Help with Final Exam Questions: QUESTION TWO (b) & SIX (a)

[uob_student]

alsalam alekum

hi

i need a help in some question

i try but i can not solve it

i have a final exam tomorrow

i need help in QUESTION TWO (b)

and QUESTION SIX (a)

if i solve another questions , i will return back to ask you

Thanks
bye

 

[uob_student]

i forget questions

this question
Q(2)(a) show the polynomail is irreducible over Q

xpower 5+8*(x power 4)+3*(x power 2)+4*x+7


question six(a): Let F subset of K be a field extension
show that |K:F|=1 if and only if K=F.

 

[HallsofIvy]

this question
Q(2)(a) show the polynomail is irreducible over Q

xpower 5+8*(x power 4)+3*(x power 2)+4*x+7

By the "rational root" theorem, what are the possible rational roots of this polynomial (there are only 4)? Are any of those actually roots? And what does "irreducible" MEAN?

question six(a): Let F subset of K be a field extension
show that |K:F|=1 if and only if K=F.

What is the DEFINITION of |K:F|?

 

[uob_student]

By the "rational root" theorem, what are the possible rational roots of this polynomial (there are only 4)? Are any of those actually roots? And what does "irreducible" MEAN?



What is the DEFINITION of |K:F|?

irreducible means that the polynomail do not need factorization

|k:F| means the degree of k over F

i need the answer today if you can

 

[uob_student]

hi

i have another question:

QUESTION:Let F:R________>S be an epimorphism of rings with unity

(a) Show that if R is a principle ideal domain ,then every ideal in S is principle

(b) Show by an example that S need not be an integral domain.

thanks

 

[matt grime]

That isn't what irreducible means,: polynomials do not 'need' or 'want' to be factored.

You were asked to define the degree of a field extension, not write out the name again. Hint: the dimension of *** over *** as a ***** *****

what have you done for the last post? Have you written out the definition of a PID and tried to see what you can show? (b) is just trivial - what is the first kind of non-integral domain you ever meet (Hint: the smallest one has 4 elements). I presume that in (b) there are some extra conditions like R ought to still be assumed a PID or even an integral domain. Otherwise the identity morphism on any non-integral domain would do, wouldn't it?

 

[Gib Z]

Actually, Irreducible means that the polynomials cannot be simplified to another of lower degree, example x^4 + x^2 + 6 can be simplified by letting x^2=a then you can reduce it to a^2 + a +6, which can be solved easily with the quadratic formula. Then once you have your a values, you set them equal to a and solve for x.

Edit: Matt beat me to it

 

[matt grime]

No. That is very very incorrect, Gib. Irreducible (over a field F) means something entirely different - cannot be written as a product of factors over strictly lower degree with coefficients in F.

The polynomial you wrote down is irreducible over Q (note all questions of reducibility have some implicit underlying field)

 

[HallsofIvy]

irreducible means that the polynomail do not need factorization

Well, not "need" factorization, "CANNOT be factored", in this case into factors with rational coefficients.

The "rational root theorem" says any rational roots of the equation "ax^n+ ...+ cx+ b= 0" (zeroes of the polynomial), with integer coefficients, must be of the form m/n where m is a factor of b and m is a factor of a. Now answer the rest of my question: What are the possible rational zeroes of that polynomial according to the rational root theorem. Are those possible zeroes actually zeroes? What does that tell you about its factorization?

|k:F| means the degree of k over F

No, that's just the how you would say it in words. What is the definition of "degree of K over F"?

 

[mathwonk]

your polynomial is irreducible if it is irreducible mod 2.

there it suffices to see if it has either an irreducible linear or quadratic factor.

halls tip let's you check for linear factors, and there is only one irreducible quadratic polynomial mod 2.

then your polynomial is irreducible over Z hence also over Q.