hello there the jacobian conjecture is false thanx to my close friend akhil for asking about it and my other close friend fable for working during the world cup final ((1+xy)^3 z + y^2 (1+xy) (4+3xy), y + 3 x (1+xy)^2 z + 3 x y^2 (4+3xy), 2 x - 3 x^2 y - x^3 z): \C^3\to \C^3, has jacobian determinant -2, and sends (0, 0, -1/4), (1, -3/2, 13/2), and (-1, 3/2, 13/2) to (-1/4, 0, 0)

Jul 20, 2026 · 2:19 AM UTC

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wolframalpha.com/input?i=Det… jacobian determinant in wolfram alpha

[Det[D{(1+x y)^3 z + y^2 (1 + x y) (4 + 3x y), y + 3x (1 + x y)^2 z + 3x y^2 (4 + 3x y), 2x - 3x^2... Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. wolframalpha.com

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wolframalpha.com/input?i=%28…wolframalpha.com/input?i=%28… evaluating at two of those points in wolfram alpha

((1+x y)^3 z + y^2 (1 + x y) (4 + 3x y), y + 3x (1 + x y)^2 z + 3x y^2 (4 + 3x y), 2x - 3x^2 y -... Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels. wolframalpha.com

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Replying to @alpoge

I asked Fable to rank the 50 most important unsolved math problems and this one ranked 23rd!

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Me asking google what the 23rd letter is for u

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Replying to @alpoge

Thank you blessed math fairy

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Listen!

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Replying to @alpoge

@grok is this true

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liz we both know @gork would know better

is this the same as in Yitang Zhang's thesis? math.purdue.edu/~ttm/Zthesis…

Yo he is so incredible. i was at his first talk on small gaps:D the story was that he relied on a result which had a gap right

My face when about to click post on this one..

investing at 0 likes in case this is true

This is the most hilarious / horrifying problem because it is like the canonical crank graveyard. Ty 4 investment🙏

maybe a stupid question, but doesnt this also imply that dixmier conjecture and poisson conjecture are false in general?

Yea i think so. i was considering tweeting it too but people know jacobian much better..

Replying to @alpoge

Eh, I remember when a Prof in my undergrad AG class discussed it at the end of the semester in a class where we were covering some fun topics after being done with coursework. He described it then as a simple to state but profound in consequences and we were all dreaming of disproving it with a simple counterexample. Now we're all the way here. Big congrats.

Replying to @alpoge

This mapping is a historic counterexample that officially disproves the Jacobian Conjecture, a famous open problem in algebraic geometry since 1939. Discovered in July 2026 by mathematicians Levent Alpoge and Akhil Mathew (with the assistance of the AI model Fable), this single function brings an end to an 87-year-old mathematical mystery.1. State the Jacobian ConjectureThe Jacobian Conjecture states that a polynomial mapping F: \C^n \to \C^n with a non-zero constant Jacobian determinant must be injective (one-to-one) and possess a polynomial inverse function.2. Verify the Constant DeterminantFor the given mapping (F(x, y, z) = (F_1, F_2, F_3)), computing the matrix of partial derivatives yields a constant determinant:(\det (J_{F})=-2)Because (-2 \neq 0), this map satisfies the exact premise required by the conjecture. According to the 1939 hypothesis, this function was supposed to be completely invertible.3. Demonstrate the Failure of InjectivityTo disprove the conjecture, a function must map multiple distinct input points to the exact same output point. Your provided function does exactly that by sending three separate coordinate sets to a single destination:(F(0, 0, -1/4) = (-1/4, 0, 0))(F(1, -3/2, 13/2) = (-1/4, 0, 0))(F(-1, 3/2, 13/2) = (-1/4, 0, 0))Because the mapping is not injective, the Jacobian Conjecture is false.4. Appreciate the Historical SignificanceThe Jacobian Conjecture was notorious for attracting hundreds of flawed or incomplete proofs over many decades. Finding a definitive counterexample in three dimensions (\C^3) represents a massive milestone in computer-assisted mathematics and modern algebraic geometry.Summary of Meaning ✅The existence of this mapping explicitly proves that the Jacobian Conjecture is false. A polynomial system can have a perfectly non-singular, non-zero constant Jacobian determinant and still fail to be injective.

Replying to @alpoge

v = y + 1/x, then cv³ − 2v² + bv − 2a = 0

Replying to @alpoge

nice work congrats !

@alpoge Question to Fable: Using the one example proof, could you find another example that proves the jacobian conjecture is false? What would the process be to do so? Yes — and I just did, live. There are two levels to this: a mechanical way that spits out infinitely many new examples, and the deeper process of understanding why the original works so you can build genuinely different ones. I did the first and reverse-engineered the second. Level 1: dress the old example in costumes. If F is a counterexample and G, H are any known polynomial automorphisms (invertible “shuffles” of space, like (x,y,z) → (x+z², y, z)), then G∘F∘H is also a Keller map (Jacobian determinants multiply: 1·(−2)·1 = −2) and still non-injective — the three colliding points just get relocated by H⁻¹. I built one above: it’s a monster with component degrees 14, 20, and 7 that looks nothing like the original, has constant Jacobian −2, and sends (−1/16, 0, −1/4), (−165/4, −3/2, 13/2), (−173/4, 3/2, 13/2) all to (−1/4, −3/16, 0). Verified symbolically. Since there are infinitely many automorphisms, this gives infinitely many counterexamples — but mathematicians would call them “equivalent”: the same example wearing different clothes. Level 2: reverse-engineer the mechanism. The tweet’s map hides a clean skeleton. Writing u = 1+xy and W = u²z + y²·q(xy) with q(v) = 4+3v, the map is exactly (u·W, y + 3x·W, x(2−3xy−x²z)). If you then hunt for all preimages of a target point (c, 0, 0), the whole system collapses to finding roots of one small polynomial P(v) in v = xy — and the coefficients were tuned so P(v) = 2 + (4/3)v, giving v = −3/2, which forces x² = −1/(4c) and hands you the mirror-image pair (±1, ∓3/2, 13/2); the point with x = 0 comes along for free. So collisions are automatic by design — roots of polynomials always exist over ℂ. The entire 87-year difficulty was the other constraint: making the Jacobian determinant exactly constant. The discovery process is therefore: pick a structured ansatz where non-injectivity is built in, write “det ≡ constant” as a system of equations on the coefficients, and solve — a search that’s brutally tedious for humans and very natural for AI plus computer algebra. How rigid is it? My second experiment: replacing q = 4+3v with a general q = t+sv and demanding constant determinant forces exactly t = 4, s = 3 — no wiggle room in that direction. So genuinely new examples require changing the skeleton itself: different building blocks than (x, 1+xy), higher-degree q to get maps that are 4-to-1 or more, analogs in dimension 4+ beyond mere padding. And there’s one guaranteed factory: a classical reduction theorem (Bass–Connell–Wright/Drużkowski) constructively converts any counterexample into one of the ultra-simple form x + (linear)³ in some higher dimension — cranking this example through that machine is now a concrete, purely mechanical project that must succeed.

Replying to @alpoge

That’s so freaking funny

Replying to @alpoge

Potential structure behind the Jacobian counterexample: with r=x/(1+xy), inversion reduces to 2ar^3-br^2+2r-c=0. Thus the map is generically 3-to-1 while \det J_F=-2. A hidden cubic covering, polynomialized by pole-zero cancellation at infinity.