Prove that there is no irreducible polynomial of degree 2 in $\mathbb C[x]$.

I know this result is true from the Fundamental Theorem of Algebra, which states that any polynomial in $\mathbb C[x]$ with degree $n\ge 1$, has at least $1$ root in $\mathbb C$.

However, this question is asked in the textbook before the FTA is introduced. How can I go about proving it?


You use the quadratic formula to find the root(s).

  • $\begingroup$ So do I just say the following: Let $f(x) = ax^2 + bx + c$, then we know $x_0= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ is a root of $f$, that is $f(x_0)=0$. We then have two cases to consider. If $b^2 - 4ac \ge 0$, then $x_0 \in \mathbb R \subset \mathbb C$ and if $b^2 - 4ac < 0$, then $x_0 \in \mathbb C$. Either way, we have found a root of $f(x)$ in $\mathbb C$ and, consequently, $f(x)$ is reducible in $\mathbb C[x]$? $\endgroup$ – user290425 Apr 5 '16 at 21:35
  • 2
    $\begingroup$ Almost. You have to deal with $a,b,c\in\mathbb C$, so it is not about the sign of $b^2-4ac$. What you need to show somehow is that any complex number has a square root; then you can use one root of $b^2-4ac$ to find the root of your polynomial. $\endgroup$ – Martin Argerami Apr 5 '16 at 21:40

It's not restrictive to assume the polynomial is monic, write it for simplicity as $x^2+2bx+c$ and consider $$ x^2+2bx+c=x^2+2bx+b^2+c-b^2=(x+b)^2-(b^2-c) $$ If you prove that $b^2-c=d^2$, for some $d\in\mathbb{C}$, then $$ x^2+2bx+c=(x+b)^2-d^2=(x+b-d)(x+b+d) $$ is not irreducible.

Now the task is to show that any complex number is a square. Consider $$ A+Bi=(X+Yi)^2 $$ with $A,B$ given real numbers and $X,Y$ unknown real numbers. This translates to $$ \begin{cases} X^2-Y^2=A \\[4px] 2XY=B \end{cases} $$ and we can divide the problem into a few cases.

Case 1: $A=0$, $B=0$. Choose $X=Y=0$.

Case 2: $B=0$, $A>0$. Choose $X=\sqrt{A}$ or $X=-\sqrt{A}$ and $Y=0$.

Case 3: $B=0$, $A<0$. Choose $X=0$ and $Y=\sqrt{-A}$ or $Y=-\sqrt{-A}$.

Case 4: $B\ne0$. Rewrite the second equation as $Y=B/(2X)$ and substitute in the first one to get $$ X^2-\frac{B^2}{4X^2}=A $$ that becomes $$ 4X^4-4AX^2-B^2=0 $$ which is a biquadratic; this gives $$ X^2=\frac{A+\sqrt{A^2+B^2}}{2}>0 $$ which has two solutions, from which we can get the corresponding values of $Y$.

Side note. Every algebra based proof of the FTA relies on the fact that every complex number is a square (equivalent to no polynomial of degree $2$ being irreducible) and on the weak form of the intermediate value theorem that odd degree polynomials with real coefficients have a root.

  • $\begingroup$ @user26857 Well, algebraic proofs. ;-) $\endgroup$ – egreg Apr 5 '16 at 22:18

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