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I am not sure if I making some very fundamental mistake. But Gallian says that $2x^2+4$ is reducible over $\mathbb C$.

If $D$ be an integral domain. A polynomial $f(x)$ from $D[x]$ is said to be irreducible over $D$ if whenever $f(x)$ is expressed as a product $f(x)=g(x)h(x)~~|~~g(x),h(x) \in D[x]$, then $g(x)$ or $h(x)$ is a unit in $D[x]$

$2x^2+4= 2(x^2+2)$ and $2$ is clearly a unit in $\mathbb C$. Hence, it should be irreducible right?

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    $\begingroup$ Er, $2x^2+4=2(x+i\sqrt{2})(x-i\sqrt{2})$ In particular, $\Bbb C$ is algebraically closed, so all polynomials factor into linear terms. $\endgroup$ Jul 22 '14 at 7:56
  • $\begingroup$ Yeah, .. But the condition says that either one of the factors should be a unit. $2$ is a unit in $\mathbb C$ ? $\endgroup$
    – MathMan
    Jul 22 '14 at 7:57
  • $\begingroup$ Yes, but the polynomial factors aren't units. In particular let $h(x)=2(x+i\sqrt{2}),\, g(x) = (x-i\sqrt{2})$ $\endgroup$ Jul 22 '14 at 7:58
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    $\begingroup$ The condition says that every factorization has one of them is a unit, not just some factorization has one factor a unit. I've given you one which fails that, hence it's not irreducible. $\endgroup$ Jul 22 '14 at 8:01
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    $\begingroup$ No problem, glad to help. It happens to the best of us. $\endgroup$ Jul 22 '14 at 8:05
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FPE's answer and Adam Hughes' comment are great. Ignoring the fact that $\mathbb{C}$ is algebraically closed (and hence all polynomials over $\mathbb{C}$ reduce into linear factors), you can look at it in this perspective:

Consider a polynomial $f(x) \in F[x]$. If $\deg(f) = 2$ or $3$, then $f$ is irreducible $\iff$ it has no roots in $F$ (Why?). From here, we can apply the quadratic formula to prove that $g(x) = 2x^2 + 4$ has roots in $\mathbb{C}$, allowing us to conclude that it is reducible by the criterion in your original post.

In particular, if $\alpha$ and $\beta$ are roots, then $g(x) = (2x - 2\alpha)(x-\beta)$, and neither of those factors are units in $\mathbb{C}[x]$.

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  • $\begingroup$ Thank you for the answer .. $\endgroup$
    – MathMan
    Jul 22 '14 at 10:08
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It has to be true for every factorisation $f(x) = g(x)h(x)$, not only for the one you give. In particular, you can always write $f(x) = 1\cdot f(x)$, and $1$ is a unit, so your argument can't work.

And remember that $\mathbb C$ is algebraically closed, so that any polynomial of degree at least $2$ is reducible, since it has roots in $\mathbb C$.

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