# Does the discriminant of an irreducible polynomial of a fixed degree determine the discriminant of the number field it generates?

In the quadratic case, it does. Given an irreducible quadratic polynomial $$f(x)=ax^2+bx+c$$, the discriminant of the quadratic number field $$\frac{\mathbb{Q}[x]}{f(x)}$$ is $$\operatorname{sqf}(d)$$ or $$4\cdot \operatorname{sqf}(d)$$, depending on if $$d \equiv 1 \mod 4$$, where $$d=b^2-4ac$$ is the polynomial discriminant of $$f(x)$$, and $$\operatorname{sqf}$$ is the squarefree part.

One can reduce this to the case of local fields, in which one can ask a similar question. Fix a positive integer $$n$$. Given a separable polynomial $$f \in \mathbb{Q}_p[x]$$ of degree $$n$$, does the discriminant of $$f$$ determine the valuation of the discriminant ideal of the associated extension? What about in the case where $$p \nmid n$$?

For the question over local fields, one should note that the dependence on the discriminant of $$f$$ does not factor through the valuation of the discriminant of $$f$$.

Another counterexample:

$$f_1(x)=x^3-9x-20, f_2(x)=x^3-6x-18$$, discriminant = $$-4*27*73$$ for both $$f_1$$ and $$f_2$$.

Both polynomials are irreducible over $$\mathbb{Q}$$ by the rational root test.

The 2-adic Newton polygons show that $$f_1$$ has 2 roots of valuation=0 and 1 root of valuation=2, and $$f_2$$ has 3 roots of valuation=1/3. So the ring of integers is ramified over 2 for $$f_2$$ but not for $$f_1$$, and their discriminants are not equal.

I found these polynomials by looking for integer solutions $$(a=-3,b=-10)$$ to $$3a^2+3a+1=-(2b+1)$$ which yield polynomials $$f_1=x^3+3ax+2b, f_2=x^3+3(a+1)x+2(b+1)$$ with same discriminant=$$-4*27*(b^2+a^3)$$.

• You made miscalculations: $x^3 + ax+b$ has discriminant $-4a^3 - 27b^2$, so $x^3+9x+20$ has discriminant $-13716 = -2^2\cdot 3^3\cdot 127$ and $x^3+6x+18$ has discriminant $-9612 = -2^2\cdot 3^3\cdot 89$. Jun 30 at 3:27
• I found your mistake: a sign error. You want $3a^2 + 3a + 1 = -(2b+1)$. Using $a = 1$ and $b = -4$, we get the polynomials $x^3 + 3x -8$ and $x^3 + 6x - 6$, which are both irreducible over $\mathbf Q$ and have discriminant $-1836 = -2^2 \cdot 3^3 \cdot 17$. The number field generated by a root of $x^3 + 3x -8$ has discriminant $-459 = -3^3 \cdot 17$, while the number field generated by a root of $x^3 + 6x - 6$ is $-1836$. Jun 30 at 3:42
The discriminants of the irreducible polynomials $$(x^2-2)^2+60 = x^4 - 4 x^2 + 64, \qquad (x^2+2)^2+60 = x^4 + 4 x^2 + 64$$ are both equal to $$58982400 = 2^{18} \cdot 3^2 \cdot 5^2$$. However, the first field $$\mathbf{Q}(\sqrt{-3},\sqrt{5})$$ is unramified at $$2$$ whereas the latter field $$\mathbf{Q}(\sqrt{3},\sqrt{-5})$$ is ramified at $$2$$.
If you want an example where both polynomials remain irreducible over $$\mathbf{Q}_2$$, take instead the minimal polynomials of $$\sqrt{5}+\sqrt{2}$$ and $$\sqrt{-5}+\sqrt{-2}$$, that is the polynomials $$(x^2 \pm 7)^2 - 40$$. Here the discriminants of the polynomials are both $$2^{14} \cdot 3^2 \cdot 5^2$$, but the first field $$\mathbf{Q}_2(\sqrt{2},\sqrt{5})$$ is an unramified degree $$2$$ extension of $$\mathbf{Q}_2(\sqrt{2})$$ and so has discriminant $$(2^3)^2 = 2^6$$, whereas $$\mathbf{Q}_2(\sqrt{-2},\sqrt{-5})$$ is a ramified degree two extension of $$\mathbf{Q}_2(\sqrt{-2})$$, and so has discriminant strictly divisible by $$2^6$$ (actually $$2^8$$).
• Thanks a lot! This is an interesting example. I wonder if one can construct a similar example in the case where $p \nmid n$? Jun 28 at 6:01
• @johng23: Some remarks: your statement "One can reduce this to the case of local fields" does not make sense since irreducible polynomials over $\mathbf{Q}$ can become reducible over $\mathbf{Q}_p$. Second, yes, one can find examples with $p \nmid n$ over $\mathbf{Q}$, and examples of irreducible polynomials over $\mathbf{Q}_p$ of degree $(n,p) = 1$ with the same discriminant (but with corresponding local fields of different discriminant). Jun 30 at 17:34