Factorize x^3+3

Question

Task: Factorize $x^3+3$ in $\mathbb{R}$ and $GF(7)$

I think the solution is $x^3+3$ in both cases, so the polynomial already is irreductible. Is my assuption is right, how do i show that?

Answer 1

To check whether a cubic polynomial is irreducible over a given field, it is sufficient to check whether $f$ has any roots in that field (why?).

In $GF(7)$ this is easy: there are only $7$ possible roots, so you can simply evaluate $f(x)$ for all $x \in GF(7)$.

Over $\mathbb{R}$, remember that every polynomial of odd degree has at least one root. You can prove this using intermediate value theorem, since polynomials are continuous functions. Given this, when does $x^3 + 3 = 0$ in $\mathbb{R}$? Once you've found a root, you can use polynomial long division to find a factorization.

Answer 2

The polynomial $x^3 + 3$ (or $x^3 -3,$ makes no difference as $-1$ is a cube) is irreducible in the field of integers $\bmod p$ for all primes that can be written in integers (not necessarily positive) as $p = 7 u^2 + 3 uv+9v^2.$ This is a theorem due to Jacobi. Such primes up to $4000$ are

     7    13    19    31    37    43    79    97   109   127
   139   157   163   181   199   211   223   229   241   277
   283   313   331   337   349   373   379   397   409   421
   433   457   463   487   541   571   601   607   631   673
   691   709   733   739   751   769   811   823   829   859
   877   883   907   937  1033  1039  1051  1063  1069  1087
  1123  1129  1153  1171  1201  1213  1231  1237  1279  1291
  1297  1327  1381  1423  1429  1447  1453  1459  1471  1483
  1489  1567  1579  1627  1657  1663  1693  1699  1723  1741
  1747  1753  1777  1789  1801  1831  1873  1951  1987  1993
  1999  2011  2017  2053  2083  2089  2113  2137  2143  2161
  2239  2281  2293  2311  2347  2371  2377  2437  2467  2473
  2503  2521  2539  2551  2557  2593  2647  2659  2677  2683
  2689  2707  2719  2731  2749  2767  2791  2797  2833  2857
  2887  2917  2953  3037  3061  3121  3163  3169  3229  3259
  3271  3301  3307  3313  3331  3361  3391  3433  3457  3463
  3469  3511  3541  3547  3559  3571  3583  3613  3637  3643
  3673  3691  3697  3709  3727  3769  3793  3823  3877  3919
  3931  3943

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To get $7,$ we take $(u=1,v=0),$ To get $13,$ we take $(u=1,v=-1),$ To get $19,$ we take $(u=1,v=1),$ and so on as needed.

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Ummm. $x^3 -3$ factors as three distinct linear factors in the field of integers $\bmod p$ for all primes that can be written in integers (not necessarily positive) as $p = u^2 + uv+61v^2.$ Such primes up to $4000$ are

    61    67    73   103   151   193   271   307   367   439
   499   523   547   577   613   619   643   661   727   757
   787   853   919   967   991   997  1009  1021  1093  1117
  1249  1303  1321  1399  1531  1543  1549  1597  1609  1621
  1669  1759  1783  1861  1867  1879  1933  2029  2131  2179
  2203  2221  2251  2269  2287  2341  2383  2389  2617  2671
  2713  2803  2851  2971  3001  3019  3049  3067  3079  3109
  3181  3187  3217  3253  3319  3343  3373  3499  3517  3529
  3607  3631  3733  3739  3847  3853  3889  3907  3967

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