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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 $f$.

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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=$-4*27*(b^2+a^3)$.

Note: The initial version of this answer had mistakes as reported in comments below by KConrad.

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    $\begingroup$ 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 $-9612 = -2^2\cdot 3^3\cdot 89$. $\endgroup$
    – KConrad
    Jun 30 at 3:27
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    $\begingroup$ 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 $-1836$. $\endgroup$
    – KConrad
    Jun 30 at 3:42
  • $\begingroup$ Thanks @KConrad. I fixed the answer according to your comments. $\endgroup$ Jun 30 at 12:37
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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 $2^8$).

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  • $\begingroup$ Thanks a lot! This is an interesting example. I wonder if one can construct a similar example in the case where $p \nmid n$? $\endgroup$
    – johng23
    Jun 28 at 6:01
  • $\begingroup$ @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). $\endgroup$
    – user484894
    Jun 30 at 17:34
  • $\begingroup$ You're right. I should have said separable instead of irreducible. I changed it. $\endgroup$
    – johng23
    Jul 1 at 5:36

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