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Can someone give me a hint why in the quotient ring $\mathbb{F}_p[x]/{\langle f\rangle}$, where $f$ is an irreducible polynomial of degree $k$ and $p$ is a prime (and $\langle f\rangle$ the ideal generated by $f$), every element is invertible ?

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Try showing that an ideal generated by an irreducible polynomial is maximal. Recall that $\mathbb{F}_p$ is a field, and thus $\mathbb F_p[X]$ is an commutitive ring.What do you know of the quotient of commtitive rings by maximal ideals? – kneidell Nov 16 '11 at 10:44
Your quotient is a nonzero ring. In a nonzero ring, $0$ is not invertible. – Pierre-Yves Gaillard Nov 16 '11 at 12:38
Here's a related question asked yesterday. – Michael Hardy Nov 16 '11 at 14:16
up vote 1 down vote accepted

As pointed out above, 0 is not invertible. However, for any field $K$, $K[x]$ is a PID. Since $f$ is irreducible and $\mathbb{F}_p[x]$ is a PID, $f$ is a prime element of $\mathbb{F}_p$ (if this isn't clear, then sit down and hammer out a simple proof).

The only thing that we need now is the following proposition.

Proposition: In a PID, every nonzero prime ideal is maximal.

Sketch of proof: Suppose that $D$ is a PID and that we have $0\subsetneq P\subseteq M$ for a maximal ideal $M$ of $D$ and a (nonzero) prime ideal $P$ of $D$. Since $D$ is a PID, we have $P=(p)$ and $M=(q)$ for some nonzero $p,q\in M$. Also, since both $M$ and $P$ are prime ideals, it follows that $p$ and $q$ are prime (and hence irreducible) elements of $D$ (if this isn't immediately clear, then sit down and write out a short proof as to why).

However, $(p)\subseteq (q)$ implies that $q \vert p$, implying that $q=up$ for some $u\in U(D)$. Therefore $M=P$ and every nonzero prime ideal of a PID is maximal. $\blacksquare$

So, going back to $f\in \mathbb{F}_p[x]$, we have that $f$ is a prime polynomial in $\mathbb{F}_p[x]$, hence $(f)$ is a maximal ideal. Therefore $\mathbb{F}_p[x]/(f)$ is a field.

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In this question, someone wanted to know how to find the multiplicative inverse of an integer modulo a prime number. (You had to assume the number whose inverse you wanted was not divisible by that prime number.)

You can do the present problem the same way.

To say that $g(x)$ is not $0$ in the quotient ring is to say that $g(x)$ is not a multiple of $f(x)$. Since $f(x)$ is irreducible, this implies that $\gcd(f(x),g(x))=1$. Applying Euclid's algorithm as in the answer to the question to which I linked above, you can find polynomials $a(x)$ and $b(x)$ such that $a(x)g(x) + b(x)f(x)=\text{the gcd}=1$. In the quotient ring, that second term, $b(x)f(x)$, is $0$, so $a(x)g(x)=1$ and so $g(x)$ is invertible.

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