Suppose we adjoin a symbol $k$ to the field $F_p$, as in $F_p(k)$. What is an intuitive understanding of the structure of this field and its elements? Since multiplication needs to be closed, all new polynomials are part of this field. But also multiplicative inverses need to be included, does that mean all rational functions over $F_p$ should also be included? What about additive inverses? Do you have a mental model of this structure that gives you a sense of what the elements look like and how they preserve the structure, or is that something you can only reasonably inspect case-by-case?

Let's consider $F_2(k)$. If my understanding above is correct, all the following are new elements added to $F_2$:


The list goes on forever, which doesn't seem right. Perhaps the structure is not well-defined if $k$ is just some arbitrary symbol, and it needs to be a root of a polynomial that doesn't have a root in $F_2$. This would mean I can only make $F_p(k)$ a field if it is an algebraic extension of $F_2$. Is that so?

If it is, then say I choose $p(x)=x^2+x+1$ in $F_2$, and I add the root $k$. How do I define $k^n$ for any power $n$, and how do I define $k+k+k$. In other words, how do I force the field structure? $k$ being the root of $p(x)$ should somehow help me fill in the multiplication and addition tables, but I don't see how.

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    $\begingroup$ Have you not yet studied polynomial rings? (rigorously - as in an abstract algebra course) $\endgroup$ – Bill Dubuque Jan 30 at 21:38
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    $\begingroup$ Your initial understanding of the field $F(t)$ is correct. It is the collection of all rational functions with coefficients in $F$. If $t$ is just a formal variable, then this is an infinite degree extension of $F$. That is entirely fine, so I’m not really sure where your issue with this is? $\endgroup$ – Adam Higgins Jan 30 at 21:39
  • $\begingroup$ @AdamHiggins That's great, that was my initial understanding of fields. This post threw me off: physicsforums.com/threads/…. It's too long so I didn't add it to the question. In here the claim is that $F_2(k)$ where $k$ is the root of $p(x)=x^2+x+1$ contains $\{0,1,k,k+1\}$, but I can't get the multiplication table to show this is a field, so I started doubting that my understanding of fields is as described above, I felt that that's "too many" elements added in with the introduction of a single symbol. $\endgroup$ – Mike Jan 30 at 21:46
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    $\begingroup$ @Mime It is a field, since it is clearly a ring, and every non-zero element has a multiplicative inverse since $k\cdot (k+1) = k^{2} + k = -1 = 1$. $\endgroup$ – Adam Higgins Jan 30 at 21:58

What we have are two closely related concepts - a field extension $F(k)$ and a ring extension $F[k]$. The difference? In the former, we adjoin rational functions of $k$ as well, so $F(k)$ is the field of quotients of $F[k]$.

The two are equal if and only if $k$ is algebraic over $F$.

So, your example, adjoining a transcendental element $k$ to $F_2$? Done as a ring extension $F_2[k]$, we get the polynomial ring in one variable over $F_2$, with such elements as $k$, $k^2$, $k^3+k+1$, and so on - $2^m$ new elements of each degree $m>0$. Done as a field extension, we get all of their quotients as well, for such elements as $\frac{(k+1)^2(k^2+k+1)}{k(k^3+k+1)}$.

If instead your $k$ is a root of the irreducible polynomial $k^2+k+1$? Then that ring extension collapses - everything of degree $2$ or higher can be reduced to an equivalent element of degree $1$ or less, by taking it mod $k^2+k+1$. For example, $k^3+k+1=(k+1)(k^2+k+1)+k\equiv k$. Moreover, everything that's not zero is invertible; we can apply the Euclidean algorithm to find the GCD of any polynomial and $k^2+k+1$, and write it as a linear combination (with polynomial coefficients) of those two. Since $k^2+k+1$ is invertible, that GCD is either $1$ or $k^2+k+1$ - and writing $1$ as a linear combination constructs the inverse mod $k+2+k+1$. That makes it a field.

Since there are only four elements, we can write out everything cleanly. We have $x^2\equiv x^2-(x^2+x+1)\equiv x+1$, and $x^3\equiv x(x+1)=x^2+x\equiv 1$. Of the four elements $0,1,x,x+1$, we've got them all accounted for, and the multiplicative group is simply cyclic of order $3$.

And we did all this with just the elements from the ring extension. There's no need to bring in the rational functions when everything's invertible already.

  • $\begingroup$ Thanks a lot, very insightful! Exactly what I was looking for. I can get $F_2[k]$ in the case where $k$ is the root of $k^2+k+1$ to be a field with 4 elements if I do all the operations mod 2, e.g. $k^2=-k-1=k+1 (mod 2)$ and $k^2(k+1)=(k+1)(k+1)=k^2+k+k+1=k+1+1=k$. Does this generally work, that if we adjoin a root and maintain the operation mod $p$ the group structure will be preserved? This seems odd, because then we can adjoin more roots of newer polynomials, and what modular arithmetic would we perform there? Finally, is the irreducibility of $p(x)$ necessary? $\endgroup$ – Mike Jan 30 at 22:24
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    $\begingroup$ Formally, what we're doing is taking quotients of the polynomial ring by an ideal. If you can show the ring structure works on that polynomial ring, then it's inherited on all those quotients. Irreducible? That's the difference between the quotient being a field and the quotient having zero-divisors. Fortunately, the minimal polynomial of any algebraic element is irreducible. Adjoining multiple elements? That gives you a chain of field extensions - and if it's finitely many algebraic elements, we can find one element (a sum or the like) that'll take us all the way in one step. $\endgroup$ – jmerry Jan 30 at 23:06

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