Is it true that a polynomial is reducible over a field only if the polynomial has a zero in the field?

Question

I am doing some practice problems for abstract algebra and have come across this idea in a couple places, but it seems fundamentally wrong. For example, in order to prove that $f(x) = x^2 + x + 1$ is irreducible over the field $\mathbb Z_2$ I came across an example where a person just plugged in $f(0) = 1$ and $f(1) = 1$ and stated that this was enough to prove it. But what about the polynomial $p(x)=21x^2 + 32x + 12=(7x+6)(3x+2)$ over the ring $\mathbb Z$? Though it's only zeros are $\frac{-6}{7}$ and $\frac{-2}{3}$, it is certainly reducible over $\mathbb Z$. I tried to solve the problem by showing that $\mathbb Z_2[x]\setminus A$, where the ideal $A=(x^2 + x +1)$, is a field, but am getting hung up on showing that every nonzero element has an inverse.

EDITED: My point about the sample polynomial talks about the ring Z, which I initially misidentified as a field, so it should be disregarded.

Thanks for all your help!

Well, $\Bbb Z$ is not a field. It is true that a quadratic polynomial over a field $K$ is reducible if and only if it has a root in $K$; this is very easy to prove (one direction is trivial and the other direction is immediate from the division algorithm). It's false for polynomials of higher degree; consider $K=\Bbb R$, $p=(x^2+1)^2$. — David C. Ullrich, May 15 '16 at 16:37
You should really correct what you said here. $\Bbb Z$ is not a field. Really. — David C. Ullrich, May 15 '16 at 17:17

score 4 · Answer 1 · answered May 15 '16 at 16:41

Assume you work over a field $K$.

Is it true that a polynomial $P$ over $K$ having a root is reducible ?

False. Example $P(X):=X$ has $0$ for root but is irreducible.

Is it true that a polynomial $P$ of degree $>1$ over $K$ having a root is reducible ?

True. Let $r$ be the root of $P$ then $P(X)=(X-r)Q(X)$ with $deg(Q)=deg(P)-1>0$.

Is it true that a polynomial $P$ without roots is irreducible ?

False. Example : $K:=\mathbb{Q}$. Then $P(X)=(X^2-2)(X^2-3)$ is clearly reducible. However its roots in $\mathbb{R}$ are $\pm\sqrt{2}$ and $\pm\sqrt{3}$, none of them belongs to $\mathbb{Q}$. It follows that $P$ has no root in $\mathbb{Q}$.

Is it true that a polynomial $P$ of degree $\leq 3$ without roots is irreducible ?

True. If $P$ were reducible then we could find $Q$ and $R$ such that $P=QR$ and $deg(Q),deg(R)>0$. Since $deg(P)\leq 3$ then we necessarily have $deg(Q)=1$ or $deg(R)=1$. Since $Q$ or $R$ is of degree $1$, at least one of them has a root and so $P$ has a root too.

score 2 · Answer 2 · answered May 15 '16 at 16:41

The argument works with polynomials of degree $2$ or $3$ over a field, but not higher.

The reason is that if a polynomial of such degree is reducible, it must have a factor of degree $1$, and in a field such a factor always has a root (in your example, the degree $1$ factors have no root in $\mathbb{Z}$, so you see why it is important that we have polynomials over a field). But as explained by David's comment it is not true for higher degree.

Thanks very much, this makes alot of sense. – Abhishek Sharma May 15 '16 at 17:50 — Abhishek Sharma, May 15 '16 at 17:50

score 2 · Answer 3 · answered May 15 '16 at 16:43

2

Counterexample:

$$x^4+1=(x^2-\sqrt 2 x+1)(x^2+\sqrt 2 x+1).$$

answered May 15 '16 at 16:43

Bernard

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Is it true that a polynomial is reducible over a field only if the polynomial has a zero in the field?

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