Let $K$ be a field and let $p(x)$ be a non-zero polynomial of $K[x]$. Suppose that deg($p(x)$)$=1$, show that $p(x)$ is irreducible.

Problem: Let $K$ be a field and let $p(x)$ be a non-zero polynomial of $K[x]$.

Suppose that deg($p(x)$)$=1$, show that $p(x)$ is irreducible.

Thoughts: My definition of an irreducible element $p$ in an integral domain is that it is non-zero, not a unit, and that it's only divisors are the associates of $1_R$ and $p$.

Clearly $p(x)$ is neither zero or a unit since it has a coefficient of $x^1$. I am not sure how to proceed here, although my intuition is to apply the euclidean algorithm of division of polynomials. Any hints or insight much appreciated.

• just use the definition: if there $q(x),t(x)$ s.t. $p(x)=q(x)t(x)$, then a simple degree argument forces one of them to be a constant. – Matematleta Oct 14 '16 at 2:01

Hint: If $p(x)=f(x)g(x)$, what can you say about the degree of $f$ and $g$?

I will try to give a hint. Write down a general formula of $p(x)$; reduce it to a monic polynomial; suppose it is not a irreducible polynomial with monic factor $f(x)$ of degree 1, then use Euclidean algorithm to show that there is a remainder when $f(x)\neq p(x)$.

Hope that helped.

SPOILER. Not a hint.

If $p(x)=f(x)g(x)$ then deg($f$)+deg($g$)= deg($p$). One of these degrees must be $1$ and the other $0$. A polynomial of degree 0 is a unit.

• The OP asked for hints... – lhf Oct 14 '16 at 2:00
• @lhf You're right I should tag as spoiler. – JKEG Oct 14 '16 at 2:02
• @JamesDickens $p$ is irreducible if $p=fg \implies f$ is a unit or $g$ is a unit. :P (Associates of $1_R$ are precisely units, so our definitions are pretty much identical.) – JKEG Oct 14 '16 at 2:05