andreas Caranti Jan 29 '13 at 7:18
  • $\begingroup$ Ah, I understand now. Thanks for explaining; this is an excellent example. Do you know if the same can be done over subfields of $\mathbb{C}$? $\endgroup$ – Julien Clancy Jan 29 '13 at 16:17
  • $\begingroup$ @JulienClancy, this works for any field $F$. $\endgroup$ – Andreas Caranti Jan 29 '13 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.