I'm preparing assignment questions for a course in ring/field theory. We'll shortly be looking at extension fields, and the students are meant to understand what notation such as $\mathbb{R}(i)$ means.

Two examples of questions that might arise:

  • If the question asked to describe the elements in $\mathbb{Z}_5(\sqrt{3})$, the student would be expected to recognise that the polynomial $x^2-3$ has no root in $\mathbb{Z}_5$, and thus is irreducible in $\mathbb{Z}_5[x]$, and thus define some element $\omega$ as a root of $x^2-3$, and set $\mathbb{Z}_5(\sqrt{3})=\mathbb{Z}_5(\omega)$, from which they can identify the elements.

  • If the question asked to describe the elements in $\mathbb{Z}_7(\sqrt{2})$, the student would be expected to recognise that the polynomial $x^2-2=(x-3)(x-4)$, and thus is reducible in $\mathbb{Z}_7[x]$, and thus recognise that $\mathbb{Z}_7(\sqrt{2})=\mathbb{Z}_7$ (via: if we define $\omega$ as a root of $x^2-2$ then $\omega \in \{3,4\}$, and so $\omega \in \mathbb{Z}_7$).

However, it is possible to have a "partially reducible" polynomial (i.e., one that does not factor entirely into linear factors in $F[x]$). Which leads to the following question:

Does $\mathbb{Z}_5(\sqrt[3]{3})$ make sense? (I.e. The extension field containing $\mathbb{Z}_5$ and $\sqrt[3]{3}$.)

If we use the above approach, we identify that the polynomial $x^3-3=(x-2)(x^2+2x+4)$, and thus is reducible in $\mathbb{Z}_5[x]$. If we define $\omega$ as a root of the polynomial $x^3-3$, then:

  • $\omega$ might be $2$, in which case $\omega \in \mathbb{Z}_5$ and $\mathbb{Z}_5(\omega)=\mathbb{Z}_5$.
  • $\omega$ might instead be a root of $x^2+2x+4$, in which case $\mathbb{Z}_5(\omega) \neq \mathbb{Z}_5$.

In order to reconcile this problem, it looks like there should be some restriction placed on the notation $F(\omega)$, such as "this is semantically incorrect, unless $\omega$ is derived from an irreducible polynomial in $F[x]$", but this would ruin the second question above.

  • 2
    $\begingroup$ Most of what you write is irrelevant to your question, no? You are simply asking if the notation $\sqrt[3]{3}$ is ambiguous. And yes, it is :-) Over the real field there are certain standard conventions used, but they do not apply muchly to other fields; and even then, $\sqrt[60]{1}$ is probably something one should not write even when talking about $\mathbb C$ without making a specific local convention. $\endgroup$ Commented Sep 10, 2012 at 2:47
  • 1
    $\begingroup$ You can talk about the ring $\mathbb{F}_5 [x]/(x^3 - 3)$, but it is not a field. $\endgroup$
    – Zhen Lin
    Commented Sep 10, 2012 at 2:54

2 Answers 2



this is not the correct usage and could be very misleading. It seems to have confused you, so it must be even more confusing for students who are just learning.

Here is the way it is used most of the time:

Let $K$ be a field contained in a larger field $L$. Let $\alpha\in L$ be an element. Then $K(\alpha)$ denotes the smallest subfield of $L$ that contains both $K$ and $\alpha$. Obviously, if $\alpha\in K$, then $K(\alpha)=K$. Otherwise, if $\alpha$ is not algebraic over $K$, then $K(\alpha)\simeq K(x)$. If $\alpha$ is algebraic over $K$, then there exists an irreducible polynomial $p\in K[x]$ such that $p(\alpha)=0$ and then $K(\alpha)\simeq K[x]/(p)$. This latter construction works even without specifying $L$ if we start by $p$. (I'm sure you know all of this, but I need to establish the background.)

Now when one says $K(\sqrt 2)$, then they actually mean (or should mean) that they want to consider the extension of $K$ with a root of $x^2-2$. In truth this is a sloppy shortcut and one has to be careful how to use it. In fact, it is not a very good notation exactly because of the issue you brought up. If the polynomial is not irreducible, then it is not clear what field one would want to define. Imagine a field in which $x^d-\lambda$ is a product of two irreducible polynomials without roots, but which are different. Extending by $\sqrt[d]\lambda$ could result in two completely different field. So, this should be avoided.

In other words there is a restriction on the meaning of $F(\omega)$ (as you suspect). And it "ruins" your problem as stated. Well, actually it's more like this is not well-phrased problem. In my opinion even the one with $\mathbb Z_7(\sqrt 2)$ is not a good problem either, because it suggests and encourages improper use.

What you could do instead is to ask for the splitting field of those polynomials. For that they would have to adjoin roots as long as there is a factor that is not linear.


No. $\mathbb{Z}_5(\sqrt[3]{3})$ does not make sense, since it is not determined uniquely.

  • 1
    $\begingroup$ Care to substantiate the claim? $\endgroup$
    – Sasha
    Commented Sep 10, 2012 at 2:47
  • $\begingroup$ @Sasha I did as you told me. $\endgroup$ Commented Sep 10, 2012 at 2:54

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