Let $\mathbb{F}$ be a field, and let $p(x)\in \mathbb{F}[x]$ be a polynomial of degree $n$ which is irreducible over $\mathbb{F}$. Let $<p(x)>$ be the principal ideal generated by $p(x)$. Then $\mathbb{F}[x]/<p(x)> $ is a field .

Is the principal ideal generated by $p(x)$ the set of all polynomials $f(x)$ such as $f(x)=p(x)k(x)$, where $k(x)\in \mathbb{F}[x]$ ? Is there a way to understand intuitively the proof of this theorem?

  • $\begingroup$ $A$ is a ring, $m$ is a maximal ideal, then $A/m$ is a field $\endgroup$
    – dcolazin
    May 25, 2019 at 17:17
  • $\begingroup$ What does the symbol / mean in this case ? $\endgroup$
    – AleWolf
    May 25, 2019 at 17:40
  • $\begingroup$ Your use of the notation $\setminus$ was wrong - the symbol $/$ means what you thought $\setminus$ meant. $\endgroup$ May 25, 2019 at 17:53
  • $\begingroup$ That is, set subtraction ? $\endgroup$
    – AleWolf
    May 25, 2019 at 18:29
  • $\begingroup$ @AleQuercia no, it means quotient $\endgroup$
    – dcolazin
    May 26, 2019 at 10:11

1 Answer 1


It isn't completely obvious what you are looking for. You are right in your identification of the principal ideal generated by $p(x)$.

Let's think about the integers for a moment. The principal ideal $I$ generated by an integer $n$ is just the multiples of $n$, and $\mathbb Z/I$ is the ring of integers modulo $n$.

If $n$ is prime, then for any $m$ which is not a multiple of $n$ we can find integers $p$ and $q$ with $pn+qm=1$ and this is the same as saying $qm\equiv 1 \bmod n$. This means that non-zero residue classes modulo $n$ are invertible and the integers modulo $n$ form a field. (And the ideal is in fact not only prime but also maximal)

If $n$ is composite, say $n=pq$, this is equivalent to $pq\equiv 0 \bmod n$, and the integers modulo $n$ are not a field because they have zero divisors.

The case of polynomials over a field runs almost precisely parallel. Suppose $p(x)$ is irreducible of degree $n$ and $f(x)$ is an arbitrary polynomial. We can use the division algorithm to write $f(x)=p(x)q(x)+r(x)$ where $r(x)$ is a polynomial of lower degree than $p(x)$ and is a representative of the residue class. (Just as $9\equiv 2 \bmod 7$, with $2\lt 7$). So the residue classes are represented by the polynomials of degree less than $n$.

Importantly, if $f(x)$ is not a multiple of $p(x)$ then they are coprime, and we can find polynomials such that $f(x)t(x)+p(x)u(x)=1$ or $f(x)t(x)\equiv 1 \bmod p(x)$.

In the field which results from this procedure, the polynomial $x$ acts as a root of $p(x)$, because $p(x)\equiv 0$.

So if our polynomial were $x^2-2$ over the field of rational numbers, the residue classes would be represented by the linear polynomials $ax+b$ and we can think of this as $a\sqrt 2 +b$. We could also have chosen the other root and gone for $-a\sqrt 2+ b$. The construction shows that this does not affect the arithmetic at all. In later work the idea that we can swap the two roots without changing the arithmetic emerges as a key idea in Galois Theory.

In fact it is pretty much the same thing as complex conjugation, which emerges if we choose the polynomial $x^2+1$ to start with and identify the linear polynomials in $x$ with $a+bi$. Taking the complex conjugate of everything leaves the arithmetic unaffected.

Quadratics are, of course, a particularly simple case, but the relationship between the structure of the field and the roots of the polynomial remains significant with polynomials of higher degree. With higher degree polynomials there are more than two roots, and some care has to be taken with how they can be swapped.

  • $\begingroup$ Your answer is illuminating. Thanks so much. Why, if the polynomial $x$ acts as a root of $p(x)$ then the powers of x (plus 0 element) form a finite field ? $\endgroup$
    – AleWolf
    May 25, 2019 at 18:39
  • 1
    $\begingroup$ @AleQuercia They don't necessarily. The powers of $x$ in fact form a basis for a finite dimensional vector space over $\mathbb F$ of dimension $n$. The least degree representatives are all polynomials of degree less than $n$. If $\mathbb F$ is finite, then there are only finitely many such polynomials, and the field is finite. If $\mathbb F$ is infinite, the polynomials of degree zero are already infinite in number. There is a lot of structure here, and it takes a bit of getting used to how everything fits together. $\endgroup$ May 25, 2019 at 18:47
  • $\begingroup$ In which cases the root of $p(x)$ acts as a primitive element ? And why it's possible to construct the field using its powers ? $\endgroup$
    – AleWolf
    May 26, 2019 at 6:42
  • $\begingroup$ @AleQuercia Indeed $x$ acts both as a root of the polynomial and as a primitive element. If we assume, as we may, that $p(x)$ is monic of degree $n$ (divide by leading coefficient otherwise) then $p(x)=x^n+q(x)$ for some polynomial $q$ of degree less than $n$ and $x^n\equiv -q(x) \bmod p(x)$. Your intuition will improve, I think, as you work through practice; examples. $\endgroup$ May 26, 2019 at 8:00

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