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TIFR 2014 Problem 26 Solution -Number of irreducible Polynomials

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TIFR 2014 Problem 26 Solution is a part of TIFR entrance preparation series. The Tata Institute of Fundamental Research is India's premier institution for advanced research in Mathematics. The Institute runs a graduate programme leading to the award of Ph.D., Integrated M.Sc.-Ph.D. as well as M.Sc. degree in certain subjects.
The image is a front cover of a book named Linear Algebra Done Right by Sheldon Axler. This book is very useful for the preparation of TIFR Entrance.

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The number of irreducible polynomials of the form (x^2+ax+b) , with (a,b) in the field (\mathbb{F}_7) of 7 elements is:

A. 7

B. 21

C. 35

D. 49


First, what is the number of polynomials of the form (x^2+ax+b) in (\mathbb{F}_7) ? (a) has 7 choices and (b) has 7 choices. So we have a total of (7 \times 7 =49) monic degree 2 polynomials in (\mathbb{F}_7).

How many of these are reducible? If a monic polynomial of degree 2 is reducible then it must break into two factors in the form (p(x)=(x-\alpha)(x-\beta) ).

We want to count the number of polynomials of the form (p(x)=(x-\alpha)(x-\beta)). This will give us the number of reducible polynomials and we will subtract it from the total number to get the number of irreducible.

For the counting purposes, we must be careful not to have repetitions. For (\alpha=0) we can allow (\beta=0,1,..,6). For (\alpha=1) we can allow (\beta=1,...,6) (NOT 0 because that will again give ((x-0)(x-0)) which we have already counted in the (\alpha=0) situation ).

For (\alpha=2) we have (\beta=2,...,6) and so on... for (\alpha=6) we have (\beta=6) only.

So in total, we have (7+6+...+1=\frac{7\times 8}{2} = 28 ) polynomials which are reducible, has degree 2 and is monic.

Therefore, the number of irreducible polynomials are (49-28=21).


  • What is this topic: Linear Algebra
  • What are some of the associated concept: Monic polynomial
  • Book Suggestions: Linear Algebra done Right by Sheldon Axler

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