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Let $K = {\mathbb Q}[t]$. Show $x^3 - tx^2 + (t-3)x + 1$ is irreducible in $K[x]$.

I tried substitution with $x-t$ and other things, hoping to use Eisenstein's criterion to finish the job. But I have not made much progress. Can I get a hint?

Thank you.

Steven Li
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    This is known as "Shanks' simplest cubic," so you could try searching for same. – Gerry Myerson Mar 10 '13 at 23:08
  • By the way, Shanks' original paper has a lot of interesting information about this polynomial. http://www.jstor.org/discover/10.2307/2005372?uid=3739696&uid=2&uid=4&uid=3739256&sid=21101786246953 – Álvaro Lozano-Robledo Mar 18 '13 at 02:22

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I’ll use freely the fact that $\mathbb Q[t]$ and $\mathbb Q[x]$ are unique factorization domains. Rewrite your polynomial as an element of $\mathbb Q[x][t]$, giving the linear (in $t$) polynomial $(x-x^2)t + (x^3-3x+1)$. Any factorization would have to be linear (in $t$) times a constant, where this last word means in $\mathbb Q[x]$. But that would say that $x-x^2$ and $x^3-3x+1$ had a common factor, which they don’t (except for genuine constants in $\mathbb Q$, of course).

Lubin
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Suppose there exists $\,p(t)\in K\,\,\,s.t.\,\,\,p(t)^3-tp(t)^2+(t-3)p(t)+1=0\,$ .

Hint: since $\,p(t)\in\Bbb Q[t],\,$ , the above implies $\,t\,$ is algebraic over the rationals...

DonAntonio
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  • Then there is a minimal polynomial $m(x)$ of $t$ such that deg $m \ge 3$ ...? – Steven Li Mar 10 '13 at 23:36
  • I don't think so...is $,t,$ algebraic over $,\Bbb Q,$? Unless something else's given, the ring of polynomials over a field is defined as an algebra over a transcendental element... – DonAntonio Mar 10 '13 at 23:38
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Did I hear Eisenstein's criterion?

Suppose for a contradiction that $f_t(x)=x^3-tx^2+(t-3)x+1$ is reducible in $(\mathbb{Q}[t])[x]$, i.e., $f_t(x)$ factors as $f_t(x)=g_t(x)h_t(x)$ where $g_t,h_t\in (\mathbb{Q}[t])[x]$ are non-constant as a function of $x$. Then $f_{t_0}(x)$ must be reducible for all values of $t_0\in\mathbb{Q}$ (because $f_{t_0}(x)=g_{t_0}(x)h_{t_0}(x)$) as a polynomial in $\mathbb{Q}[x]$. In particular, when $t_0=0$, the polynomial $$f_0(x)=x^3-3x+1$$ is reducible. But Eisenstein's criterion (for $p=3$) says that $f_0(x+2)=\tilde{f}_0(x)= x^3 + 6x^2 + 9x + 3$ is irreducible, and we have reached a contradiction.

Álvaro Lozano-Robledo
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  • I don't understand why $f_{t_0}(x)$ must be reducible for all values of $t_0\in\mathbb{Q}$. Why can we pick a value for $t_0$? – Steven Li Mar 12 '13 at 22:07
  • @StevenLi I added a little more explanation on this. Essentially, being reducible over $(\mathbb{Q}[t])[x]$ is a very strong condition, as it gives you a factorization $f_t(x)=g_t(x)h_t(x)$ which you can, in turn, evaluate at any value of $t$. – Álvaro Lozano-Robledo Mar 12 '13 at 23:43