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### Why is PolynomialRing over SR much slower than SR?

In my code, I was trying to invert a matrix whose entries are symbolic expressions that depend on a single variable (called "tau").

Because of the fact that in my code the dependency on tau is polynomial, at some point I realized, that, in principle, it would be better to use the FractionField over the PolynomialRing in one variable (tau) defined over the symbolic ring (SR) instead of using the symbolic ring with a new variable defined as var("tau"). I was expecting that this would have improved the performances because of the fact that now SageMath knows that there can not be any log(tau) or exp(tau) or something like that.

But, instead, everything became hundreds of times slower. In particular, the method solve_right of the matrix, for a matrix 4 x 4, requires more than half an hour instead of 0.1 seconds. Also, the simplify method seems to be a lot slower.

Do you know why this happens? I have tried to find which algorithm SageMath uses for solving a linear system, but I did not find anything. Does it uses a different algorithm for polynomial and for symbolic expressions? I suspect that for polynomial over a "standard" ring (like for example QQ), SageMath uses Singular while, when solving a system with a matrix on the symbolic ring, it uses Maxima. Therefore, maybe the problem is related with the fact that, when I use a polynomial ring over SR, Sage must use an internal implementation. Do you know if this is the case? And why this also affect the method simplify?

Thank you!

### Why is PolynomialRing over SR much slower than SR?

In my code, I was trying to invert a matrix whose entries are symbolic expressions that depend on a single variable (called "tau").

Because of the fact that in my code the dependency on tau is polynomial, at some point I realized, that, in principle, it would be better to use the FractionField over the PolynomialRing in one variable (tau) defined over the symbolic ring (SR) instead of using the symbolic ring with a new variable defined as var("tau"). I was expecting that this would have improved the performances because of the fact that now SageMath knows that there can not be any log(tau) or exp(tau) or something like that.

But, instead, everything became hundreds of times slower. In particular, the method solve_right of the matrix, for a matrix 4 x 4, requires more than half an hour instead of 0.1 seconds. Also, the simplify method seems to be a lot slower.

Do you know why this happens? I have tried to find which algorithm SageMath uses for solving a linear system, but I did not find anything. Does it uses a different algorithm for polynomial and for symbolic expressions? I suspect that for polynomial over a "standard" ring (like for example QQ), SageMath uses Singular while, when solving a system with a matrix on the symbolic ring, it uses Maxima. Therefore, maybe the problem is related with the fact that, when I use a polynomial ring over SR, Sage must use an internal implementation. Do you know if this is the case? And why this also affect the method simplify?

The following snippet generate two matrices with the same content (but using a different ring):

def build_matrix(poly_ring=False):
if poly_ring:
tau_ring = PolynomialRing(SR, 'tau')
tau_field = FractionField(tau_ring)
tau = tau_ring.gen(0)

M = matrix(tau_field, 2, 2)
f = vector(tau_field, [1, 1])
else:
tau = var('tau')
M = matrix(SR, 2, 2)
f = vector(SR, [1, 1])

M[0, 0] = (e^(2/ 3) + 1) * tau + e^2
M[0, 1] = e^-1 * tau + 3
M[1, 0] = tau + 2
M[1, 1] = 3 * tau + e^(1/ 2) + e^4

return M, f


And here there are the benchmark that I see:

m1, f1 = build_matrix(True)
m2, f2 = build_matrix(False)

%timeit m1.inverse()
>> 345 ms ± 14.4 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit m2.inverse()
>> 2.57 ms ± 52.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit m1.solve_right(f1)
>> 370 ms ± 72.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit m2.solve_right(f2)
>> 2.5 ms ± 21.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)


Thank you!