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Let $V$ be a real vector space over $\mathbb{R}$ with $\dim(V) \geq 3$. Show that $\exists W \subset V : W \ne V, W \ne \{0\}$ and $T(W) \subset W$ with $T \in End_{\mathbb{R}}(V)$ arbitrary. The solution considers $V$ as a $\mathbb{R}[T]$ module, with the operation of $f(T)\cdot v = f(T(v)) \in V$. They then show $V$ cannot be a simple module because if it were, then it is isomorphic to $\mathbb{R}[T] / (f(T))$, where $f(T)$ has degree at least $3$ and is an irreducible polynomial, which they state cannot exist. Can someone give me a reason why a degree at least $3$ irreducible polynomial cannot exist in $\mathbb{R}[T]$?

user100101212
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    Every odd degree polynomial over $\mathbb R$ has a real root (just look at what happens at $\pm \infty$). – lulu Jan 03 '20 at 18:06
  • Sorry, I meant to say degree at least $3$. The degree of polynomial is the dimension of $V$ as a real vector space. What about the case when the degree is even but greater than 3? – user100101212 Jan 03 '20 at 18:10
  • Even degree polynomials over $\mathbb R$ factor as products of linear and quadratic terms. this is the Fundamental Theorem of Algebra. – lulu Jan 03 '20 at 18:11
  • I see. So this is because if $f(x)$ is a polynomial with real coefficients, then $\overline{f(x)} = f(\overline{x})$, so $a+bi$ being a root of f implies $a-bi$ is a root of $f$ and multiplying these linear factors in the complex numbers gives an irreducible quadratic over the reals. – user100101212 Jan 03 '20 at 18:16
  • Exactly correct. – lulu Jan 03 '20 at 18:16

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